The velocity-time graph of a ball of mass $20 \ g$ moving along a straight line on a level ground is given below. How much force does the ground exert on the ball to bring it to rest?

(N/A) Given:
Mass of the ball,$m = 20 \ g = 0.02 \ kg$.
From the velocity-time graph:
Initial velocity,$u = 20 \ m \ s^{-1}$ (at $t = 0 \ s$).
Final velocity,$v = 0 \ m \ s^{-1}$ (at $t = 10 \ s$).
Time taken,$t = 10 \ s$.
Using Newton's second law of motion,the force $F$ exerted by the ground on the ball is given by:
$F = m \times a = m \times \frac{(v - u)}{t}$
Substituting the values:
$F = 0.02 \times \frac{(0 - 20)}{10}$
$F = 0.02 \times (-2) = -0.04 \ N$.
The negative sign indicates that the force exerted by the ground (frictional force) is in the direction opposite to the motion of the ball. Thus,the magnitude of the force is $0.04 \ N$.