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Question: Answered & Verified by Expert
A force of F=(5y+20)j^ N acts on a particle. The work done by this force when the particle is moved from y=0 m to y=10 m is ________J.
PhysicsWork Power EnergyJEE MainJEE Main 2021 (25 Jul Shift 2)
Solution:
2985 Upvotes Verified Answer
The correct answer is: 450

F=(5y+20)j^

ω=Fdy=010(5y+20)dy

=5y22+20y010

=52×100+20×10

=250+200=450 J

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