id int64 1 14M | domain stringclasses 6
values | topic stringclasses 23
values | subtopic stringclasses 37
values | difficulty int64 1 8 | unit_type stringclasses 3
values | title stringlengths 14 86 | content stringlengths 203 553 | key_equations stringclasses 23
values | prerequisites stringclasses 29
values | learning_objective stringclasses 37
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3,801 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 6.694 kg, acceleration 3.257 m/s² | A net force acting on a mass of 6.694 kg produces an acceleration of 3.257 m/s². By Newton's second law, F_net = m a = 6.694 × 3.257 = 21.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,802 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 31.14 kg, acceleration 3.362 m/s² | A net force acting on a mass of 31.14 kg produces an acceleration of 3.362 m/s². By Newton's second law, F_net = m a = 31.14 × 3.362 = 104.7 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,803 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 9.593 kg, acceleration 11.22 m/s² | A net force acting on a mass of 9.593 kg produces an acceleration of 11.22 m/s². By Newton's second law, F_net = m a = 9.593 × 11.22 = 107.6 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,804 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.19 kg, acceleration 12.64 m/s² | A net force acting on a mass of 28.19 kg produces an acceleration of 12.64 m/s². By Newton's second law, F_net = m a = 28.19 × 12.64 = 356.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,805 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 40.92 kg, acceleration 7.08 m/s² | A net force acting on a mass of 40.92 kg produces an acceleration of 7.08 m/s². By Newton's second law, F_net = m a = 40.92 × 7.08 = 289.7 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,806 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 16 kg, acceleration 9.664 m/s² | A net force acting on a mass of 16 kg produces an acceleration of 9.664 m/s². By Newton's second law, F_net = m a = 16 × 9.664 = 154.7 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,807 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 41.45 kg, acceleration 2.556 m/s² | A net force acting on a mass of 41.45 kg produces an acceleration of 2.556 m/s². By Newton's second law, F_net = m a = 41.45 × 2.556 = 105.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,808 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 40.27 kg, acceleration 9.735 m/s² | A net force acting on a mass of 40.27 kg produces an acceleration of 9.735 m/s². By Newton's second law, F_net = m a = 40.27 × 9.735 = 392 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,809 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 10.14 kg, acceleration 1.715 m/s² | A net force acting on a mass of 10.14 kg produces an acceleration of 1.715 m/s². By Newton's second law, F_net = m a = 10.14 × 1.715 = 17.39 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,810 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 4.352 kg, acceleration 9.088 m/s² | A net force acting on a mass of 4.352 kg produces an acceleration of 9.088 m/s². By Newton's second law, F_net = m a = 4.352 × 9.088 = 39.54 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,811 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 42.41 kg, acceleration 0.5179 m/s² | A net force acting on a mass of 42.41 kg produces an acceleration of 0.5179 m/s². By Newton's second law, F_net = m a = 42.41 × 0.5179 = 21.96 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,812 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 14.47 kg, acceleration 8.169 m/s² | A net force acting on a mass of 14.47 kg produces an acceleration of 8.169 m/s². By Newton's second law, F_net = m a = 14.47 × 8.169 = 118.2 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,813 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 3.733 kg, acceleration 8.997 m/s² | A net force acting on a mass of 3.733 kg produces an acceleration of 8.997 m/s². By Newton's second law, F_net = m a = 3.733 × 8.997 = 33.59 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,814 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.95 kg, acceleration 4.706 m/s² | A net force acting on a mass of 37.95 kg produces an acceleration of 4.706 m/s². By Newton's second law, F_net = m a = 37.95 × 4.706 = 178.6 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,815 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 27.81 kg, acceleration 9.612 m/s² | A net force acting on a mass of 27.81 kg produces an acceleration of 9.612 m/s². By Newton's second law, F_net = m a = 27.81 × 9.612 = 267.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,816 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 2.102 kg, acceleration 4.111 m/s² | A net force acting on a mass of 2.102 kg produces an acceleration of 4.111 m/s². By Newton's second law, F_net = m a = 2.102 × 4.111 = 8.641 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,817 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 48.84 kg, acceleration 8.515 m/s² | A net force acting on a mass of 48.84 kg produces an acceleration of 8.515 m/s². By Newton's second law, F_net = m a = 48.84 × 8.515 = 415.9 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,818 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 28.41 kg, acceleration 11.87 m/s² | A net force acting on a mass of 28.41 kg produces an acceleration of 11.87 m/s². By Newton's second law, F_net = m a = 28.41 × 11.87 = 337.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,819 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 2.761 kg, acceleration 4.787 m/s² | A net force acting on a mass of 2.761 kg produces an acceleration of 4.787 m/s². By Newton's second law, F_net = m a = 2.761 × 4.787 = 13.22 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,820 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 47.2 kg, acceleration 0.2263 m/s² | A net force acting on a mass of 47.2 kg produces an acceleration of 0.2263 m/s². By Newton's second law, F_net = m a = 47.2 × 0.2263 = 10.68 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,821 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 18.15 kg, acceleration 4.308 m/s² | A net force acting on a mass of 18.15 kg produces an acceleration of 4.308 m/s². By Newton's second law, F_net = m a = 18.15 × 4.308 = 78.2 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,822 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 20.21 kg, acceleration 3.004 m/s² | A net force acting on a mass of 20.21 kg produces an acceleration of 3.004 m/s². By Newton's second law, F_net = m a = 20.21 × 3.004 = 60.7 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,823 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.72 kg, acceleration 10.91 m/s² | A net force acting on a mass of 37.72 kg produces an acceleration of 10.91 m/s². By Newton's second law, F_net = m a = 37.72 × 10.91 = 411.6 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,824 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 48.71 kg, acceleration 10.57 m/s² | A net force acting on a mass of 48.71 kg produces an acceleration of 10.57 m/s². By Newton's second law, F_net = m a = 48.71 × 10.57 = 514.