Question

A 0.60-kg basketball is dropped out of a window that is 6.1mabove the ground. The ball is caught by a person whose handsare 1.5m above the gound.

1. How much work is done on e the ball by its weight?

What is the gravitational potential energy of the basketball,relative to the ground, when it is ... 2. released and 3. caught?4. How is the change (PEf- PE0) in the ball'sgravitational potential energy related to the workdone by its weight?

Answer 1

(1)

Work Done = mg(h₁-h₂)
                   =0.60*9.8*(6.1-1.5)
                   =27.048 J.

(2)

gravitational potential energy of the basketball, relative tothe ground when it is released is P.E. = mgh₁.

           =0.60*9.8*6.1

           =35.868 J.
(3)

gravitational potential energy of the basketball, relative tothe ground when it is released is P.E. = mgh₂.

           =0.60*9.8*1.5

           = 8.82J.
(4) Work Done = mg(h₁-h₂)
    Work Done = PE_I -PE_f