Work Energy Power

JEE Main level, numerical answer, medium

16 of 183

Question 16 of 183

A small particle moves to position 5i^−2j^+k^5\hat{i}-2\hat{j}+\hat{k} from its initial position 2i^+3j^−4k^2\hat{i}+3\hat{j}-4\hat{k} under the action of force 5i^+2j^+7k^N5\hat{i}+2\hat{j}+7\hat{k}N. The value of work done will be ______ JJ.

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Answer: 40

Work done by a constant force is given by W=Force(F⃗)⋅Displacement(r⃗)W=Force\left(\vec{F}\right)\cdot Displacement\left(\vec{r}\right), where, r⃗=r⃗2−r⃗1\vec{r}={\vec{r}}_{2}-{\vec{r}}_{1}
Given here, r1⃗=5i^−2j^+k^\vec{{r}_{1}}=5\hat{i}-2\hat{j}+\hat{k} and r⃗2=2i^+3j^−4k^{\vec{r}}_{2}=2\hat{i}+3\hat{j}-4\hat{k}
So, W=F⃗⋅(r⃗2−r⃗1)W=\vec{F}\cdot \left({\vec{r}}_{2}-{\vec{r}}_{1}\right)
=(5i^+2j^+7k^)⋅[(5i^−2j^+k^)−(2i^+3j^−4k^)]=(5\hat{i}+2\hat{j}+7\hat{k})\cdot \left[(5\hat{i}-2\hat{j}+\hat{k})-(2\hat{i}+3\hat{j}-4\hat{k})\right]
⇒W=40J\Rightarrow W=40J

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