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A rubber ball is dropped and bounces vertically from a horizontal concrete floor. If the ball has a speed of 3 m/s just before striking the floor, and a speed of 6.5 m/s just after bouncing, find the average force of the floor on the ball. Assume that the ball is in contact with the floor for 0.32 s, and that the mass of the ball is 0.42 kg

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davidlcastiblanco

Answer:

F=12.5N

Explanation:

Net force = rate of change of momentum

[tex]F = m*a[/tex]

so find the change of momentum P

Pdown

[tex]P=m*v_1=0.42kg*3m/s[/tex]

Pup

[tex]P=m*v_1=0.42kg*6.5m/s[/tex]

dP = change in P

[tex]dP= 0.42kg (3- -.6.5)m/s =3.99 kg m/s[/tex]

[tex]dT = 0.32 s[/tex]

so

[tex]F = \frac{dP}{dt}=\frac{3.99Kg*m/s}{0.32s} =12.468 N[/tex]

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