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A particle moves from position $\overrightarrow{\mathbf{r}}_1=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$ to position $\overrightarrow{\mathbf{r}}_2=14 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$ under action of force $\mathbf{F}=4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \mathrm{N}$. The work done will be:
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Verified Answer
The correct answer is:
100 J
Using the formula for work done
$W=\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{F}} \cdot\left(\overrightarrow{\mathbf{r}_2}-\overrightarrow{\mathbf{r}_1}\right)$
$\begin{array}{r}W=(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(14 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}) -3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})\end{array}$
$\begin{aligned} & W=(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(11 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}+15 \hat{\mathbf{k}}) \\ & W=44+11+45=100 \text { joule }\end{aligned}$
$W=\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{F}} \cdot\left(\overrightarrow{\mathbf{r}_2}-\overrightarrow{\mathbf{r}_1}\right)$
$\begin{array}{r}W=(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(14 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}) -3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})\end{array}$
$\begin{aligned} & W=(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(11 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}+15 \hat{\mathbf{k}}) \\ & W=44+11+45=100 \text { joule }\end{aligned}$
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