Child sliding down playground slide with friction losses

[PhysicsPhun]

A 33.0 kg child slides down a long slide in a play ground. she starts from rest at a height h1 of 19.00 m. When she is partway down the slide, at a height h2 of 9.00 m, she is moving at a speed of 8.90 m/s. Calculate the mechanical energy lost due to friction.

I know that Conservation of mechanical energy is K2 + U2 = K1 + U1. I don't really know how to start this problem though.

Any help would be appreciated.

 

[kuba]

Well, the conservation of energy means this:
if you slide on a slide long enough and the slide is steep enough, your butt will eventually be set on fire. We all know this, right?

Well, the energy that was "set free" by coming down those 10 meters had to go somewhere. It went into two places: heat generated at the butt-slide interface and your kinetic energy. The "set free" is quoted since it's the most awkward and unscientific way of saying it. But I'm trying to get to the point quickly

So,
Eg=Q+Ek
Eg-difference in potential energy at endpoints of the motion
Q-heat generated to keep your butt warm
Ek-kinetic energy difference at endpoints of the motion
_f = final, _0 = (read "not") is initial

Ek=Ek_f-Ek_0=Ek_f since you start at rest and thus Ek_0=0. Then Ek=m*v_f^2=33kg*(8.90m/s)^2

Eg=m*g*h_f-m*g*h_0 = m*g*(h_f-h_0) = 33kg*9.81m/s^2*(19m-9m)

Q is the unknown, mostly butt-absorbed heat that you need to solve for.

Kuba

 

[PhysicsPhun]

After reading over that i got 623.37 Joules.. that's not right is it?

Hehe. It's an entertaining way to put it btw :)

 

[kuba]

I don't know. You should have something like

$$Q=E_g-E_k=m(g\,\Delta h-{v^2\over2})$$

 

[PhysicsPhun]

I got 1928.035 by Adding Initial Kinetic Energy and Initial Potential Energy and then subtracting Final Kinetic Energy and Initial Kinetic Energy.

I think that's right, and you were probably saying the same thing with different symbols, hehe.

Thanks a lot.

 

[kuba]

Initial kinetic energy was zero
Kuba