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Q.Show that the mechanical energy of a body falling freely is always conserved.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2024Subjective· 3mImportance★★★★★
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For a body falling freely from height hh, the sum KE+PEKE + PE is the same (mghmgh) at every point of the fall.

Consider a body of mass mm released from rest at height hh above the ground, falling freely under gravity (no air resistance). Take a point at height xx above the ground, where the body has already fallen a distance (h−x)(h-x).

At the starting point (height hh):

KEi=0,PEi=mgh,Ei=0+mgh=mghKE_i = 0, \qquad PE_i = mgh, \qquad E_i = 0 + mgh = mgh

At an intermediate height xx: using v2=u2+2asv^2 = u^2 + 2as with u=0u=0, a=ga=g, and distance fallen s=(h−x)s = (h-x):

v2=2g(h−x)v^2 = 2g(h-x)

KE=12mv2=12m×2g(h−x)=mg(h−x)KE = \dfrac{1}{2}mv^2 = \dfrac{1}{2}m \times 2g(h-x) = mg(h-x)

PE=mgxPE = mgx

E=KE+PE=mg(h−x)+mgx=mgh−mgx+mgx=mghE = KE + PE = mg(h-x) + mgx = mgh - mgx + mgx = mgh

So at this arbitrary intermediate point, the total mechanical energy is still mghmgh — exactly the same as at the start.

At the ground (height 0), just before impact: using v2=2ghv^2 = 2gh,

KE=12mv2=mgh,PE=0,E=mgh+0=mghKE = \dfrac{1}{2}mv^2 = mgh, \qquad PE = 0, \qquad E = mgh + 0 = mgh …

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