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Physics
A force vecF=(5 hati+3 hatj) newton is applied over a particle which displaces it from its origin to the point vecr=2 hati- hatj metres. The work done on the particle is
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Q. A force $\vec{F}=(5 \hat{i}+3 \hat{j})$ newton is applied over a particle which displaces it from its origin to the point $\vec{r}=2 \hat{i}-\hat{j}$ metres. The work done on the particle is
A
- 7 joules
6%
B
+ 13 joules
0%
C
+ 7 joules
89%
D
+ 11 joules
6%
Solution:
$W=\vec{F} \vec{s}=(5 \hat{i}+3 \hat{j}) \cdot(2 \hat{i}-\hat{j})$
$=10-3$
$=7 \,J$