Question

A force $\overrightarrow{F}=(3t\hat{i}+5\hat{j})$ N acts on a particle whose position vector varies as $\overrightarrow{S}=(2t^2\hat{i}+5\hat{j})$ m, where $t$ is time in seconds. The work done by this force from $t=0$ to $t=2s$ is:

• A. 23 J
• B. 32 J
• C. zero
• D. can't be obtained

Answer 1

Answer: (B)

32 J

The work done by the force $\overrightarrow{F}=(3t\hat{i}+5\hat{j})$ is

$W=\int \overrightarrow{F}.d\overrightarrow{s}=\int (3t\hat{i}+5\hat{j}).\left(4tdt\hat{i}\right)$

$={\int }_0^{2}12t^{2}dt=\frac{12[t^3{]}_0^2}{3}=\frac{12(8-0)}{3}=32J$