Question

A smooth track in the form of a quarter circle of radius $6\text{m}$ lies in the vertical plane. A particle moves from ${\text{P}}_1$ to ${\text{P}}_2$ under the action of forces $F_1,F_2$ and $F_3.$ Force $F_1$ is always towards ${\text{P}}_2$ and is always $20\text{N}$ in magnitude. Force $F_2$ always acts horizontally and is always $30\text{N}$ in magnitude. Force $F_3$ always tangential to the track and is of magnitude $15\text{N}.$
Select the correct alternative(s).

• A. Work done by $F_1$ is $120\text{J}$.
• B. Work done by $F_2$ is $180\text{J}$.
• C. Work done by $F_3$ is $45\pi \text{J}$.
• D. $F_1$ is conservative in nature.

Answer 1

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

$F_1$ is conservative in nature.

Workdone by the variable force $F$ is,

$W={\int }_{P_1}^{P_2}F\cos \theta ds.......\left(1\right)$

Workdone by the constant force $F$ is,

$W=\overrightarrow{F}\cdot \overrightarrow{s}=Fs\cos \theta .......\left(2\right)$ [$s$=Displacement in the direction of force ]

$W=Fs[\text{if}\theta =0^{\circ }]$

Since, here all three forces are constant, hence we can find work done by all three forces using eqn $\left(2\right)$.

Work done by force $F_1$ is

$W_1=F_{1}s=20\times \sqrt{2}R=120\sqrt{2}\text{J}[\because R=6\text{m}]$

And $F_1$ is a conservative force because it is always directed toward fixed point ${\text{P}}_2.$

Work done by force $F_2$ is

$W_2=F_{2}s=30\times R=180\text{J}$

Workdone by force $F_3$ is

Here, $F_3$ is acts along the tangential, then displacement along the tangential is,

$s=\frac{2\pi R}{4}=3\pi [\because R=6\text{m}]$

$\Rightarrow W_3=F_{3}s=15\times 3\pi =45\pi \text{J}$

Hence, options (b), (c) and (4) are the correct answer.