Question

A small body of mass $500g$ moves on a rough horizontal surface before finally stops. The initial velocity of the body is $2m/s$ and coefficient of friction is $0.3$. Then, find absolute value of the average power developed by the frictional force during the time of motion. (Take, $g=10m/s^2$ )

• A. 1 W
• B. 1.5 W
• C. 2 W
• D. 2.5 W

Answer 1

Answer: (B)

As, retardation is constant, so average velocity
$v_{av}=\frac{u+v}{2}=\frac{2+0}{2}=1m/s$
Now, friction force, $f=\mu N=\mu mg$
Given, coefficient of friction, $\mu =0.3$
Mass of body, $m=500g=0.5kg$
$\Rightarrow f=0.3\times 0.5\times 10$
$f=1.5N$
Average power by friction,
$P_{av}=f{v}_{av}$
$P_{av}=1.5\times 1=1.5W$