Question

While leaning out a window that is 6.1 m above the ground, you drop a 0.60 kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.3 m above the ground. Determine the following.

(a)

the amount of work done (in J) by the force of gravity on the ball

( ) J

(b)

the gravitational potential energy (in J) of the ball-earth system, relative to the ground when it is released

( ) J

(c)

the gravitational potential energy (in J) of the ball-earth system, relative to the ground when it is caught

( ) J

(d)

the ratio of the change (PE_f − PE₀) in the gravitational potential energy of the ball-earth system to the work done on the ball by the force of gravity

ΔPE

W

= ( )

Answer 1

the amount of work done (in J) by the force of gravity on the ball= gravitational force*distance travelled= 0.6*9.8*(6.1-1.3)= 28.2J

the gravitational potential energy (in J) of the ball-earth system, relative to the ground when it is released= 0.6*9.8*6.1= 35.87J

the gravitational potential energy (in J) of the ball-earth system, relative to the ground when it is caught= 0.6*9.8*1.3= 7.64J

So, the ratio of the change (PEf − PE0) in the gravitational potential energy of the ball-earth system to the work done on the ball by the force of gravity= (35.87-7.64)/28.2= 1.001