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,825 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 5.32 kg, acceleration 6.169 m/s² | A net force acting on a mass of 5.32 kg produces an acceleration of 6.169 m/s². By Newton's second law, F_net = m a = 5.32 × 6.169 = 32.82 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,826 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 37.2 kg, acceleration 1.605 m/s² | A net force acting on a mass of 37.2 kg produces an acceleration of 1.605 m/s². By Newton's second law, F_net = m a = 37.2 × 1.605 = 59.71 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,827 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 40.47 kg, acceleration 2.338 m/s² | A net force acting on a mass of 40.47 kg produces an acceleration of 2.338 m/s². By Newton's second law, F_net = m a = 40.47 × 2.338 = 94.61 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,828 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 29.57 kg, acceleration 5.793 m/s² | A net force acting on a mass of 29.57 kg produces an acceleration of 5.793 m/s². By Newton's second law, F_net = m a = 29.57 × 5.793 = 171.3 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,829 | physics | mechanics | newton_second_law | 3 | worked_example | Newton's second law: mass 29.52 kg, acceleration 12.8 m/s² | A net force acting on a mass of 29.52 kg produces an acceleration of 12.8 m/s². By Newton's second law, F_net = m a = 29.52 × 12.8 = 377.8 N. Direction of F_net is the same as the direction of the acceleration. This relation defines the inertial mass and is the foundation of classical dynamics. | F_net = m a | kinematics_1d | Compute net force from mass and acceleration using Newton's second law. |
3,830 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 6.486 m | An object of mass 5.944 kg is released from rest at height 6.486 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 11.28 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,831 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 35.47 m | An object of mass 9.37 kg is released from rest at height 35.47 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 26.37 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,832 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 28.88 m | An object of mass 2.047 kg is released from rest at height 28.88 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.8 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,833 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 32.4 m | An object of mass 17.71 kg is released from rest at height 32.4 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 25.21 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,834 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 30.87 m | An object of mass 13.15 kg is released from rest at height 30.87 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 24.61 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,835 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 38.56 m | An object of mass 3.389 kg is released from rest at height 38.56 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 27.5 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,836 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 29.28 m | An object of mass 10.42 kg is released from rest at height 29.28 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.97 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,837 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 37.24 m | An object of mass 13.1 kg is released from rest at height 37.24 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 27.03 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,838 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 31.52 m | An object of mass 6.96 kg is released from rest at height 31.52 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 24.86 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,839 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 14.73 m | An object of mass 6.913 kg is released from rest at height 14.73 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 17 m/s at the reference level.... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,840 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 17.52 m | An object of mass 17.15 kg is released from rest at height 17.52 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 18.54 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,841 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 31.08 m | An object of mass 17.84 kg is released from rest at height 31.08 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 24.69 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,842 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 8.748 m | An object of mass 7.875 kg is released from rest at height 8.748 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 13.1 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,843 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 3.462 m | An object of mass 12.29 kg is released from rest at height 3.462 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 8.241 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,844 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 25.3 m | An object of mass 3.218 kg is released from rest at height 25.3 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 22.27 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,845 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 32.29 m | An object of mass 10.05 kg is released from rest at height 32.29 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 25.16 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,846 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 34.88 m | An object of mass 3.024 kg is released from rest at height 34.88 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 26.16 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,847 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 38.77 m | An object of mass 11.03 kg is released from rest at height 38.77 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 27.58 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,848 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 3.943 m | An