Q81 / 90 · JEE Main 2021 (26 Feb Shift 1) · Physics

A boy pushes a box of mass $2 kg$ with a force $\vec{F} = (20 \hat{i} + 10 \hat{j}) N$ on a frictionless surface. If the box was initially at rest, then _______ $m$ is displacement along the $x$ -axis after $10 s$

Correct answer: 500

Solution

$\vec{F} = 20 \hat{i} + 10 \hat{j}$

$\vec{a} = \frac{\vec{F}}{m} = \frac{20 \hat{i} + 10 \hat{j}}{2} \Rightarrow 10 \hat{i} + 5 \hat{j}$

$∴ \vec{s} = \frac{1}{2} \vec{a} t^{2} = \frac{1}{2} (10 \hat{i} + 5 \hat{j}) \times {(10)}^{2}$

$\Rightarrow 50 (10 \hat{i} + 5 \hat{j}) m$

$∴$ Displacement along $x$ -axis

$\Rightarrow 50 \times 10 \Rightarrow 500 m$

Hence, the correct answer is 500.