Question

A body of mass 4 kg moves under the action of a force $\overrightarrow{F}=(4\hat{i}+12t^2\hat{j})N$, 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 1 s, work done is :

• A. 4 J
• B. 8 J
• C. 12 J
• D. 16 J

Answer 1

The correct option is D 16 J

$\int \overrightarrow{F}dt=\Delta \overrightarrow{p}$

${\int }_0^1(4\hat{i}+12\hat{t})dt=m(\overrightarrow{V_f}-\overrightarrow{V_i})$

$⟹4\hat{i}+\frac{12}{3}{t}^3{|}_0^1\hat{j}=4(\overrightarrow{V_f}-\overrightarrow{V_i})$

$\hat{i}+\hat{j}=\overrightarrow{V_f}-(2\hat{i}+\hat{j}+2\hat{k})$

$\overrightarrow{V_f}=3\hat{i}+2\hat{j}+2\hat{k}$

$|\widehat{V_f}|=\sqrt{9+4+4}=\sqrt{17}$

$|\widehat{V_i}|=\sqrt{4+1+4}=3$

$\Delta K.E=\frac{1}{2}4(17-9)$

$=2\times 8=16J$