Question

A smooth track in the form of a quarter circle of radius $6m$ lies in a vertical plane. A particle moves from $P_1$ to $P_2$ under the forces ${\overrightarrow{F}}_1,{\overrightarrow{F}}_2$ and ${\overrightarrow{F}}_3\cdot$ Force ${\overrightarrow{F}}_1$ is always always towards $P_2$ and is always $20N$ in magnitude force $F_2$ always acts tangentiallyand is always $15N$ in magnitude. Force ${\overrightarrow{F}}_3$ always acts horizontally is of magnitude $30N$. Select the correct alternative(s):

• A. Work done by ${\overrightarrow{F}}_1$ is $120J$
• B. Work done by ${\overrightarrow{F}}_2$ is $45\pi$
• C. Work done by ${\overrightarrow{F}}_3$ is $180J$
• D. ${\overrightarrow{F}}_1$ is conservative in nature

Answer 1

Answer: (B), (C), (D)

The correct options are
B Work done by ${\overrightarrow{F}}_2$ is $45\pi$
C Work done by ${\overrightarrow{F}}_3$ is $180J$
D ${\overrightarrow{F}}_1$ is conservative in nature

Work done $=\int \overrightarrow{F}.\overrightarrow{dr}$

Work done by $F_1=\int \overrightarrow{F_1.}\overrightarrow{dr}$

$=F_1\left(P_1{P}_2\right)$

$=20\times 6\sqrt{2}J$

And is conservative in nature as the distance is decided by the end points.

Work done will only depend on endpoints.

Work done by $F_2=\int F.dr$

$=15\times \frac{\pi R}{4}=15\times \frac{\pi }{4}\times 6=45\pi J$

But $F_2$ is non conservative as work done will depend on the arc of motion.

Work done by $F_3=30$ (Horizontal distance)

$=30\times 6=180J$

$F_3$ work done also depend on end points.

$F_3$ is conservative.