# A boy pushes a box of mass 2 kg

Question:

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 $10 \mathrm{~s}$.

Solution:

(500)

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

$\mathrm{F}_{\mathrm{x}}=20 \mathrm{~N}$

$F_{y}=10 N$

$a_{x}=\frac{F_{x}}{M}=\frac{20}{2}=10 \mathrm{~m} / \mathrm{s}^{2}$

$a_{y}=\frac{F_{y}}{M}=\frac{10}{2}=5 \mathrm{~m} / \mathrm{s}^{2}$

displacement on $\mathrm{x}$ axis is

$\mathrm{S}_{\mathrm{x}}=\mathrm{u}_{\mathrm{x}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2}$

$S=0 \times 10+\frac{1}{2} \times 10 \times(10)^{2}$

$S=500 \mathrm{~m}$