Work Energy Power

JEE Main level, numerical answer, medium

22 of 183

Question 22 of 183

A force of F=(5y+20)j^NF=(5y+20)\hat{j}N acts on a particle. The work done by this force when the particle is moved from y=0my=0m to y=10my=10m is ________JJ.

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Answer: 450

F=(5y+20)j^F=(5y+20)\hat{j}
ω=∫Fdy=∫010(5y+20)dy\omega =\int Fdy={\int }_{0}^{10}(5y+20)dy
=(5y22+20y)010={\left(\dfrac{5{y}^{2}}{2}+20y\right)}_{0}^{10}
=52×100+20×10=\dfrac{5}{2}\times 100+20\times 10
=250+200=450J=250+200=450J

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