Question

A block of mass m is placed at point A on a rough track as shown in the figure. If slightly pushed towards right, it stops at point B of the track. Calculate the work done by the frictional force on the block during its transit from A to B.

• A. $-mg(H-h)$
• B. $-mg(H+h)$
• C. $mg\frac{h}{H}$
• D. $-mg\frac{h}{H}$

Answer 1

The correct option is A $-mg(H-h)$

Kinetic energy of block at point $A,(K.E.{)}_A=0$
(Initially, the block is at rest)
Kinetic energy of block at point $B,(K.E.{)}_B=0$
(Block stops at point B)
Using work energy theorem:
Work done by gravitational force + Work done by frictional force = Change in kinetic energy
$\Rightarrow W_g+W_f=(K.E.{)}_B–(K.E.{)}_A=0\Rightarrow W_f=–W_g\dots ..\left(i\right)$
Work done by gravitational force,$W_g=Fscos\theta$
Here, Gravitational force, $F=mg$
Displacement, s = H – h
Angle between $Fands,\theta =0^{\circ }\Rightarrow W_g=mg(H–h)\Rightarrow W_f=–W_g\left[From\right(i\left)\right]\Rightarrow W_f=–mg(H–h)$
Hence, the correct answer is option (a).