# Calculating the work done by friction

1. Dec 10, 2005

### NutriGrainKiller

I am having a hard time figuring out how to calculate the work done by friction with varying acceleration. I will post the entire problem just for reference, but I know how to do everything EXCEPT how to calculate the work done by non conservative forces, i.e. friction in this case. Here is the problem:

I know the work done by friction is 0.24J, but I don't know how to calculate it.

Help?

2. Dec 10, 2005

### Staff: Mentor

You calculate the work done by friction exactly as you'd calculate the work done by any force: Work = force X distance. What's the friction force? What's the distance over which it acts?

(What makes you think the work done by friction is 0.24 J?)

3. Dec 10, 2005

### NutriGrainKiller

for a spring, Fnet=-kx, so the force is 1.2N and the distance is .1m so the answer should be .12...

I'm obviously missing something, so I went ahead and took a screenshot of the reader file. Here is the page.

4. Dec 10, 2005

### Staff: Mentor

The force of the spring has nothing to do with friction.

Friction acts between a mass and the surface on which it is moving.

Normally, friction is defined as the product of the coefficient for friction $\mu$ and the weight of the mass (mg). Then the work is the integral of the force over the distance. The spring is pulled 0.2 m from equilibrium.

$$Work, W\,=\,\int\,F\,dx$$

If F is constant, then Work, W = F*d, where d is the distance over which the force is applied, not the differential operator in dx.

Last edited: Dec 10, 2005
5. Dec 10, 2005

### NutriGrainKiller

I actually thought of this, because I saw that by integrating from 0 to .2 xdx and moving the spring constant outside I got the right answer.

Here is what I don't understand, and maybe you can help explain it to me: why .2 and not .1? Aren't we talking about half the distance between x and 0? why .2? why why why?

6. Dec 10, 2005

### Staff: Mentor

The spring force has nothing to do with friction!

The friction is applied over the distance that the glider travels on the track, and that force is completely independent from the force of the spring. The friction is proportional to the weight of the glider.

Now, try to calculate the friction force based on the weight of the glider and apply it over the distance traveled.

Half-way along the track is 0.1 m, and that is part of one of the questions

To solve this equation, one would have to subtract the friction force from the spring force to solve the problem.

7. Dec 10, 2005

### NutriGrainKiller

Astronuc: i am still confused. Here is my thought process, maybe you can see where I am going wrong.

Friction = .08[muK] * 2.45[mg]. How do I apply it to the distance traveled...?

K - Ff => k - (muk)(normal force) => 12 - (.08)(2.45) = 11.804

8. Dec 10, 2005

### Staff: Mentor

The weight is mg, the friction is $\mu$mg, and the work is friction force applied over distance.

W = $\int_o^d\,F\,dx$, which if F is constant is F*d.

So W = $\mu$mg*d = 0.08 * 0.25 kg * 9.8 m/s2 * 0.2 m, which is the work of force friction force F applied over d = 0.2 m.

9. Dec 10, 2005

### NutriGrainKiller

that makes sense, but i got 0.0392, and he said the answer was .24J

10. Dec 10, 2005

### Staff: Mentor

Referring to the image of the page you posted, I believe the 0.24J has to do with the stored energy in the compressed spring.

1/2 * k * x2, with k = 12 N/m and x = 0.2 m,

1/2 * 12 N/m * (0.2m)2 = 0.24 J

Hence Doc Al's question in his post.

11. Dec 10, 2005

### lightgrav

Energy is a conditional quantity ; initial condition of the stretched spring is ½ k s^2 = 24 [Joules] contained, since stretched .2 [m] .

Work is a process quantity ; it depends on the (average) Force component parallel to the displacement.

The Energy in the later configuration (spring only stretched 0.1 [m])
is less than in the initial condition because the Work by friction
transferred some of the Energy to the track.
Because the Potential Energy function is not linear, you have to measure its stretch distance from the its unstretched condition (minimum PE).
That is, ½ k (.2[m])^2 - ½ k (.1[m])^2 is NOT equal ½ k (.1[m])^2 .
gravitational PE near Earth surface is UNusual, being almost linear.

12. Dec 11, 2005

### gunblaze

Sry, not really sure it helps

$$EPE_i = EPE_f + KE$$ + Energy loss in overcoming friction

assuming u noe how to calculate the rest.. Energy in overcoming friction will be $$E_f= u_f * F_normal$$ * distance travelled

By using KE, u shld be able to solve for v.

Last edited: Dec 11, 2005