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Q.Prove that the mechanical energy of a ball of mass 'm' dropped from a height 'H' is conserved.

Kerala DhseKerala DHSE Plus One Board 2021Subjective· 4mImportance★★★★★
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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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