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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.

Asked in JEE Main 25th July 2nd Shift 2021 · Integrating a position-dependent force

Answer: 450

Step-by-step solution

W=∫₀¹⁰(5y+20) dy=[(5y²)/2+20y]₀¹⁰.

W=250+200=450 J.

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