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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AcademicPhysicsNCERTClass 9

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 $0.05\ N$.

raja
Updated on 10-Oct-2022 13:28:50

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