Work Energy Power

JEE Main level, numerical answer, medium

13 of 183

Question 13 of 183

A block of mass 10kg10kg is moving along xx-axis under the action of force F=5xNF=5xN. The work done by the force in moving the block from x=2mx=2m to 4m4m will be _______ JJ.

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Answer: 30

Work done is W=∫FdxW=\int Fdx
It is given that
F=5xNF=5xN
So, work done is
W=∫245xdx=5x2224⇒W=5(16−4)2=30JW={\int }_{2}^{4}5xdx={\dfrac{5{x}^{2}}{2}}_{2}^{4} \Rightarrow W=\dfrac{5(16-4)}{2}=30J

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