A cricket ball of mass 90 g moving with a velocity of 0.5 m/s is stopped by a player in 0.45 s. what is the force applied by the player to stop the ball?
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Answer:
The answer will be 0.099 N
Explanation:
According to the problem the mass of the ball , m = 90 g = 0.09 kg
The initial velocity of the ball ,u = 0.5 m/s and the player has stopped the ball is 0.45 s
Therefore the acceleration of the ball , a = v-u/t =- 0.5/0.45 = - 1.11 m/s^2
Let the force applied is f
Therefore f = ma = 0.09 x (-1.1) = - 0.099 N
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