Question

A body of mass $4kg$ moves under the action of a force $\overrightarrow{F}=(\widehat{4i}+12t^2\hat{j})N$, where $t$ is the time in second. The initial velocity of the particle is $(2\hat{i}+\hat{j}+2\hat{k})m{s}^{-1}$. If the force is applied for $1s$, work done is:

• A. $4J$
• B. $8J$
• C. $12J$
• D. $16J$

Answer 1

The correct option is A $4J$

Given Mass $m=4kg$

Force $\overrightarrow{F}=(4\hat{i}+12t^2\hat{j})N$

Initial velocity $\overrightarrow{v}=(2\hat{i}+\hat{j}+2\hat{k})m/s$

Magnitude of $\overrightarrow{v}=\sqrt{2^2+I^2+I^2}=\sqrt{9}=3m/s$

Net force $=\left|(4i+\left|2t^2\hat{j}\right|)N\right|\Rightarrow ma=\sqrt{{16}^2+144t^4}\Rightarrow 4a=4\sqrt{1+9t^4}\Rightarrow a=\sqrt{1+9t^4}\Rightarrow \frac{dv}{dt}=\sqrt{1+9t^4}\Rightarrow v=\sqrt{1^2+{\left({3t}^2\right)}^2}at(t=1)=4$

Work $=$ Force $\times$ displacement

$=12.6\times \frac{1}{3}=4J$