# Coefficient of Friction / Speed on incline.

1. May 23, 2010

### NoobertNoClue

1. The problem statement, all variables and given/known data

A waterslide ride consists of a straight guided ramp, at an angle theta (to the horizontal plane), with a transition to a short horizontal section at the bottom of the slide (1m long), just above the surface of a splace pool of length L. Patrons queue on a platform of height h (edge of platform is 18m from start of horizontal section of slide) before climbing onto a 10kg toboggan and sliding down slide under the force of gravity until they leave the slide and skip along surface of splash pool.

Water slide is lubricated by 12L/s of water

When theta = 53degrees
h = 7.6m

1. Determine Rider speed
2. and stopping distance if;
average weight of rider is 100kg and toboggan is 10kg
coefficient of friction between toboggan and slide (us) is;
0.2 (Q 20)/50 (Where Q is in L/s)
coefficient of friction between toboggan and pool water uw is;
0.6
****Taking into account dynamic friction only*****
3. Determine length of splash pool needed L

2. Relevant equations

3. The attempt at a solution

I have been trying to figure this out using various equations, none of which seems correct... someone please help guide me onto the right track?!

I think i need to use the equation a= netforce / mass = g(sin theta - cos theta)
then use the formula V(final)^2 = V(initial)^2 + 2 as
to find net force I just used netforce= mass x g <<<<<<< I think this is incorrect because I am supposed to factor in the angle?! but how?

then use formula Stopping distance (D) = (V(final)^2 / 2x us x g) - (V(final)^2 / 2x uw x g)

i cannot figure out where to use the flow rate in this at all... i think i have totally stuffed up.... i really need help

Last edited: May 23, 2010
2. May 24, 2010

### tiny-tim

Welcome to PF!

Hi NoobertNoClue! Welcome to PF!

(have a mu: µ and a theta: θ and try using the X2 and X2 tags just above the Reply box )

Don't use acceleration.

Use ∆(KE) + ∆(PE) = work done by friction.

(The only relevance of the flow rate is that you need it in the formula for the friction on the slide)

3. May 24, 2010

### NoobertNoClue

∆KE = 1/2mV(final)² - 1/2mV(initial)² ???
∆PE = mg∆h
∆EE is N/A
∆KE + ∆PE = 0

∆PE = 110kg x 9.81 x 7.6m
=8201.16

.·. Should ∆KE = -8201.16 ???

V(initial) = 0
∆KE = 0 - 1/2mV(final)²
V(final) = √(2x∆KE/ -m)
V(final) = 12.2114.... m/s
at the end of the angular part of slide...

velocity at staright part of slide would be;
∆PE = 110kg x 9.81 x 0
=0
so ∆KE = 0 ???

0 = 12.21 - 1/2mV(final)²
-12.21 = -1/2mV(final)²

V(final) = 0.4712 m/s

am i correct so far??

Then velocity on pool =
0 = 0.4712 - 1/2mV(final)²
-0.4712 = -1/2mV(final)²
V(final)= 0.092559....m/s

how do i determine stopping distance from all of this?
do i use the coefficient of friction to determine force?
then divide?
if so do i use the formula

F= µR where R is normal force = mg <<<<<<<<<At this point was I supposed to use F= µN where N = mg(sin$$\vartheta-cos\vartheta$$)

F on slide = 0.2 x (110 x 9.81)
= 0.2 x 1079.1
= 215.82 N

then on the pool
= 0.6 x 1079.1
= 647.46

Stopping distance =
work = distance x force

so d = w/f
Total work (∆KE) = 1/2mV(final)² - 1/2mV(initial)²
= 1/2x110x0² - 1/2x110x0.09255²
=-0.4711026...

at this point i know i have gone terribly wrong...
should i just be using the formula
d=V²/2µg
where V would be the velocity at the end of the straight part of the slide
and µ would be the coefficient of friction of the pool water.?????

Last edited: May 24, 2010
4. May 25, 2010

### tiny-tim

Hi NoobertNoClue!

(just got up :zzz: …)
No!! As I said …

Start again.