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Question

A particle of mass 1 kg is initially at rest. A force starts acting on it in one direction whose magnitude changes with time. The (F - t) graph is shown in figure. The velocity of the particle at the end of 10 s is

A
150 m/s
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B
120 m/s
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C
100 m/s
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D
112.5 m/s
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Solution

The correct option is D 112.5 m/s
Given:
Velocity of particle at (t=0); vi=0
Let velocity of particle at (t=10 s) is vf.
Using Impulse(J) = change in momentum ΔP

As we know area under (F -t) graph gives change in momentum.

we have, m(vfvi)= Area under (F - t) graph from t=0 to t=10 s

m(vfvi)= Area under (F - t) graph from (t=0 to t=5 s)+Area under (F - t) graph from (t=5 to t=10 s)
mvf=12×5×15+5×15
1.vf=12×5×15+5×15
vf=112.5 m/s

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