A 4.33-kg soccer ball rolling eastward at a speed of 2.74 m/s is kicked so that it reverses direction and attains a speed of 6.35 m/s. If the duration of the interaction is 55.33 ms, what is the average force on the ball by the player's foot, in N

0.71 N

Explanation:

The inertia of a body, I, is the product of the force applied on it and the time which it acts. Also, it is equal to the change in  the momentum, P, of a body.

So that:

I = Ft

I = ΔP

Ft = m(v - u)

Where F is the force, t is the time, m is the mass of object, v is the final velocity and u is the initial velocity.

Given that: m = 4.33 kg, u = -2.74 m/s, v = 6.35 m/s and t = 55.33 s

F x 55.33 = 4.33(6.35 - (-2.74))

F x 55.33 = 4.33(6.35 + 2.74)

55.33 F = 4.33 x 9.09

55.33 F = 39.3597

F =

  = 0.711363

The average force on the ball by the player is 0.71 N.

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