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Question: Answered & Verified by Expert
A small object of mass of 100 g moves in a circular path. At a given instant velocity of the object is 10i^ m s-1 and acceleration is (20i^+10j^) m s-2. At this instant of time, the rate of change of kinetic energy of the object is
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Options:
  • A 200 kg m2s-2
  • B 300 kg m2 s-2
  • C 10000 kg m2 s-2
  • D 20 kg m2 s-2
Solution:
2866 Upvotes Verified Answer
The correct answer is: 20 kg m2 s-2
Given, the mass of the object m=100 g

=100×10-3 kg

Velocity of object v=10i^ m s-1 

Acceleration of object a=(20i^+10j^) m s-2

We know that,

d(KE)dt=F.v=ma.v  [K.E.=12mv2]

=(100×10-3)(20i^+10j^).(10i^)

=100×10-3×200

=100×2001000=20 kg ms-3 

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