Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
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Energy Conservation for a Free Falling Body
The mechanical energy of a free falling body under the effect of gravity is constant. This is shown in the given figure.
Let us take an object of mass m at a point A, above the ground at a height of h from the ground.
At A, the body is at rest. Thus,
K.E. of the body = 0
P.E. of the body = m g h
Total energy of the body at
A = K.E. + P.E. = 0 + m g h = m g h
Suppose, the body falls freely from A to the ground and at a point B on its path, its height above the ground will become (h - x).
Therefore, potential energy of the body = m g (h - x)
If vB is the velocity at B, then

Total energy of the body at
B = K.E. + P.E. = m g (h - x) + m g x = m g h
At point C,
If vC be the velocity of the body at point C (just before striking the ground), then

Therefore, total energy at point C= K.E. + P.E. = m g h + 0 = m g h

Fig. Variation of kinetic energy and potential energy at different points on the path
Thus, the total mechanical energy of the body remains the same at all points during the downward journey. As the body falls from height h, its potential energy decreases while the kinetic energy begins to increase; the total mechanical energy at any point being the same i.e. m g h. When the body reaches the ground, P.E. = 0 and K.E. is equal to the initial potential energy m g h at the height h. The figure below shows the variation of P.E. and K.E. with height. Similar case can be applied when a body is thrown vertically upward.
hope it will help you
The mechanical energy of a free falling body under the effect of gravity is constant. This is shown in the given figure.
Let us take an object of mass m at a point A, above the ground at a height of h from the ground.
At A, the body is at rest. Thus,
K.E. of the body = 0
P.E. of the body = m g h
Total energy of the body at
A = K.E. + P.E. = 0 + m g h = m g h
Suppose, the body falls freely from A to the ground and at a point B on its path, its height above the ground will become (h - x).
Therefore, potential energy of the body = m g (h - x)
If vB is the velocity at B, then

Total energy of the body at
B = K.E. + P.E. = m g (h - x) + m g x = m g h
At point C,
If vC be the velocity of the body at point C (just before striking the ground), then

Therefore, total energy at point C= K.E. + P.E. = m g h + 0 = m g h

Fig. Variation of kinetic energy and potential energy at different points on the path
Thus, the total mechanical energy of the body remains the same at all points during the downward journey. As the body falls from height h, its potential energy decreases while the kinetic energy begins to increase; the total mechanical energy at any point being the same i.e. m g h. When the body reaches the ground, P.E. = 0 and K.E. is equal to the initial potential energy m g h at the height h. The figure below shows the variation of P.E. and K.E. with height. Similar case can be applied when a body is thrown vertically upward.
hope it will help you
Answered by
13
ANSWER:-
The mechanical energy of a free falling body under the effect of gravity is constant. This is shown in the given figure.
Let us take an object of mass m at a point A, above the ground at a height of h from the ground.
At A, the body is at rest. Thus,
K.E. of the body = 0
P.E. of the body = m g h
Total energy of the body at
A = K.E. + P.E. = 0 + m g h = m g h
Suppose, the body falls freely from A to the ground and at a point B on its path, its height above the ground will become (h - x).
Therefore, potential energy of the body = m g (h - x)
If vB is the velocity at B, then
Total energy of the body at
B = K.E. + P.E. = m g (h - x) + m g x = m g h
At point C,
If vC be the velocity of the body at point C (just before striking the ground), then
Therefore, total energy at point C= K.E. + P.E. = m g h + 0 = m g h
Thus, the total mechanical energy of the body remains the same at all points during the downward journey. As the body falls from height h, its potential energy decreases while the kinetic energy begins to increase; the total mechanical energy at any point being the same i.e. m g h. When the body reaches the ground, P.E. = 0 and K.E. is equal to the initial potential energy m g h at the height h. The figure below shows the variation of P.E. and K.E. with height. Similar case can be applied when a body is thrown vertically upward.
The mechanical energy of a free falling body under the effect of gravity is constant. This is shown in the given figure.
Let us take an object of mass m at a point A, above the ground at a height of h from the ground.
At A, the body is at rest. Thus,
K.E. of the body = 0
P.E. of the body = m g h
Total energy of the body at
A = K.E. + P.E. = 0 + m g h = m g h
Suppose, the body falls freely from A to the ground and at a point B on its path, its height above the ground will become (h - x).
Therefore, potential energy of the body = m g (h - x)
If vB is the velocity at B, then
Total energy of the body at
B = K.E. + P.E. = m g (h - x) + m g x = m g h
At point C,
If vC be the velocity of the body at point C (just before striking the ground), then
Therefore, total energy at point C= K.E. + P.E. = m g h + 0 = m g h
Thus, the total mechanical energy of the body remains the same at all points during the downward journey. As the body falls from height h, its potential energy decreases while the kinetic energy begins to increase; the total mechanical energy at any point being the same i.e. m g h. When the body reaches the ground, P.E. = 0 and K.E. is equal to the initial potential energy m g h at the height h. The figure below shows the variation of P.E. and K.E. with height. Similar case can be applied when a body is thrown vertically upward.
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