# Homework Help: How to Solve Friction Problem

1. Oct 2, 2009

### r_swayze

When your sled starts down from the top of a hill, it hits a frictionless ice slick that extends all the way down the hill. At the bottom, the ground is dry and level. The effective coefficient of friction between the sled runners and the ground is 0.62. If the hill is 50 m high, how far will your sled travel once it reaches the bottom?

How do I solve this problem? I have no clue where to begin, because I thought you would need to know at least the mass and angle of the hill to figure out the velocity and then the distance. Any help?

2. Oct 2, 2009

### Delphi51

You could leave m and A in your work and they will cancel out, so the same answer regardless of the mass and the angle of the hill. It is probably quicker to get the answer using
PE at top = KE at bottom
Fill in the detailed formulas for those energies and you'll see the m's cancel out right away.

3. Oct 3, 2009

### shootingrubbe

Ok, so what you have to do is use the equations:

Eg=mgh and Ek=1/2mv^2

You know that when the sled starts at the top of the hill there is no kinetic energy acting on the sled and at the bottom there is no gravitational potential energy acting on the sled. Since there is no friction, every object on this hill will end up at the same speed at the bottom of the hill.

So we can just make up a mass...

Let's say m=1.0 kg
We know the height of the hill is 50m
We know gravity is 9.8 N/kg

At the top of the hill Eg=Total Energy... At the bottom of the hill Ek=Total Energy.

Top of the hill:
Eg=(1)(9.8)(50)
= 490

Bottom of the hill:
Ek=1/2(1)(v^2)
490 = 0.5(v^2)
v=33.305 m/s

Because 1 Joule equals a newton x metre... (Newton x metre)/kg = m^2/s^2

Now you know the initial speed when it reaches the bottom of the hill. If you draw an FBD of the sled at the bottom of the hill you will notice there are only 3 forces acting on it. Fg and Fn, which obviously cancel out because there is no angle, and Ff... **Remember** You cannot use the mass from the first step for this step.

Now Ff=∑F, so you can calculate acceleration. We know ∑F=ma, Fn=Fg=mg, μ=Ff/Fn and μ=0.62. Now you can make up the equations that will cancel mass out in the end.

∑F=Ff=ma
Fn=Fg=mg
μ=0.62
g=9.8 N/kg

We can say that:
μ=Ff/Fn
μ=ma/mg
0.62(m)(9.8) = ma
a=(0.62)(9.8)
= 6.076 m/s^2 [backwards]

With the acceleration you now have V1=33.305 m/s, V2=0 m/s and the acceleration which you previously calculated. You can now use your normal motion equations (the one without time) to calculate the distance.

Hope this helps...

Last edited: Oct 3, 2009
4. Jun 24, 2010

### govols

Shootingrubbe:

That's nearly flawless. Thanks for the help. I believe the velocity at the ground (y=0m) should actually 31.305m/s. Using the equation V^2=Vo^2+2a(X) We know that Vo is zero.

V^2= 2(9.8)(50m)
V^2= (19.6)(50)
V^2= 980
V= SQRT(980) = 31.3049m/s

Using V1=31.305 in the final equation: V^2=Vo^2+2a(X) gives the correct answer.

Cheers,