Question

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

Answer 1

let retardation caused by the player be a m/s^2
v=u+at. (where v is final velocity=0, u is initial velocity=0.5m/s and t is time=0.5s)

0=0.5-a×0.5 (here a is -ve for retardation)
or, 0.5×a=0.5
or, a=0.5/0.5=1 m/s^2

putting the formula, F=ma,
F=70g×1m/s^2
=0.07kg×1m/s^2
=0.07N (ans)

Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\pink{Answer:-}$

Here, m= 70 g= 0.070 kg, u= 0.5 m/s, v= 0, t= 0.5 s

Force, F= m [v-u/t] = 0.070 [0-0.5/0.5] = -0.07 N.

[Negative sign shows the retarding nature of the force]

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