Question

A particle of mass m is released in a smooth hemispherical bowl from shown position A. The work done by gravity as it reaches the lowest point B is [R is radius of bowl]

• A. $2mgR$
• B. $mgR$
• C. $\frac{mgR}{\sqrt{2}}$
• D. Zero

Answer 1

The correct option is B $mgR$

We know that, gravitational force, F = mg
Displacement from A to B = s
Angle between F and $s=\theta$


In $\Delta ABC:cos\theta =\frac{AC}{AB}AC=scos\theta Also,AC=OB=R\Rightarrow R=scos\theta$

Now, work done by gravity, $W=Fscos\theta \Rightarrow W=mgR(\because scos\theta =R)$

Hence, the correct answer is option (b).