Question

A 0.537-kg basketball is dropped out of a window that is 5.88 m above the ground. The ball is caught by a person whose hands are 1.98m 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 (
`PE_f - PE_0`) in the ball's gravitational potential energy?

Answer 1

a) 20.54 J

b) 30.97 J

c) 10.43 J

d) -20.54 J

Step-by-step explanation:

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

h = Height

`W=mgh\\\Rightarrow W=0.537* 9.81* (5.88-1.98)\\\Rightarrow W=0.537* 9.81* 3.9\\\Rightarrow W=20.54\ J`

Work done by the ball's weight is 20.54 J

`W=mgh\\\Rightarrow W=0.537* 9.81* (5.88)\\\Rightarrow W=30.97\ J`

Gravitational potential energy of the basketball, relative to the ground, when it is released is 30.97 J

`W=mgh\\\Rightarrow W=0.537* 9.81* (1.98)\\\Rightarrow W=10.43\ J`

Gravitational potential energy of the basketball, relative to the ground, when it is released is 10.43 J

Change in gravitational potential energy

`\Delta U=10.43-30.97=-20.54\ J`

Change in gravitational potential energy is given by -20.54 J