Q.Prove that the mechanical energy of a ball of mass 'm' dropped from a height 'H' is conserved.
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Start your 14-day free trial to unlock the full solution →Compute KE + PE at the point of release, at an intermediate height h, and just before impact, using v² = 2g(H−h) for free fall. All three totals equal mgH, proving mechanical energy is conserved.
Consider a ball of mass m dropped from rest from height H above the ground. Take the ground as the reference (zero) level for gravitational potential energy. Only gravity acts (air resistance neglected), so the ball is in free fall with acceleration g.
Point 1 — at the point of release (height H, speed v = 0):
KE₁ = 0
PE₁ = mgH
Total mechanical energy, E₁ = KE₁ + PE₁ = 0 + mgH = mgH
Point 2 — at an intermediate height h above the ground (the ball has fallen a distance x = H − h):
Using v² = u² + 2as with u = 0, a = g, s = (H − h):
v² = 2g(H − h)
KE₂ = ½mv² = ½m·2g(H−h) = mg(H − h)
PE₂ = mgh
Total mechanical energy, E₂ = KE₂ + PE₂ = mg(H − h) + mgh = mgH
Point 3 — just before striking the ground (h = 0):
v² = 2gH
KE₃ = ½m(2gH) = mgH
PE₃ = mg(0) = 0 …
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