A particle of mass 2 kg is initially at rest. A fore acts on it whose magnitude changes with time. The force - time graph is shown below. The velocity of the particle after 10s is
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deepas05122001:
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see, first we should understand, what is the area of force-time graphs shows. It's the impulse. And impulse is equals to change in momentum.
first find impulse,
impluse = area
=((1/2)×2×10)+ (2×10)+ ((1/2)×(10+20)×2)+((1/2)×4×20)
=10+20+30+40
=100 N-s
now, impulse = change in momentum
=m×v - m×u
where v= final velocity and u =initial velocity =0
so, 100 = 2×v
⇒v = 50 m/s
hope this will help you
first find impulse,
impluse = area
=((1/2)×2×10)+ (2×10)+ ((1/2)×(10+20)×2)+((1/2)×4×20)
=10+20+30+40
=100 N-s
now, impulse = change in momentum
=m×v - m×u
where v= final velocity and u =initial velocity =0
so, 100 = 2×v
⇒v = 50 m/s
hope this will help you
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