Conservation of energy of a basketball

[Amar.alchemy]

Homework Statement

A child rolls a 0.600-kg basketball up a long ramp. The basketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp. it has a speed of 8.0 m/s. When the ball returns to her after rolling up the
ramp and then rolling back down, it has a speed of 4.0 m/s. Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase
of the ball as it rolls up the ramp.

Homework Equations

K1 + U1 + Wf = K2 + U2

The Attempt at a Solution

1)(If i consider the work done static friction force is zero)
From base of the hill to the uphill:
K1= 5/6 MV², U1=0
k2=0, U2=Mgh and Wf=0

Then 32 = Mgh
h=5.4m...

2)If work done by static friction force is considered then,
From base of the hill to the uphill:
K1= 5/6 MV², U1=0
k2=0, U2=Mgh and Wf

32+wf = mgh ... 1

from uphill to base of the hill:
Mgh + wf = 8...2

If i solve the equations 1 and 2 then i will get h =3.4m as given in the textbook.

My doubt is, The ball is rolling both up and down the hill due to static friction force. So the workdone by a static friction force is zero. Is this correct??Kindly explain me...

 

[kuruman]

Conservation of energy is not an equation. Please write down an equation expressing conservation of energy.

 

[Amar.alchemy]

@Kuruman
updated :)

 

[kuruman]

Let me make sure I understand exactly what is going on. The child gives the ball a psh up the incline with an initial velocity 8 m/s. When the ball returns to her, it has velocity 4 m/s going down the incline. Is that correct?

 

[Amar.alchemy]

ya... and also in both ways(ie up and down) ball is rolling without slipping...

 

[kuruman]

The answer 3.4 m is correct. The loss of energy that you used to calculate this height is not the work done by static friction as the ball rolls. It is energy lost to dissipative forces such as air-resistance and the like. The statement of the problem is somewhat confusing.