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Q.Show that the total mechanical energy of an object falling freely under gravity is conserved.

Nagaland NbseNagaland Board of School Education (Class XI) 2023Subjective· 3mImportance★★★★★
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Computing KE + PE at the top, at an intermediate point, and just before landing all give the same value mghmgh, proving conservation.

Consider a body of mass mm released from rest at height hh above the ground, falling freely under gravity gg (no air resistance, so only the conservative gravitational force acts).

At the starting point (height hh, velocity 0): KEA=0KE_A = 0, PEA=mghPE_A = mgh. Total energy EA=0+mgh=mghE_A = 0+mgh = mgh.

At an intermediate point, after falling a distance xx (so at height h−xh-x above the ground), using v2=2gxv^2=2gx: KEB=12mv2=12m(2gx)=mgxKE_B = \tfrac12mv^2 = \tfrac12m(2gx) = mgx, and PEB=mg(h−x)PE_B = mg(h-x). Total energy EB=mgx+mg(h−x)=mghE_B = mgx + mg(h-x) = mgh.

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