Work Energy Power

JEE Main level, numerical answer, medium

69 of 183

Question 69 of 183

A body of mass 5kg5kg is moving with a momentum of 10kgms−110kgm{s}^{-1}. Now a force of 2N2N acts on the body in the direction of its motion for 5s5s. The increase in the Kinetic energy of the body is _____ JJ.

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Answer: 30

The formula to calculate the change on momentum of the object is given by
pf−pi=∫Fdt...(1){p}_{f}-{p}_{i}=\int Fdt...\left(1\right)
Substitute the values of the known parameters into equation (1) and solve to calculate the final momentum of the object.
pf−10kgms−1=2N×5s⇒pf=(10+10)kgms−1=20kgms−1\begin{matrix}{p}_{f}-10kgm{s}^{-1} & = & 2N\times 5s \\ & \Rightarrow & {p}_{f}=\left(10+10\right)kgm{s}^{-1} \\ & = & 20kgm{s}^{-1}\end{matrix}
The formula to calculate increase in kinetic energy (∆K)\left(∆K\right) of the particle is given by
∆K=12m(pf2−pi2)...(2)∆K=\dfrac{1}{2m}\left({{p}_{f}}^{2}-{{p}_{i}}^{2}\right)...\left(2\right)
Substitute the values of the known parameters into equation (2) to calculate the required increase in kinetic energy.
∆K=12×5kg×(202−102)kg2m2s−2=30J\begin{matrix}∆K & = & \dfrac{1}{2\times 5kg}\times \left({20}^{2}-{10}^{2}\right){kg}^{2}{m}^{2}{s}^{-2} \\ & = & 30J\end{matrix}

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