The initial velocity of the ball is 20 cm `s^(-1)`. Due to the frictional force exerted by the table, the velocity of the ball decreases down to zero in 10 s. Thus, u = 20 cm `s^(–1)`, v = 0 cm `s^(-1)` and t = 10 s. Since the velocity-time graph is a straight line, it is clear that the ball moves with a constant acceleration. The acceleration a is
`a=(v-u)/(t)`
=0 cm `s^(-1)` -20 cm `s^(-1)//10 s`
=-2 cm `s^(-2)`=-0.02 m`s^(-2)`
The force exerted on the ball F is , `F=ma=(20//1000) kg xx (-0.02ms^(-2))`
`= -0.0004 N`. The negative sign implies that the fricitonal force exerted by the table is opposite to the direction of motion of the ball.
