A 0.579 kg basketball is dropped out of a window that is 6.22 m above the ground. The ball is caught by a person whose hands are 1.31 m above the ground. How much work is done on the ball by its weight?
To calculate the work done on the ball by its weight, we need to determine the change in potential energy of the ball as it falls from the window to the person's hands.
The potential energy of an object near the surface of the Earth is given by the formula: PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the object above a reference level.
In this case, the initial potential energy of the ball when it is at the window is PE_initial = m * g * h_initial, where h_initial is the height of the window above the ground.
The final potential energy of the ball when it is caught by the person is PE_final = m * g * h_final, where h_final is the height of the person's hands above the ground.
The change in potential energy is then given by: ΔPE = PE_final - PE_initial.
To calculate the work done on the ball by its weight, we use the equation: Work = ΔPE.
Let's substitute the given values into the equations:
m = 0.579 kg (mass of the ball)
g = 9.8 m/s^2 (acceleration due to gravity)
h_initial = 6.22 m (height of the window above the ground)
h_final = 1.31 m (height of the person's hands above the ground)
PE_initial = m * g * h_initial = 0.579 kg * 9.8 m/s^2 * 6.22 m
PE_final = m * g * h_final = 0.579 kg * 9.8 m/s^2 * 1.31 m
Now we can calculate the change in potential energy and the work done on the ball:
ΔPE = PE_final - PE_initial
Work = ΔPE
Work = (0.579 kg * 9.8 m/s^2 * 1.31 m) - (0.579 kg * 9.8 m/s^2 * 6.22 m)
Work = 6.029 J
Therefore, the work done on the ball by its weight is 6.029 Joules.