Work Energy Power

JEE Main level, numerical answer, medium

15 of 183

Question 15 of 183

A force F=(5+3y2)F=(5+3{y}^{2}) acts on a particle in the yy-direction, where FF is newton and yy is in meter. The work done by the force during a displacement from y=2my=2m to y=5my=5m is ______ JJ.

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Answer: 132

If an object is moved by certain distance by the application of an external force along the direction of the force, then the force is said to perform some work.
Given, the force exerted on the object is
F=5+3y2F=5+3{y}^{2}
The formula to calculate the work done (W)\left(W\right) by the force in moving the object from a position y1{y}_{1} to another position y2{y}_{2} can be written as
W=∫y1y2Fdy..........................(1)W={\int }_{{y}_{1}}^{{y}_{2}}Fdy..........................(1)
Substitute the values into equation (1) and solve to calculate the required work done by the force.
W=∫25(5+3y2)dy=[5y+3y33]25=132\begin{matrix}W & = & {\int }_{2}^{5}\left(5+3{y}^{2}\right)dy \\ & = & {\left[5y+\dfrac{3{y}^{3}}{3}\right]}_{2}^{5} \\ & = & 132\end{matrix}

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