Coefficient of Friction / Speed on incline.

[NoobertNoClue]

Homework Statement

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

Homework Equations

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

 

[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)

 

[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.?

 

[tiny-tim]

Hi NoobertNoClue!

(just got up :zzz: …)

∆KE + ∆PE = 0

No! As I said …

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

Start again.

 

[rdl]

Firstly, it's important to note that the coefficient of friction is different for the toboggan and the pool water. The coefficient of friction between the toboggan and the slide is 0.2 when the water flow rate is less than 20L/s, and 50 when the water flow rate is greater than 20L/s. Similarly, the coefficient of friction between the toboggan and the pool water is 0.6.

To determine the rider's speed, you can use the equation for conservation of energy:

mgh = 1/2mv^2 + 1/2Iw^2

Where m is the mass of the rider and toboggan (110kg), g is the acceleration due to gravity (9.8 m/s^2), h is the vertical height (7.6m), v is the final speed, I is the moment of inertia (assuming a cylindrical toboggan, I = 1/2mr^2), and w is the angular velocity (which is assumed to be zero in this case).

Solving for v, we get v = √(2gh). Plugging in the given values, we get v = 12.5 m/s.

To determine the stopping distance, we can use the equation for work done by friction:

Wfr = μmgd

Where μ is the coefficient of friction, m is the mass, g is the acceleration due to gravity, and d is the distance.

For the toboggan and slide, we can use the value of μ = 0.2 since the water flow rate is less than 20L/s. For the toboggan and pool water, we can use the value of μ = 0.6.

Setting the work done by friction equal to the change in kinetic energy (since the final speed is zero), we get:

μmgd = 1/2mv^2

Solving for d, we get d = v^2/2μg. Plugging in the given values, we get d = 78.1 m.

To determine the length of the splash pool needed, we can use the equation for projectile motion:

y = y0 + v0t + 1/2at^2

Where y is the vertical position, y0 is the initial vertical position, v0 is the initial vertical velocity (which is