# Work Energy Principle

1. Oct 30, 2008

### tater08

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

On an essentially frictionless horizontal ice-skating rink, a skater moving at 3.00 m/s encounters a rough patch that reduces her speed by 45.0 % to a friction force that is 25.0 % of her weight.

Use the work-energy principle to find the length of the rough patch.

2. Relevant equations
KE initial + Potential Energy Initial + Work= KE Final + PE Final

3. The attempt at a solution

0.5 MV^2 + mgj + mgd = 0.5 mv^2 + mgh mass will cancel out so and h is 0

0.5 v^2 + 0.25 gd= 0.5v^2

0.5 (3^2) + 0.25 * 9.8 d= 0.5 (3^2) *.45

4.5+2.45d=2.025
d=1.01 m

i guess i do not know what im doing wrong in this problem besides the reducing the final velocity.

thanks for the help!

2. Oct 30, 2008

### Rake-MC

You have your work on the wrong side of the equation.
Look at it like this: initially there's no friction so there's no work being done by the surface.
Finally there is friction, so there is work being done. Therefore:

$$\frac{1}{2}mv^2 = \frac{1}{2}m(0.55v)^2 + 0.25mgd$$

3. Oct 30, 2008

### alphysicist

Hi Rake-MC,

I have to disagree with your reasoning that the work is on the wrong side of the equation (although I agree that your equation is mathematically correct). An important fact is that the work being done by friction is negative. When we consider the work done by friction in the energy formula (as opposed to the thermal energy created by friction), the correct formula is:

Wnc = Ef-Ei

Or am I just misunderstanding your post?

Tater08,

In addition to the sign error that rake-mc pointed out, when you get to these two lines in your solution:

0.5 v^2 + 0.25 gd= 0.5v^2

0.5 (3^2) + 0.25 * 9.8 d= 0.5 (3^2) *.45

you have made two errors in the second line. First they say the speed drops by 45% (not that the speed drops to 45% of what it was).

Second, the entire final velocity needs to be squared.

4. Oct 30, 2008

### Rake-MC

Perhaps I am misunderstanding you alphysicist, but are you saying that correctly, it should be written:

$$\frac{1}{2}mv^2 - 0.25mgd = \frac{1}{2}m(0.55v)^2$$? I can see how that would work, from the work being negative. Thanks for pointing that out.