Question

a cricket ball of mass 70 g moving with a velocity of 0.5 m/s^2is stopped by a player in 0.5 sec . what is the force applied by the player to stop the ball ?

Answer 1

mass=70g=70/1000=0.07kg
u=0.5m/s
v=0
a=?
a=v-u/t=0-0.5/0.5=-1
F=ma
0.07×-1
-0.07N Answer

Answer 2

Given ➪

• Mass of the cricket ball(m) = 70g  (0.07kg)
• Initial velocity(u) = 0.5m/s
• Final velocity(v) = 0

To Find ➪

Force exerted by the player (F)

Answer ➪

We know →

⇒ a = v-u/t

⇒ a = 0-0.5/0.5

⇒ a = -0.5/0.5

⇒ a = -1m/s²

by Newton's second law of motion →

⇒ F = ma

⇒ F = 0.07×(-1)

⇒ F = -0.07N

∴ The force applied by the player is -0.07N