JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 24)
A small particle moves to position $$5 \hat{i}-2 \hat{j}+\hat{k}$$ from its initial position $$2 \hat{i}+3 \hat{j}-4 \hat{k}$$ under the action of force $$5 \hat{i}+2 \hat{j}+7 \hat{k} \mathrm{~N}$$. The value of work done will be __________ J.
Answer
40
Explanation
The given expression calculates the work done by a force vector $\vec{F} = 5\hat{i} + 2\hat{j} + 7\hat{k}$ when it acts on an object that moves from an initial position vector $\vec{r}_i = 2\hat{i} + 3\hat{j} - 4\hat{k}$ to a final position vector $\vec{r}_f = 5\hat{i} - 2\hat{j} + \hat{k}$.
To find the work done, we use the dot product of the force and displacement vectors :
$$ \begin{aligned} & W=\vec{F} \cdot\left(\vec{r}_f-\vec{r}_{\mathrm{i}}\right) \\\\ & =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot((5 \hat{i}-2 \hat{j}+\hat{k})-(2 \hat{i}+3 \hat{j}-4 \hat{k})) \\\\ & =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}+5 \hat{k}) \\\\ & =15-10+35 \\\\ & =40 \mathrm{~J} \end{aligned} $$
To find the work done, we use the dot product of the force and displacement vectors :
$$ \begin{aligned} & W=\vec{F} \cdot\left(\vec{r}_f-\vec{r}_{\mathrm{i}}\right) \\\\ & =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot((5 \hat{i}-2 \hat{j}+\hat{k})-(2 \hat{i}+3 \hat{j}-4 \hat{k})) \\\\ & =(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}+5 \hat{k}) \\\\ & =15-10+35 \\\\ & =40 \mathrm{~J} \end{aligned} $$
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