Question

If a tennis ball of mass 50 g is moving with a velocity of 0.2 m s^–1 is stopped by player in 0.8 s, what is the force applied by the player to stop the ball ?

• A. 0.00135 N
• B. 0.125 N
• C. 0.0155 N
• D. - 0.0125 N

Answer 1

The correct option is D

- 0.0125 N

here m = 50 g = 0.050 kg

u = 0.2 m/s

v = 0

t = 0.8 s

F = $\frac{m\times (v-u)}{t}$

F = $\frac{0.05\times (0-0.2)}{0.8}$

F = - 0.0125 N