Question

An object of mass $m$ is tied to a string of length $l$ and a variable force $F$ is applied on it which brings the string gradually to angle $\theta$ with the vertical. Find the work done by the force $F$. ( Assume initial and final kinetic energy to be zero)

• A. $mgl(1-\sin \theta )$
• B. $mgl\left(\sin \theta \right)$
• C. $mgl\left(\cos \theta \right)$
• D. $mgl(1-\cos \theta )$

Answer 1

The correct option is D $mgl(1-\cos \theta )$

In this case three forces are acting on the object:
1. Tension $\left(T\right)$
2. Weight $\left(mg\right)$ and
3. Applied force $\left(F\right)$

Using work-energy theorem
$W_{\text{net}}=\Delta KE$
or $W_T+W_{mg}+W_F=\mathrm{0....}\left(i\right)$
as $\Delta KE=0$
because $K_i=K_f=0$
Further, $W_T=0$, as tension is always perpendicular to displacement.
$W_{mg}=-mgh$ or $W_{mg}=-mgl(1-\cos \theta )$

As by the figure the height reached by the mass$\left(m\right)$ is $h=l(1-\cos \theta )$, where $l$ is the length of the string

Substituting these values in Eq. $\left(i\right)$, we get

$W_F=mgl(1-\cos \theta )$

Hence option $\left(d\right)$ is the correct answer