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. $1.96mJ$
• B. $3.92mJ$
• C. $4.90mJ$
• D. $5.88mJ$

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 ${\upsilon }^2=2gh=2\times 9.8\times 0.20$
$\left[\because h=radius=20cm=0.2m\right]$
According to work energy theorem work done on the ball= change in kinetic energy
$=\frac{1}{2}\times \frac{2}{1000}\times 9.8\times 0.2$
$=3.92\times 10^{-3}J=3.92mJ$