Question

A SOCCER PLAYER GIVES A 2KG BALL A HORIZONTAL VELOCITY OF 20M/S. THE BALL SLIDES ALONG THE TURF AGAINST A FRICTIONAL FORCE OF 4N,WHAT IS THE VELOCITY OF THE BALL AFTER 25 M?

Answer 1

According to the work-energy theorem, the work done by the frictional force is equal to the negative ball's kinetic energy change. By definition, the work is:

$$W = Fs$$

where $F = 4N$ is the frictional force, and $s = 25m$ is the length of the ball's path.

The initial kinetic energy is:

$$K_i = \dfrac{mv_i^2}{2}$$

where $m = 2kg$ is the mass of the ball, and $v_i = 20m/s$ is its initial speed. Thus, obtain:

$$K_i - K_f = W\\ K_f = K_i -W = \dfrac{mv_i^2}{2}-Fs$$

On the other hand, the final kinetic energy $K_f$ is:

$$K_f = \dfrac{mv_f^2}{2}$$

where $v_f$ is the final speed of the ball. Thus, obtain:

$$\dfrac{mv_f^2}{2}=\dfrac{mv_i^2}{2}-Fs\\ v_f^2 = v_i^2-\dfrac{2Fs}{m}\\ v_f = \sqrt{ v_i^2-\dfrac{2Fs}{m}}\\ v_f = \sqrt{ 20^2-\dfrac{2\cdot 4\cdot 25}{2}} \approx 17.3 m/s$$

Answer. 17.3 m/s.