Question

A cricket ball of mass 150g moving at a speed of 25 ms^-1 is bought to rest by a player in 0.03s. Find the average force applied by the player.

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Answer 1

Given :-

A cricket ball of mass 150g moving at a speed of 25 ms^-1 is bought to rest by a player in 0.03s

To Find :-

Average force

Solution :-

We know that

v = u + at

0 = 25 + a(0.03)

0 - 25 = 0.03a

-25 = 0.03a

-25/0.03 = a

-2500/3 = a

Now

1 kg = 1000 g

150 g = 150/1000 = 0.15 kg

F = ma

F = 0.15 × -2500/3

F = 15/100 × (-2500)/3

F = -125 N

Answer 2

Given :-

• Mass of cricket ball , m = 150 g = 0.15 kg  

• initial velocity of ball, u = 25 ms⁻¹

• final velocity of ball, v = 0 (rest)

• time taken for velocity change, t = 0.03 s

To find :-

• Average force applied by player to bring the ball to rest.

Formula used :-

Formula to calculate acceleration

$\boxed{\red{\bigstar}\sf{acceleration=\dfrac{final \:velocity-initial\:velocity}{time\:taken}}}$

Formula to calculate Force

$\boxed{\red{\bigstar}\sf{Force =mass\times acceleration}}$

Solution :-

Calculating acceleration of ball while coming to rest

$\longmapsto\sf{acceleration,a=\dfrac{v-u}{t}}\\\\\\\longmapsto\sf{a=\dfrac{0-25}{0.03}}\\\\\\\longmapsto \overbrace{\underbrace{\sf{a=\dfrac{-2500}{3}\;\;ms^{-2}}}}$

Calculating Force applied by Player on the ball to stop it

$\longmapsto\sf{Force,F=m\times a}\\\\\\\longmapsto\sf{F=0.15\times \dfrac{-2500}{3}}\\\\\\\longmapsto\red{\overbrace{\underbrace{\underline{\underline{\large{\sf{F=-125\:N}}}}}}}$

Hence,

125 Newton of force is applied by Player to bring the ball to rest . and the negative sign with the magnitude shows that Force is applied in the direction opposite to motion of ball.