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. $ 100N $ backward
B. $ 100N $ forward
C. $ 60N $ forward
D. $ 60N $ backward

Answer 1

Hint :To solve this question, first we rewrite the given facts that are given in the question and then we will find the maximum acceleration needed to slide the box. And then compare the acceleration of the truck and the maximum acceleration to find the friction force between the box and the truck.

Complete Step By Step Answer:
Firstly, rewrite the given information about the question:
Mass of the box which is on the floor of the truck $ = 20kg $
Static friction between the box and the truck is $ 0.3 $ .
Sliding Friction of the box is $ 0.2 $ .
And, the frictional force $ = 2m{s^{ - 2}} $ .
Now,
The box is sliding, so we can calculate the maximum acceleration needed to slide the box:
 $ \therefore m \times {a_{\max }} = {\mu _s} \times N = {\mu _s} \times mg $
here, $ m $ is the given mass of the box.
 $ {a_{\max }} $ is the maximum acceleration.
 $ {\mu _s} $ is the static friction between the box and the truck.
 $ g $ is the gravitational force because the box is sliding downward.
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, i.e. in the forward direction.
Hence, the correct option is (C.) $ 60N $ forward.

Note :
Friction always reduces acceleration. Friction occurs between the interaction of an object against a surface. Its magnitude depends on the characteristics of both the surface and the object, and whether the object is moving or not.