A 0.620 kg basketball is dropped out of a window that is 6.40 m above the ground. The ball is caught by a person whose hands are 1.61 m above the ground. How much work is done on the ball by its weight?
To find the work done on the basketball by its weight, we need to calculate the gravitational potential energy difference between the initial position (when the ball is dropped) and the final position (when the ball is caught).
The work done on an object is defined as the change in its potential energy. In this case, the potential energy is gravitational potential energy, given by the equation:
PE = m * g * h
Where,
PE is the potential energy,
m is the mass of the object,
g is the acceleration due to gravity,
h is the height of the object.
Now, let's calculate the potential energy difference for the basketball.
Given:
Mass of the basketball, m = 0.620 kg
Height of the window, h = 6.40 m
Height of the person's hands, h' = 1.61 m
First, let's calculate the potential energy at the initial position (window height):
PE_initial = m * g * h
Substituting the values:
PE_initial = 0.620 kg * 9.8 m/s^2 * 6.40 m
Next, let's calculate the potential energy at the final position (hands height):
PE_final = m * g * h'
Substituting the values:
PE_final = 0.620 kg * 9.8 m/s^2 * 1.61 m
Now, we can calculate the work done as the difference between the initial and final potential energies:
Work done = PE_final - PE_initial
Substituting the values:
Work done = (0.620 kg * 9.8 m/s^2 * 1.61 m) - (0.620 kg * 9.8 m/s^2 * 6.4 m)
Simplifying the expression gives us the work done on the basketball by its weight.