Answer to Question #171881 in Atomic and Nuclear Physics for Nema

Question #171881

What was the force being applied by the pitcher's hand if the ball is in contact with it for 0.2s?


1
Expert's answer
2021-03-18T20:11:55-0400

Let's assume that the mass of the ball is m=150g=0.15kgm = 150g = 0.15kg, and the final speed is vf=40m/sv_f = 40m/s.

Then the accleration of the ball during the contact is:


a=vfvita = \dfrac{v_f - v_i}{t}

where vi=0m/sv_i = 0m/s is the initial speed of the ball, and t=0.2st = 0.2s is the time of the contact. Thus, obtain:


a=40m/s0.2s=200m/s2a = \dfrac{40m/s}{0.2s} = 200 m/s^2

According to the second Newton's law, the force applied by the pitcher's hand is:


F=maF=0.15kg200m/s2=30NF = ma\\ F = 0.15kg\cdot 200 m/s^2 = 30N

Answer. 30N.


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