Question

A conveyor belt is moving with a constant velocity 2 m/s. Block is projected with speed of 4 m/s in
opposite direction of velocity of belt, if mass of block is '4 kg' and coefficient of friction is 0.2. Find work
done against friction on conveyor belt up to the instant when slipping between block and belt ceases :
(A) 24 Joule
(B) 48 Joule
(C) 12 Joule
(D) 36 Joule
​

Answer 1

Answer:

The answer is 16 J.

Explanation:

Normal force on belt due to block is

F= m×a

F=4 ×g (Since g = 10 m/s^2)

F= 4 X 10

F= 40 N

Kinetic frictional force on package = Coefficient of friction  x Force on belt due to block = 8N

Acceleration due to block

a=F / m

a=8 / 4=2 m/s2

The velocity with which body is projected is 4m/s

The friction will cause the body to decelerate from 4m/s to 2m/s

From  

v=u+at

t=v−u /a

t=2 sec

Work done = Force x displacement

               = =8×V / T=8×4 / 2=16 J