A barefoot field-goal kicker imparts a speed
of 44 m/s to a football initially at rest.
If the football has a mass of 0.49 kg and the
time of contact with the ball is 0.028 s, what
is the magnitude of the force exerted by the
ball on the kicker’s foot?
a = (V-V)/t = (44-0)/0.028 = 1571 m/s^2
F = m*a = 0.49 * 1571 = 770 N.
To find the magnitude of the force exerted by the ball on the kicker's foot, we can use Newton's second law of motion, which states that the force is equal to the rate of change of momentum.
Momentum (p) is given by the equation:
p = mass × velocity
In this case, the mass of the football (m) is 0.49 kg and the initial velocity (v) of the football is 44 m/s.
So, the initial momentum (p_initial) of the football is:
p_initial = 0.49 kg × 44 m/s
During the contact time (t) of 0.028 s, the football imparts this momentum to the kicker's foot. The change in momentum (Δp) is equal to the initial momentum (p_initial).
Using the equation for force (F) as the rate of change of momentum (F = Δp / t), we can find the magnitude of the force exerted by the ball on the kicker's foot.
F = p_initial / t
Substituting the given values, we get:
F = (0.49 kg × 44 m/s) / 0.028 s
Now, let's calculate the magnitude of the force:
F = (0.49 kg × 44 m/s) / 0.028 s = 764 Newtons
Therefore, the magnitude of the force exerted by the ball on the kicker's foot is 764 Newtons.