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Question

# A body of mass 5 kg has momentum of 10 kg m/s. When a force of 0.2N is applied on it for 10 seconds, what is the change in its kinetic energy?

A
1.1J
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B
2.2J
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C
3.3J
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D
4.4J
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Solution

## The correct option is A 4.4J%Increase in momentum(P)=n.Then,% Increase in kinetic energy(E)=n2+200n100.\quad Initial Momentum =P=10N−s Mass=5kg, Force =0.2N. Time for which force acted =10s. Now, Initial Kinetic Energy=E=12mv2=12(m)(pm)2=12(5)(105)2=10J Now, ΔP=F×tP′−P=2P′=P+2. we see that increase in momentum is (210)×100=20 Increase in Kinetic Energy=44=4410Increase in Kinetic Energy =4.4J.

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