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Question

# A body of mass 2 kg having a velocity of 2 ms−1 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 NGiven: Initial velocity, u=2 ms−1 Stopping distance, s=2 m As the object comes to rest, final velocity v=0ms−1 Mass, m=2 kg Let the acceleration be a From the third equation of motion, we have v2−u2=2as a=v2−u22s=02−222×2=−1 ms−2 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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