Question

A soccer ball with mass 0.420 kg is initially moving with speed 2.00 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 40.0 N in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 m/s?​

Answer 1

Using the work energy theorem:

m = 0.42 kg

vi = 2 m/s

vf = 6 m/s

F = 40 N

work = F.d = delta_KE

delta_KE = (1 / 2) * m * vf ^ 2 - (1 / 2) * m * vi ^ 2

delta_KE = 6.72 J

d = delta_KE / F

d=6.72/40

d = 0.168 m

Answer 2

Given : mass m = 0.42 kg

$\[{v_i}\]$ = 2 m/s

$\[{v_f}\]$ = 6 m/s

F = 40 N

To Find : Distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 m/s.

Solution :

Using the work energy theorem:

The work W done by the net force on a particle equals the change in the particle's kinetic energy KE

W=ΔKE

As,

mass m = 0.42 kg

$\[{v_i}\]$ = 2 m/s

$\[{v_f}\]$ = 6 m/s

F = 40 N

From Work- Energy Theorem,

W=ΔKE

$\[\Delta KE = \frac{1}{2}m{v_f}^2 - \frac{1}{2}m{v_i}^2\]$

Putting value in equation,

$\[\Delta KE = \frac{1}{2} \times 0.42 \times {6^2} - \frac{1}{2} \times 0.42 \times {2^2}\]$

ΔKE = 6.72 J

As,

W = ΔKE = $\[F \times d\]$                          

where,

d = Player foot be in contact with the ball to increase the ball's speed

$\[d = \frac{W}{F}\]$

$\[d = \frac{{6.72}}{{40}}\]$

 d = 0.168 m.

Hence, player's foot be in contact with the ball to increase the ball's speed to 6.00 m/s is 0.168 m.