Question

1 ball of mass 50 gram moving with a velocity of 0.75 metre per second is stopped by a player in 0.75 second what is the force applied by the player to stop the board ball​

Answer 1

Answer:

-0.05N

Explanation:

given the mass of ball is =50 gram. 50/1000= 0.05kg

the velocity is =0.75 m/s^2

time taken by player to stop the ball is =0.75 second

w we know that F = m×v-u/t

$f = m \times \frac{v - u}{t}$

so F = 0.05×0-0.75 / 0.75

therefore, F = 0.05×-1

so,F= -0.05 N

so , the force applied by the player to stop the ball is - 0.05 Newton