Work Energy Power

JEE Main level, single correct, medium

132 of 183

Question 132 of 183

A particle gets displaced by Δrˉ=(2i^+3j^+4k^)m\Delta \bar{r}=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \mathrm{m} under the action of a force F⃗=(7i^+4j^+3k^)\vec{F}=(7 \hat{i}+4 \hat{j}+3 \hat{k}). The change in its kinetic energy is

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Answer: (A) 38 J38 \mathrm{~J}

According to work-energy theorem, Change in kinetic energy == work done =F⃗⋅Δr⃗=(7i^+4j^+3k^)⋅(2i^+3j^+4k^)=\vec{F} \cdot \Delta \vec{r}=(7 \hat{i}+4 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k}) =14+12+12=38 J=14+12+12=38 \mathrm{~J}

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