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Question

A particle of mass 5 kg is initially at rest. A force starts acting on it along a fixed direction. The magnitude of the force changes with time as shown in the figure. Find the velocity of the particle at the end of 20 s.


A
10 m/s
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B
50 m/s
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C
100 m/s
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D
20 m/s
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Solution

The correct option is D 20 m/s
Impulse imparted by force is given by J=F.dt
i.e Impulse = area under the Ft curve
J=(5×2)+(5×4)+(5×6)+(5×8)
J=100 N-s

Initial velocity v1=0 m/s
particle is at rest initially.
Let the velocity of the particle at the end of 20 s be v2
Impulse imparted J=m Δv
J=m(v2v1)
100=5(v20)
v2=20 m/s

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