Work and enegry theoretical problem

[chazgurl4life]

Ok so my problem was: A 80 kg hiker starts at an elevation of 1600 m and climbs to the top of a 3600 m peak.

(a) What is the hiker's change in potential energy?
J

(b) What is the minimum work required of the hiker?
J

(c) Can the actual work done be more than this? YES!

so i figured out a) Change in PE = mg(y2-y1)
=80kg(9.8 m/s^2)(3600m-1600m)
= 1.56 x 10^6 Joules
B) Part b is the same answer cause with your PE you can perform minimum work

But for C i don't understand how the actual work be more than delta PE?

 

[PPonte]

Are you considering friction?

 

[chazgurl4life]

Ok ..if we consider friction t the opposite force ( energy used to climb up the peak) has to be more than 1.5 e6 because it has to overcome the force of friction to kepp climbing up. is that right?

 

[PPonte]

Yes. That's what I was thinking.

 

[chazgurl4life]

thanx so much for your help

 

[PPonte]

It was a pleasure, but you should ask a second and better opinion.

 

[chazgurl4life]

im going to ask my professor in class thanks anyway