Work Energy Power

JEE Main level, single correct, medium

33 of 183

Question 33 of 183

A force F⃗=(5i^+3j^+2k^)N\vec{F}=(5 \hat{i}+3 \hat{j}+2 \hat{k}) N is applied over a particle which displaces it from its origin to the point r⃗=(2i^−j^)m\vec{r}=(2 \hat{i}-\hat{j}) m. The work done on the particle in joules is

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Answer: (B) +7+7

Work done, W =F→⋅s→=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{s}}
Here s⃗=r⃗f−r⃗i=(2i^−j^)\vec{s}=\vec{r}_f-\vec{r}_i=(2 \hat{i}-\hat{j})
W=(5i^+3j^+2k^)(2i^−j^)=10−3=7 JW=(5 \hat{i}+3 \hat{j}+2 \hat{k})(2 \hat{i}-\hat{j})=10-3=7 \mathrm{~J}

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