object of mass 15.67 kg is released from rest at height 3.943 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 8.794 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,849 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 12.89 m | An object of mass 2.063 kg is released from rest at height 12.89 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 15.9 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,850 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 6.878 m | An object of mass 18.69 kg is released from rest at height 6.878 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 11.61 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,851 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 14.85 m | An object of mass 5.746 kg is released from rest at height 14.85 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 17.07 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,852 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 7.578 m | An object of mass 1.967 kg is released from rest at height 7.578 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 12.19 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,853 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 10.69 m | An object of mass 15.48 kg is released from rest at height 10.69 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 14.48 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,854 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 10.61 m | An object of mass 9.571 kg is released from rest at height 10.61 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 14.43 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,855 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 26.19 m | An object of mass 17.56 kg is released from rest at height 26.19 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 22.66 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,856 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 5.397 m | An object of mass 2.981 kg is released from rest at height 5.397 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 10.29 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,857 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 2.064 m | An object of mass 11.46 kg is released from rest at height 2.064 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 6.362 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,858 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 16.34 m | An object of mass 17.21 kg is released from rest at height 16.34 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 17.9 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,859 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 36.73 m | An object of mass 17.17 kg is released from rest at height 36.73 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 26.84 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,860 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 14.99 m | An object of mass 2.754 kg is released from rest at height 14.99 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 17.14 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,861 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 38.11 m | An object of mass 16.93 kg is released from rest at height 38.11 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 27.34 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,862 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 27.45 m | An object of mass 7.599 kg is released from rest at height 27.45 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.2 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,863 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 10.88 m | An object of mass 10.22 kg is released from rest at height 10.88 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 14.61 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,864 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 27.46 m | An object of mass 17.54 kg is released from rest at height 27.46 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.21 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,865 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 25.93 m | An object of mass 12.49 kg is released from rest at height 25.93 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 22.55 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,866 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 10.17 m | An object of mass 14.87 kg is released from rest at height 10.17 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 14.12 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,867 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 0.8137 m | An object of mass 16.46 kg is released from rest at height 0.8137 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 3.995 m/s at the reference le... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,868 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 11.98 m | An object of mass 17.09 kg is released from rest at height 11.98 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 15.33 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,869 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 16.27 m | An object of mass 3.332 kg is released from rest at height 16.27 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 17.86 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,870 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 15.83 m | An object of mass 1.645 kg is released from rest at height 15.83 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 17.62 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,871 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 4.193 m | An object of mass 9.353 kg is released from rest at height 4.193 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 9.069 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,872 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 25.88 m | An object of mass 9.424 kg is released from rest at height 25.88 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 22.53 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,873 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 18.86 m | An object of mass 18.02 kg is released from rest at height 18.86 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 19.23 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,874 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 29.41 m | An object of mass 1.899 kg is released from rest at height 29.41 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 24.02 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,875 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 36.94 m | An object of mass 18.37 kg is released from rest at height 36.94 