Mechanical energy of a basketball

[texasgrl05]

Help! I don't understand how to do this problem:

A basketball of mass 0.50 kg is dropped from rest from a height of 1.22 m. It rebounds to a height of 0.40 m.

(a) How much mechanical energy was lost during the collision with the floor?

(b) A basketball player dribbles the ball from a height of 1.22 m by exerting a constant downward force on it for a distance of 0.13 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.22 m, what is the magnitude of the force?

 

[WhirlwindMonk]

1. Mechanical energy is potential energy + kinetic energy. You need to calculate the mechanical energy before and after the bounce.

2. Work = Change in KE (which = change in ME in this case)
W = Fd

 

[thepatientmental]

(a) To solve this problem, we can use the formula for potential energy, which is PE = mgh, where m is the mass of the object (in this case, the basketball), g is the acceleration due to gravity (9.8 m/s^2), and h is the height. We can calculate the initial potential energy of the basketball by plugging in the given values: PE = (0.50 kg)(9.8 m/s^2)(1.22 m) = 6.05 J. The final potential energy can be calculated the same way, using the height of 0.40 m: PE = (0.50 kg)(9.8 m/s^2)(0.40 m) = 1.96 J. The difference between these two values is the mechanical energy lost during the collision: 6.05 J - 1.96 J = 4.09 J. Therefore, 4.09 J of mechanical energy was lost during the collision with the floor.

(b) To find the magnitude of the force exerted by the player, we can use the formula for work, which is W = Fd, where F is the force and d is the distance. The work done by the player must be equal to the mechanical energy lost during the bounce, which we calculated in part (a) to be 4.09 J. Therefore, we can set up the equation: 4.09 J = F(0.13 m). Solving for F, we get F = 31.46 N. Therefore, the magnitude of the force exerted by the player is 31.46 N.