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Question

A body of mass 2 kg having a velocity of 2 ms1 is brought to rest within a distance of 2 m. Find the magnitude of force acting on the body.

A
8 N
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B
1 N
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C
4 N
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D
2 N
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Solution

The correct option is D 2 N
Given:
Initial velocity, u=2 ms1
Stopping distance, s=2 m
As the object comes to rest, final velocity v=0ms1
Mass, m=2 kg

Let the acceleration be a
From the third equation of motion, we have
v2u2=2as
a=v2u22s=02222×2=1 ms2

From Newton's second law of motion, force on the body is:
F=m×a
F=2×1=2 N

It may be noted that the negative sign implies that the force is acting in a direction opposite to the motion of the object.

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