Conservation of Energy with Friction

[SpaXe]

Homework Statement

A 179 g block is launched by compressing a spring of constant k=200 N/m a distance of 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has coefficient of friction \mu = 0.27. This frictional surface extends 85 cm, followed by a frictionless curved rise, as shown in the figure:

                        |
                       /
|xxxx|[ ]             /
---------=========---/
         ^           ^---- Imagine this as a bottom right quarter of a circle.
         |                     Things go up and slide down with no friction
         |-- Friction
               85 cm

After launch, where does the block finally come to rest? Measure from the left end of the frictional zone.

Homework Equations

Conservation of Energy:
KE_i + PE_i + WE_i= KE_f + PE_f + WE_f

The Attempt at a Solution

So, values:

m = 0.179 kg

k = 200 N/m
x = -0.15 m

\mu = 0.27
d_{friction} = 0.85 m
F_{normal} = mg = 0.179 * 9.8

----

The initial total energy is the spring's potential energy. All other initial energies are zero. So:

PE_i = 1/2kx^2 = 1/2(200)(0.15) = 2.25 J

Then, I proceeded to calculate the energy loss due to friction, by the amount of work done:

WE_{per slide} = F_{friction} d_{friction} = \mu F_{normal} d_{friction} = 0.27 \times (0.179)(9.8) \times 0.85 = 0.40258 J

I found out that, if I divide the first by the second, it takes the 6th slide (back and forth) to actually "stop" the box from sliding, but at this point I'm really lost. The remaining energy might not be enough, for example, to overcome the static friction, and I'm unsure as to how to approach this problem anymore.

Any help would be appreciated. This problem is a bonus problem in one of my assessments, and I don't really need it solved to pass. But I'd really love to know how to solve it.

 

[SpaXe]

Actually, never mind me. I solved it with a friend's help. Yay!

If you divide the first by the second, the remainder is 0.237 J. Using that information, I was able to calculate the distance travelled, 50cm. So, subtract that from 85cm, it stopped at 35cm from the left of the frictional surface.

(Work = Force * distance, or 0.237 = Friction Force * distance.)

 

[Doc Al]

The initial total energy is the spring's potential energy. All other initial energies are zero. So:

PE_i = 1/2kx^2 = 1/2(200)(0.15) = 2.25 J

Then, I proceeded to calculate the energy loss due to friction, by the amount of work done:

WE_{per slide} = F_{friction} d_{friction} = \mu F_{normal} d_{friction} = 0.27 \times (0.179)(9.8) \times 0.85 = 0.40258 J

Good.

I found out that, if I divide the first by the second, it takes the 6th slide (back and forth) to actually "stop" the box from sliding, but at this point I'm really lost. The remaining energy might not be enough, for example, to overcome the static friction, and I'm unsure as to how to approach this problem anymore.

You found that it makes 5 complete trips across the friction patch and one partial trip. Find the length of that partial trip.

 

[SpaXe]

You found that it makes 5 complete trips across the friction patch and one partial trip. Find the length of that partial trip.

Thanks! That's exactly what I had to do.

As a side question, though, how does this problem relate to the concept of static friction and dynamic friction?

 

[Doc Al]

Static friction doesn't play a role in this problem.