Prove the law of conservation of energy taking a stone moving vertically down.
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Let the stone of mass m be at a height h and be dropped from rest at t = 0 sec.
Initial potential energy = PE1 = m g h
initial kinetic energy = 1/2 m 0^2 = 0
Let the stone drop by a distance s in time t. As per the laws of kinetics we have
2 g s = (v^2 - u^2 ) = v^2
Potential energy at time t and height of (h - s) is PE2 = m g (h - s)
Kinetic energy = KE2 = 1/2 m v^2 = 1/2 m (2 g s) = m g s
Total energy = m g (h - s) + m g s = m g h
All along the path , from t = o to t = root(2h/g) , and height h to 0, the total energy remains as m g h.
The other forms of energy in the object are assumed to be constant and not varying.
Initial potential energy = PE1 = m g h
initial kinetic energy = 1/2 m 0^2 = 0
Let the stone drop by a distance s in time t. As per the laws of kinetics we have
2 g s = (v^2 - u^2 ) = v^2
Potential energy at time t and height of (h - s) is PE2 = m g (h - s)
Kinetic energy = KE2 = 1/2 m v^2 = 1/2 m (2 g s) = m g s
Total energy = m g (h - s) + m g s = m g h
All along the path , from t = o to t = root(2h/g) , and height h to 0, the total energy remains as m g h.
The other forms of energy in the object are assumed to be constant and not varying.
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Answer:
Let the stone of mass m be at a height h and be dropped from rest at t = 0 sec.
Initial potential energy = PE1 = m g h
initial kinetic energy = 1/2 m 0^2 = 0
Let the stone drop by a distance s in time t. As per the laws of kinetics we have
2 g s = (v^2 - u^2 ) = v^2
Potential energy at time t and height of (h - s) is PE2 = m g (h - s)
Kinetic energy = KE2 = 1/2 m v^2 = 1/2 m (2 g s) = m g s
Total energy = m g (h - s) + m g s = m g h
All along the path , from t = o to t = root(2h/g) , and height h to 0, the total energy remains as m g h.
The other forms of energy in the object are assumed to be constant and not varying.
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