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

Academic Physics NCERT Class 9

Here, mass of the ball $m=70\ g=\frac{70}{1000}\ kg.=\frac{7}{100}\ kg.$

The initial velocity of the ball $u=0.5\ m/s$

The final velocity of the ball $v=0$

Time taken $t=0.5\ s$

Therefore, acceleration $a=\frac{v-u}{t}$

$=\frac{0-0.5}{0.5}=-1\ ms^{-2}$

Therefore, force applied by the player to stop the ball $F=ma$

$=\frac{7}{100}\ kg.\times -1\ ms^{-2}$

$=-0.07\ N$

Therefore, the $0.07\ N$ force is applied by the player to stop the ball in the opposite direction which is indicated by $-ve$ sign.

Simply Easy Learning

Updated on: 10-Oct-2022

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