Question

An object of mass m is tied to a string of length l and a variable force Fis applied on it which brings the string gradually to an angle with the vertical. The work done by the force F is

• A. mgl sin θ
• B. mgl (1 – sin mgl sin θ)
• C. mgl cosθ
• D. mgl (1 – cos θ)

Answer 1

The correct option is D

mgl (1 – cos θ)

$\text{The forces acting on the bob are : mg, F and T}{\text{W}}_T=0[\therefore Tisalwaysperpendiculartods]W_{mg}=-mgh=-mgl(1-cos\theta )K{E}_i=0andK{E}_f=0,\text{Since the bob is moved slowly.}Accordingtoworkenergytheorem{W}_{Total}=\Delta KE\therefore W_r+W_{mg}+W_F=K{E}_f-K{E}_i=00-mgl(1-cos\theta )+W_F=0\Rightarrow W_F=mgl(1-cos\theta )$