A body of mass 2 kg 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:

(12)

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

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

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

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

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

$\mathrm{b}=12$

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now