A soccer player kicks a soccer ball of mass 0.50 kg that is initially at rest. T

Question

A soccer player kicks a soccer ball of mass 0.50 kg that is initially at rest. The player's foot is in contact with the ball for 1.8 10-3 s. The force of the kick is given by the following formula for 0 t 1.8 10-3 s, where t is in seconds.

F(t) = [(7.1 106)t - (2.8 109)t2] N

(a) Find the magnitude of the impulse imparted to the ball. Ns

(b) Find the magnitude of the average force exerted by the player's foot on the ball during the period of contact. N

(c) Find the magnitude of the maximum force exerted by the player's foot on the ball during the period of contact. N

(d) Find the magnitude of the ball's velocity immediately after it loses contact with the player's foot. m/s

Explanation / Answer

a)

Impulse = Ft = (7.1E6t – 2.8E9t2)t = [3.55E6t² - 0.93E9t³]

Using t = 1.8E-3 gives Impulse = 6.08 N.sec

b)

Fave = Impulse/t = 6.08/1.8E-3 = 3376.8 N

c)

Differentiate the Force function and set equal to zero: 7.1E6 – 5.6E9*t = 0 t = 1.268E-3 when the force is max. Plug this value back into the force equation to get Fmax = 4500.89 N

d)

Since Impulse = m*V, V = Impulse/m = 6.08/0.50 = 12.16 m/sec

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