Question

25. A body of mass M falls from a height $H_1$ to a height $H_2$, above the ground ($H_1>H_2$). Then, the loss in its potential energy is given by ____.
A) $Mg(H_1-H_2)$
B) $Mg(H_1+H_2)$
C) $Mg(H_2-H_1)$
D) $Mg(H_1.H_2)$

Answer 1

Answer: (A) $Mg(H_1-H_2)$.

Given,

Mass of body = M

Initial height = $H_1$

Final height = $H_2$

As we know, Potential Energy= $MgH$
Therefore, Loss in the P.E. = Initial P.E. - Final P.E. = $Mg{H}_1-Mg{H}_2$

$△U=Mg(H_1-H_2)$.