# Work-Energy Theorem question

1. Nov 10, 2004

### pulau_tiga

Hi.

I was wondering if anyone could help me out in this question:

A mass M is sliding down the inclined plane at a speed of 2.13 m/s. The mass stops a distance of 2.45 m away from the base of the inclinde, along a flat surface. The mass slides 1.24 m down the incline before reaching the horizontal surface. The inclide has an angle theta of 31.1 degrees above the horizontal. Calculate the coefficient of kinetic friction for the mass on the surface.

I said that
Mechanical energy@ topof the incline = Mechanical energy@bottom
KE(top) + PE(top) = KE(bottom) + PE(bottom)
0.5m(Vtop^2) + mgh = 1/2m(Vbottom^2) + 0 J
Solving for Vbottom = 4.14 m/s
**I said that h = sin(theta) x 1.24 m; h=0.641 m

Then by using Vf^2 = Vo^2 + 2ad, and solving for accleration, given that Vf=0 m/s (when it comes to a stop), and Vo = 4.14 m/s(the velocity at the bottom of the incline), and d = 2.45 m
I got an acceleration of -3.50 m/s^2

then looking at a freebody diagram, I said that the sum of all forces in the x-direction is = ma.
I only have friction acting in the x direction?? Is there another force?
I said. -f=ma
-(mu)FN = ma where mu=coefficient of friction, FN = normal force
-(mu)mg= ma mg=FN, because, FN=Fg
masses cancel, and solving for mu....
mu = -0.357,
but, mu cannot be -?? and 0.357 is not the answer.

If anyone could help me out, it would be greatly appreciated thanks!!

2. Nov 10, 2004

### Staff: Mentor

Is there friction on the incline? Is 2.13 m/s the initial speed of the mass at its starting point on the incline?

You assume no friction on the incline. If so, your method is good, but check your arithmetic.

As far as $\mu$ being negative, I'm glad you realize that that makes no sense. The magnitude of the kinetic friction force = $\mu N = \mu mg$.

3. Jan 9, 2005

### Eus

Hi Ho! ^_^

Assuming that there's no frictional force on the incline, you got the right answer.
You forgot to include the minus sign for the acceleration.
Thus, -f = ma
-(mu)mg = m(-3.50)
The sign cancels and you'll get mu = 0.357 (I found it to be 0.356).
If that's not the answer, then there's frictional force on the incline.

Next time, better you use isolated system method.
Like what I have learned from Fundamental of Physics 6th edition by Halliday/Resnick/Walker.
So you doesn't need to do kinematics to find the acceleration and drawing the free body diagram.
I'll show you how.

Let's take an isolated system consists of: ramp-surface-earth-mass.
Because there's no external force acts on it, Work by external force = (delta)Energy of the isolated system = 0.
Thus, W = (delta)K + (delta)Ugravitation + (delta)Ethermal
0 = 0.5*m*(0 - 2.13^2) + m*g*(0 - 1.24*sin(31.1)) + (muk)*m*g*2.45
(delta)Ethermal = fk*d = (muk)*m*g*d
Solving for muk yields 0.3559084970
Same result with less effort for sure.

Now I will solve for muk if there's frictional force on the incline.

Let's take an isolated system consists of: ramp-surface-earth-mass.
Because there's no external force acts on it, Work by external force = (delta)Energy of the isolated system = 0.
Thus, W = (delta)K + (delta)Ugravitation + (delta)Ethermal
0 = 0.5*m*(v^2 - 2.13^2) + m*g*(0 - 1.24*sin(31.1)) + (muk)*m*g*cos(31.1)*1.24 ...(1)
(delta)Ethermal = fk*d = (muk)*N*d -> For inclined surface, N = m*g*cos(theta). Where fk is kinetic frictional force.
0 = 0.5*m*(0 - v^2) + 0 + (muk)*m*g*2.45 ...(2)
Substituting (2) to (1) for v yields:
0 = 0.5*m*(2*(muk)*g*2.45 - 2.13^2) + m*g*(0 - 1.24*sin(31.1)) + (muk)*m*g*cos(31.1)*1.24
Solving for (muk) yields:
muk = 0.248

So, any comments for this? If 0.248 is not the answer either, I think the problem or its data is wrong.