Velocity versus time graph of a ball of mass 50 g rolling on a concrete floor is shown in Fig. Calculate the acceleration and frictional force of the floor on the ball.
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From the given velocity-time graph,
Initial velocity $v_1=80\ ms^{-1}$
Final velocity $v_2=0$
Time $t_1=0$
Time $t_2=8\ s$
Acceleration of the ball $a=\frac{change\ in\ velocity(v_2-v_1)}{time(t_2-t_1}$
$=\frac{0-80}{8-0}$
$=\frac{-80}{8}$
$=-10\ m/s^2$
As given, mass of the ball $m=50\ gm=0.05\ kg$ [because $1\ kg=1000\ g$]
Therefore, frictional force offered by the floor on the ball $F=ma$
$=0.05\ kg\times -10\ m/s^2$
$=-0.5\ N$ [$-ve$ sign indicates opposite direction]
Therefore, friction force of the floor is $0.05\ N$.
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