Gravitation potential energy without hieght

[ncote]

1.
A 55.5 kg skateboarder starts out with a speed of 1.75 m/s. He does +80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 5.90 m/s.
(a) Calculate the change (PE = PEf - PE0) in the gravitational potential energy.



i also have to calculate the change in the vertical hieght. i have absolutely no clue how to do this without h...

 

[PhanthomJay]

1.
A 55.5 kg skateboarder starts out with a speed of 1.75 m/s. He does +80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 5.90 m/s.
(a) Calculate the change (PE = PEf - PE0) in the gravitational potential energy.



i also have to calculate the change in the vertical hieght. i have absolutely no clue how to do this without h...

You might want to consider the work energy theorem W_c + W_{nc} = \Delta KEand go from there, if you're familar with it.

 

[ncote]

i got W using (1/2)(m)(vf^2)-(1/2)(m)(vo^2). not sure where to go from here.

 

[PhanthomJay]

i got W using (1/2)(m)(vf^2)-(1/2)(m)(vo^2). not sure where to go from here.

Correct, that's the total (or net) work done. Some of that work is the work done by the non conservative forces, which is given. The rest is the work done by conservative forces, which is the gravity force in this case. Work done by gravity is just -mgh. If this equation is confusing to you, perhaps you should instead use the conservation of total energy equation, which is W_{nc} = \Delta KE + \Delta PE, which might be a bit easier to understand.

 

[ncote]

i got it!

i used W-Wnc=Wc and got the correct answer for the change

for the hieght i just did W-Wnc/mg.

 

[im2hope4]

I was looking at the different formulas and stuff for this, and I'm stuck too, especially when I have taken the formula -Ffrd= (1/2)mvf^2-(1/2)mvi^2+mgyfinal-mgyinitial. I seriously am stuck on how to solve for the change in gravitational potential energy without the the final y-value or even distance.