# A body of mass 2kg moves under a force

Question:

A body of mass $2 \mathrm{~kg}$ moves under a force of $(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \mathrm{N}$. It starts from rest and was at the origin initially. After $4 \mathrm{~s}$, its new coordinates are $(8, b, 20)$. The value of $b$ is_________

(Round off to the Nearest Integer)

Solution:

Ans. (12)

$\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{m}}=\frac{2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{2}$

$=\hat{\mathrm{i}}+1.5 \hat{\mathrm{j}}+2.5 \hat{\mathrm{k}}$

$\vec{\tau}=\vec{u} t+\frac{1}{2} \vec{a} t^{2}$

$=0+\frac{1}{2}(\hat{\mathrm{i}}+1.5 \hat{\mathrm{j}}+2.5 \hat{\mathrm{k}})(16)$

$=8 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}+20 \hat{\mathrm{k}}$

$\mathrm{b}=12$