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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?
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.
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