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Q.Show that the total mechanical energy of a freely falling body is constant.

Kerala DhseKerala DHSE Plus One Board 2022Subjective· 3mImportance★★★★★
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Compute KE + PE at an arbitrary height during the fall using the equations of motion, and show the sum always equals mgh, independent of position — this constancy is the proof.

Let a body of mass m be released from rest at height h above the ground (taking the ground as the reference level for PE, and neglecting air resistance).

At the point of release (height h, speed 0):

KE = 0

PE = mgh

Total mechanical energy, E = KE + PE = 0 + mgh = mgh ... (i)

After falling through a height x (so the body is now at height h′ = h − x above the ground), using v² = u² + 2as with u = 0, a = g, s = x:

v² = 2gx

KE at this point = ½mv² = ½m(2gx) = mgx

PE at this point (height h − x above ground) = mg(h − x)

Total mechanical energy at this point:

E = KE + PE = mgx + mg(h − x) = mgx + mgh − mgx = mgh ... (ii)

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