Question

A 20kg box is on the floor of a truck. The coefficient of static friction between the box and the truck floor is 0.3 and the coefficient of sliding friction is 0.2. The magnitude and direction of the frictional force acting on the box when the truck is accelerating at $2m{s}^{-2}$ is:

• A. 100 N backward
• B. 100 N forward
• C. 60 N forward
• D. 60 N backward

Answer 1

Answer: (C)

60 N forward

From the free body diagram of the box on the floor (fig-a), we can calculate the maximum acceleration needed to slide the box.

$m\times a_{max}$=${\mu }_s\times N$= ${\mu }_s\times mg$.

Using, $g=10m/s^2$, we can get $a_{max}=3m/s^2$.

Now here the acceleration of the truck is $2m/s^2$ which is smaller than $a_{max}$ so the maximum friction force acting on the box can be $0.3\times 10\times 20$ $=60N$, along the direction of motion of the truck.