Question

A force $\overrightarrow{F}=(2\hat{1}+5\hat{j}+k)$ is acting on a particle. The particle is first displaced from $(0,0,0)$ to $(2m,2m,0)$ along the path $x=y$ and then from $(2m,2m,0)$ to $(2m,2m,2m)$ along the path $x=2,,y=2m.$ The toal work done in the complete path is:

• A. $12J$
• B. $8J$
• C. $16J$
• D. $10J$

Answer 1

The correct option is C $16J$

$\overrightarrow{F}=2\hat{i}+5\hat{j}+\hat{z}$

position is ${\overrightarrow{r}}_1=0\hat{i}+0\hat{j}+0\hat{z}$

position is ${\overrightarrow{r}}_2=2\hat{i}+2\hat{j}+2\hat{z}$

displacement $\overrightarrow{d}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1=2\hat{i}+2\hat{j}+2\hat{z}$

work $=\overrightarrow{F}.\overrightarrow{d}=(2\hat{i}+5\hat{j}+\hat{z}).(2\hat{i}+2\hat{j}+2\hat{z})$

$w=2\times 2+5\times 2+1\times 2=9+10+2=16J$

No need of intermediate paths