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 26.92 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,876 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 8.379 m | An object of mass 15.41 kg is released from rest at height 8.379 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 12.82 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,877 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 13.03 m | An object of mass 6.709 kg is released from rest at height 13.03 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 15.99 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,878 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 19.71 m | An object of mass 19.93 kg is released from rest at height 19.71 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 19.66 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,879 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 19.31 m | An object of mass 14.11 kg is released from rest at height 19.31 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 19.46 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,880 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 2.012 m | An object of mass 5.712 kg is released from rest at height 2.012 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 6.282 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,881 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 19.06 m | An object of mass 3.284 kg is released from rest at height 19.06 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 19.34 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,882 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 13.15 m | An object of mass 12.88 kg is released from rest at height 13.15 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 16.06 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,883 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 24.24 m | An object of mass 0.8202 kg is released from rest at height 24.24 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 21.81 m/s at the reference le... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,884 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 37.49 m | An object of mass 7.73 kg is released from rest at height 37.49 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 27.12 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,885 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 17.49 m | An object of mass 7.358 kg is released from rest at height 17.49 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 18.52 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,886 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 2.636 m | An object of mass 9.574 kg is released from rest at height 2.636 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 7.19 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,887 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 38.13 m | An object of mass 17.39 kg is released from rest at height 38.13 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 27.35 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,888 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 11.16 m | An object of mass 13.74 kg is released from rest at height 11.16 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 14.79 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,889 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 3.802 m | An object of mass 12.79 kg is released from rest at height 3.802 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 8.635 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,890 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 16.5 m | An object of mass 1.571 kg is released from rest at height 16.5 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 17.99 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,891 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 28.63 m | An object of mass 6.107 kg is released from rest at height 28.63 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.7 m/s at the reference leve... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,892 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 21.23 m | An object of mass 17.45 kg is released from rest at height 21.23 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 20.41 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,893 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 10.75 m | An object of mass 13.46 kg is released from rest at height 10.75 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 14.52 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,894 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 4.151 m | An object of mass 16.64 kg is released from rest at height 4.151 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 9.023 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,895 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 14.01 m | An object of mass 18.05 kg is released from rest at height 14.01 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 16.58 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,896 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 6.999 m | An object of mass 15.11 kg is released from rest at height 6.999 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 11.72 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,897 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 17.05 m | An object of mass 7.116 kg is released from rest at height 17.05 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 18.29 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,898 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 27.71 m | An object of mass 9.615 kg is released from rest at height 27.71 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.31 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,899 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 28.02 m | An object of mass 3.382 kg is released from rest at height 28.02 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.44 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
3,900 | physics | mechanics | mechanical_energy | 4 | worked_example | Conservation of mechanical energy: drop from height 27.87 m | An object of mass 7.223 kg is released from rest at height 27.87 m above a reference level. Taking gravitational potential energy as m g h and kinetic energy as (1/2) m v², conservation of mechanical energy (neglecting non-conservative work) yields (1/2) m v² = m g h, so v = sqrt(2 g h) = 23.38 m/s at the reference lev... | K + U = constant (conservative systems); U_g = m g h; K = (1/2) m v^2 | newton_second_law; work-energy theorem | Apply conservation of mechanical energy to free-fall motion. |
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