Work Energy and Power - Result Question 3
5. A uniform force of $(3 \hat{i}+\hat{j})$ newton acts on a particle of mass $2 kg$. The particle is displaced from position $(2 \hat{i}+\hat{k})$ meter to position $(4 \hat{i}+3 \hat{j}-\hat{k})$ meter. The work done by the force on the particle is
[2013]
(a) $6 J$
(b) $13 J$
(c) $15 J$
(d) $9 J$
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Answer:
Correct Answer: 5. (d)
Solution:
- (d) Given : $\overrightarrow{{}F}=3 \hat{i}+\hat{j}$
$ \vec{r} _1=(2 \hat{i}+\hat{k}), \quad \vec{r} _2=(4 \hat{i}+3 \hat{j}-\vec{k})$
$\vec{r}= \vec{r} _2- \vec{r} _1=(4 \hat{i}+3 \hat{j}-\vec{k})-(2 \hat{i}+\hat{k})$
or $\vec{r}=2 \hat{i}+3 \hat{j}-2 \hat{k}$
So work done by the given force, $w=\vec{f} \cdot \vec{r}$
$=(3 \hat{i}+\hat{j}) \cdot(2 \hat{i}+3 \hat{j}-2 \hat{k})=6+3=9 J$
