Question

A pendulum of mass $m\text{kg}$ has velocity $v$ at point $A$. What is the change in potential energy when the pendulum reaches point $B$ as shown in figure ?

• A. $\frac{mgl}{2}$
• B. $\frac{2mgl}{5}$
• C. $\frac{mgl}{5}$
• D. $\frac{4mgl}{5}$

Answer 1

The correct option is C $\frac{mgl}{5}$

Taking position $A$ as reference point.
Therefore potential energy at point $A$ is zero.
$\Rightarrow U_A=0$
Potential energy at point $B$ is $U_B=mgx$
and $x=l-l\cos \theta =l(1-\cos \theta )$
Therefore,
$U_B=mgl(1-\cos \theta )$

Change in potential energy between $A$ and $B$ is :
$\Delta U=U_B-U_A=mgl(1-\cos \theta )-0$
$\Delta U=mgl(1-\cos \theta )$
$\theta ={37}^{\circ }\Rightarrow \cos {37}^{\circ }=\frac{4}{5}$ $\Rightarrow \Delta U=\frac{mgl}{5}$