Work Energy Power

JEE Main level, single correct, medium

10 of 183

Question 10 of 183

A particle of mass mm moves on a straight line with its velocity increasing with distance according to the equation v=αxv=\alpha \sqrt{x}, where α\alpha is a constant. The total work done by all the forces applied on the particle during its displacement from x=0x=0 to x=dx=\mathrm{d}, will be :

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Answer: (D) mα2 d2\frac{m \alpha^2 \mathrm{~d}}{2}

v=αx at x=0:v=0& at x=d;v=αd W.D =Kf−Ki W.D =12 m(αd)2−12 m(0)2⇒ W.D =mα2 d2\begin{aligned} & \mathrm{v}=\alpha \sqrt{\mathrm{x}} \\ & \text { at } \mathrm{x}=0: \mathrm{v}=0 \\ & \& \text { at } \mathrm{x}=\mathrm{d} ; \mathrm{v}=\alpha \sqrt{\mathrm{d}} \\ & \text { W.D }=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}} \\ & \text { W.D }=\frac{1}{2} \mathrm{~m}(\alpha \sqrt{\mathrm{d}})^2-\frac{1}{2} \mathrm{~m}(0)^2 \\ & \Rightarrow \text { W.D }=\frac{\mathrm{m} \alpha^2 \mathrm{~d}}{2}\end{aligned}

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