Q.Draw the variation of kinetic energy and potential energy of a freely falling body, with height.
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Start your 14-day free trial to unlock the full solution →Plotted against height above the ground, PE is a straight line increasing with height (through the origin) and KE is a straight line decreasing with height — the two lines cross, and at every height their sum equals the constant total energy mgh.
Let a body of mass m be released from rest at height h above the ground, and let h′ be its height above the ground at any instant during the fall (0 ≤ h′ ≤ h).
Potential energy at height h′ (taking the ground as reference):
PE = mgh′
This is a straight line through the origin when plotted against h′ — it is zero at the ground (h′ = 0) and rises linearly to its maximum value mgh at the point of release (h′ = h).
Kinetic energy at height h′, using v² = 2g(h − h′) from the equations of motion (falling from rest):
KE = ½mv² = ½m·2g(h − h′) = mg(h − h′)
This is also a straight line in h′, but a decreasing one — it is zero at the top (h′ = h, where the body is still at rest) and rises linearly to its maximum value mgh at the ground (h′ = 0).
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