prove that energy remains constant in the case of freely falling bodies.
Answers
Explanation:
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ANSWER:
In this case we have to show that the total energy (potential energy + kinetic energy) of the body at A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy.
EXPLANATION:
At A,
Potential energy = mgh
Kinetic energy = 1/2 mv² = 1/2 * m* 0Kinetic energy = 0 [the velocity is zero as the object is initially at rest]
Total energy at A = Potential energy + Kinetic energy
= mgh + 0
Total energy at A = mgh …(1)
At B,
Potential energy = mgh
= mg(h - x) [Height from the ground is (h - x)]Potential energy = mgh - mgx
Kinetic energy = 1/2 mv²
The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body.
v2 - u2 = 2aSHere, u = 0, a = g and S = x
v² - 0 = 2gx
Here, u = 0, a = g and S = x
v² - 0 = 2gx
v² = 2gx
kinetic energy = 1/2 mv²
= 1/2 m2gx
Kinetic energy = mgx
Total energy at B = Potential energy + Kinetic energy
= mgh - mgx + mgx
Total energy at B = mgh …(2)
At C,
Potential energy = m x g x 0 (h = 0)
Potential energy = 0Kinetic energy = 1/2 mv²
The distance covered by the body is hv2 - u2 = 2aS
Here, u = 0, a = g and S = h
v² - 0 = 2gh
v² = 2gh
kinetic energy = 1/2 mv²
= 1/2 m * 2gh
Kinetic energy = mgh
Total energy at C = Potential energy + Kinetic energy= 0 + mgh
Total energy at C = mgh …(3)It is clear from equations 1, 2 and 3 that the total energy of the body remains constant at every point. Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body.