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Question

A box of mass 20 kg is moving with a velocity of 2 ms1 on a rough floor. The box moves 5 m on the floor before coming to rest. What must be the frictional force acting on the box?


Your Answer
A
4 N
Your Answer
B
2 N
Your Answer
C
20 N
Correct Answer
D
8 N

Solution

The correct option is D 8 N
Given: mass of box , m=20 kg, initial velocity, u=2 ms1, final velocity, v=0 ms1 and distance travelled, s=5 m
Let frictional force be F.
From the third equation of motion, v2u2=2as, we get a=v2u22s=0222×5=0.4 ms2
Frictional force, F=m×a=20×0.4=8 N

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