# Homework Help: Girl on Slide problem help

1. Oct 17, 2015

### King_Silver

1. The problem statement, all variables and given/known data
A girl whose weight is 294 N slides down a 6.5 m long playground slide that makes an angle of 20◦ with the horizontal. The coefficient of kinetic friction between the slide and the girl is 0.23.
(a) What is the normal force of the slide acting on the girl?
(b) How much energy is transferred to thermal energy during her slide?
(c) If she starts at the top with a speed of 0.46 m/s, what is her speed at the bottom?

2. Relevant equations
N = mgCosQ
Ffr = Coefficient of Ke * N
Wfr = Ffr * d

where....
Ke =0.23
Q =20 degrees
m = 294 N
g = 9.81m/s^2
d (length) = 6.5m

3. The attempt at a solution

(a) N = mgCosQ as I am looking for the normal force.
-Sub in values given...
N = 294Cos(20)
N = 276.27N

(b) Ffr = 0.23 * 276.27N
Wfr = Ffr *d
Wfr = Ffr *6.5
Ffr = 63.54J
Wfr = 413.023 Joules

(c) I think I have gotten the last two parts right but its the 3rd part that I don't understand. I drew a diagram and because it is a slide (I presume it has a ladder going straight up) which would make it a right angled triangle.
20,70,90 degree angles with a length of 6.5m. Using this I calculated the height to be 2.22m.

I know have this formula: Eo = mgh + mv^2 but also V^2 = 2gh
I'm given an initial speed but am confused how it relates to anything I have. Am I approaching this question in a correct manner? Thanks in advance to anyone willing to guide me along and help me understand this.

2. Oct 17, 2015

### mcairtime

I think the solution is as follows:

She starts off with potential energy mgh (where mg is the girl's weight (294 N) and the h = 2.22 m) and with kinetic energy 0.5mu^2 (where u = 0.46 m/s is her initial speed). All these values we can calculate explicitly.

By the time she reaches the bottom, 413 joules of the total that she started off with has been converted into heat because of the work done against friction, but whatever is left over will be the girl's kinetic energy at the bottom.

Does that help?

3. Oct 17, 2015

### King_Silver

Ok yea that helps a lot, I'm just wondering about two things. When mgh and 1/2 mu^2 are calculated, mg = 294N. So m alone would have a value of roughly 30kg? would this be used instead of the 294N for the 1/2mu^2 part of the equation?
Also I am looking for the speed at the bottom (presuming it means when she reaches the bottom of the slide and not when she comes to a stop).
Would this be achieved by saying 413 J = mgh + 1/2mu^2 ?

4. Oct 17, 2015

### mcairtime

Yes, assuming we take g to be 9.8, the girls mass will be 30kg. In fact the equation is really 0.5mu^2 + mgh = 413 + 0.5mv^2 where v is her final speed at the bottom of the slide. The left hand side of the equation is the total energy that she starts with at the top of the slide, the right hand side is where that energy goes to by the time she reaches the bottom; some is lost as heat (413J) and the rest is the girl's kinetic energy.

This is, in effect, the principle of conservation of energy.

5. Oct 17, 2015

### King_Silver

Perfect! :) I understand it now thanks for that huge help!