Question

a cricket ball of mass 70 g moving with a velocity of 0.5 m/s is topped by a player in 0.5 s. what is the force applied by the player to stop the ball?

Answer 1

Answer:

0.07 N

Explanation:

Given:

mass of  ball =m=70g=0.070kg

Initial velocity=u= 0.5 m/s

Final velocity =v=0m/s

time=t= 0.5 sec

a= v-u/t

⇒0-0.5/0.5

⇒-1 m/s²

F=m*a

⇒0.07x-1

⇒-0.07 N