A boy pushes a box of mass


A boy pushes a box of mass $2 \mathrm{~kg}$ with a force

$\overrightarrow{\mathrm{F}}=(20 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}) \mathrm{N}$ on a frictionless surface. If the

box was initially at rest, then $\mathrm{m}$ is displacement along the x-axis after $10 \mathrm{~s}$.



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

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

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

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

$\therefore$ Displacement along $x$-axis

$\Rightarrow 50 \times 10 \Rightarrow 500 \mathrm{~m}$

$\therefore$ Ans. 500

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