A 0.582-kg basketball is dropped out of a window that is 5.98 m above the ground. The ball is caught by a person whose hands are 1.58 m above the ground. (a) How much work is done on the ball by its weight? What is the gravitational potential energy of the basketball, relative to the ground, when it is (b) released and (c) caught? (d) What is the change (PEf - PE0) in the ball's gravitational potential energy?
To answer these questions, we need to understand the concepts of work done, gravitational potential energy, and the equation for calculating them.
(a) How much work is done on the ball by its weight?
Work done is defined as the force applied over a certain distance. In this case, the force applied on the basketball is its weight, which can be calculated using the equation:
Weight = mass * gravity
where the mass of the basketball is given as 0.582 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.
So, Weight = 0.582 kg * 9.8 m/s^2 = 5.7 N
Now, let's calculate the work done on the ball. The work done by a constant force can be calculated using the equation:
Work = force * distance * cos(theta)
In this case, theta is the angle between the direction of the force applied and the direction of motion. Since the ball is dropped vertically, theta is 0 degrees.
The distance is the vertical distance the ball travels, which is the height of the window (5.98 m) minus the height of the person's hands (1.58 m). So, the distance is 4.4 m.
Work = 5.7 N * 4.4 m * cos(0°) = 25.08 J
Therefore, the work done on the ball by its weight is 25.08 Joules.
(b) What is the gravitational potential energy of the basketball, relative to the ground, when it is released?
Gravitational potential energy is given by the equation:
Potential Energy = mass * gravity * height
where height is the vertical distance from the reference point (in this case, the ground) to the object.
When the basketball is released, the height is 5.98 m, and the mass and gravity are the same as before.
Potential Energy = 0.582 kg * 9.8 m/s^2 * 5.98 m = 34.43 J
Therefore, the gravitational potential energy of the basketball, relative to the ground, when it is released is 34.43 Joules.
(c) What is the gravitational potential energy of the basketball, relative to the ground, when it is caught?
When the basketball is caught, its height from the reference point (the ground) is 1.58 m.
Potential Energy = 0.582 kg * 9.8 m/s^2 * 1.58 m = 9.07 J
Therefore, the gravitational potential energy of the basketball, relative to the ground, when it is caught is 9.07 Joules.
(d) What is the change (PEf - PE0) in the ball's gravitational potential energy?
The change in gravitational potential energy can be calculated by subtracting the initial potential energy from the final potential energy.
Change in Potential Energy = PEf - PE0
= (9.07 J) - (34.43 J)
= -25.36 J
Therefore, the change in the ball's gravitational potential energy is -25.36 Joules.