Question

A 2 g ball of glass is released from the edge of a hemispherical cup whose radius is 20 cm. How much work is done on the ball by the gravitational force during the ball’s motion to the bottom of the cup?

• A. $1.96\,mJ$
• B. $3.92\,mJ$
• C. $4.90\,mJ$
• D. $5.88\,mJ$

Answer 1

Answer: (B)

According to the conservation of energy, kinetic energy at A + potential energy at A = kinetic energy at B + potential energy at B
$0+mgh=\frac{1}{2}m\upsilon^{2}+0$
or $\upsilon^{2}=2gh=2\times9.8\times0.20$
$\left[\because h=radius = 20\, cm = 0.2 \,m\right]$
According to work energy theorem work done on the ball= change in kinetic energy
$=\frac{1}{2}\times\frac{2}{1000}\times9.8\times0.2$
$=3.92\times10^{-3}\,J=3.92\,mJ$