Gravitational Potential Energy Loss

[tquiva]

Homework Statement

Assume no friction. A 58-kg teenager at a water park slides down a long, winding waterslide of varying slope. The slide has a net height difference of 30. m from start to finish, and the teenager starts from rest. Throughout the problem, let y=0 and Ugrav = 0 at the end (bottom) of the slide.
a) What is the teenager's net loss of gravitation potential energy?
b) If the slide is frictionless, what is the teenager's net gain of kinetic energy?
c) What is the teenager's total mechanical energy?

Homework Equations

I know that
W = Kf - Ko + Uf - Uo

The Attempt at a Solution

Since there is no friction, W = 0
And since Ugrav = 0, Uf and Ko cancels out
this gives me Kf = Uo (KE gained is PE lost)
=> (1/2) *m*v^2 = mgh

After calculating for Uo, I get (58 kg) * (9.8 m/s^2) * (30. m) = 17040 J = 17.04 kj

My given choices are
A. -490 J D. -6.8 kJ
B. -860 J E. -12 kJ
C. -2.9 kJ F. -17 kJ

My guess is that for a, Uo = -17 kj (negative since it's lost) & for part b, Kf = 17 kJ (positive since it's gained). Is that right?

But I need help in doing part C also. Is mechanical energy the sum of both Uo and Kf?

 

[lightgrav]

yes, Mechanical Energy is just the sum of
Kinetic Energy , Gravitational Potential Energy , and Elastic Potential Energies
(usually NOT including Pressure Potential Energy, or Electric, or Magnetic ...).

technically, the answer should be the same for all of these questions (all positive)
... if you lose 75 cents , did you lose 3 quarters , or -3 quarters ?

 

[tquiva]

thank you. I got 17 kJ for the total mechanical energy. is this right?

I have a second part to this question, and the only difference is that friction now exists.

I found that the net gain of kinetic energy is 1.9 kJ

But what is the teenager's total mechanical energy at the top of slide?
My given choices are:
A. 190 J D. 4.5 kJ
B. 460 J E. 8.3 kJ
C. 1.9 kJ F. 17 kJ

My answer is D since mgh = -17 kJ
is this right?

And the total mechanical energy at the end of the slide is Kf = 1.9 kJ ?

 

[lightgrav]

mgh is a _positive_ 17 kJ at the top of the slide (just like before).

If they're going 8 m/s at the bottom , okay.

 

[tquiva]

thanks again, I appreciate it.
one last question...
I discovered that the percentage of initial mechanical energy dissipitated by friction is 89%
the last question asked be for the average power, given that the total time down the slide is 30s.
My choices are:
A. 38 W D. 340 W
B. 79. W E. 510 W
C. 190 W F. 3.3 kW

Here's how I did it:
Wtot = (1/2)mv^2 - 1/2mvo^2
Wtot = (1/2)mv^2 = (1/2)(58 kg)(8.0 m/s) = 1859 J
P = W/t = 1859 J / 30s = 62 W
it doesn't match any of the choices. could someone help?