A particle of mass 5kg 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 20s.
A
10m/s
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B
50m/s
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C
100m/s
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D
20m/s
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Solution
The correct option is D20m/s Impulse imparted by force is given by J=∫F.dt i.e Impulse = area under the F−t curve ⇒J=(5×2)+(5×4)+(5×6)+(5×8) ∴J=100N-s
Initial velocity v1=0 m/s ∵ particle is at rest initially. Let the velocity of the particle at the end of 20s be v2 Impulse imparted J=mΔv ⇒J=m(v2−v1) ⇒100=5(v2−0) ∴v2=20m/s