Question #285116

A basketball of mass 0.0400 kg is dropped from a height of 5.00 meters to the ground and bounces back to a height of 3.00 meters .


A. On its way down, how much potential energy does the ball lose?


B. On its ways, how much potential energy does the ball regain?

Expert's answer

A. The potential energy is defined as follows:


E1=mgh1E_1 = mgh_1

where m=0.0400kgm =0.0400kg is the mass of the ball, g=9.81m/s2g = 9.81m/s^2 is the gravitational acceleration, h1=5.00mh_1 = 5.00m is the initial height. Thus, obtain:


E1=0.049.8151.96JE_1 = 0.04\cdot 9.81\cdot 5 \approx 1.96J

B. Similarly, on the way up:


E2=mgh2=0.049.8131.18JE_2 = mgh_2 = 0.04\cdot 9.81\cdot 3 \approx 1.18J

Answer. A. 1.96J, B. 1.18J.


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