Work Energy Power

JEE Main level, single correct, medium

166 of 183

Question 166 of 183

A body of mass mm, accelerates uniformly from rest to v1v_1 in time t1t_1. The instantaneous power delivered to the body as a function of time tt is

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Answer: (B) mv12tt12\frac{\mathrm{mv}_1^2 \mathrm{t}}{\mathrm{t}_1^2}

Power P=F⃗⋅v⃗=mav⁡=m(v1t1)(v1t1t)=m12tt12P=\vec{F} \cdot \vec{v}=\operatorname{mav}=m\left(\frac{v_1}{t_1}\right)\left(\frac{v_1}{t_1} t\right)=\frac{m_1^2 t}{t_1^2}

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