# Physics Problem with Spring Constant and Kinetic Friction

1. Jan 3, 2015

### Q7heng

1. The problem statement, all variables and given/known data
A person attaches a spring to an cubic object that weighs 36kg and pulls this object along a table made of material X horizontally with a steady speed of 1.3m/s. The spring stretches a distance of 3.4cm. Find the ratio of the spring constant to the kinetic friction coefficient.

2. Relevant equations
I suppose FkkN
Fspring=-kx
F=ma
Not sure if any other ones are needed, but I couldn't find a way to solve this and get a reasonable answer.

2. Jan 3, 2015

### Bystander

What does this information tell you?

3. Jan 4, 2015

### Q7heng

I'm not sure, but I tried to apply it to the F=MA formula... Am I on track?

4. Jan 4, 2015

### Bystander

Okay, we'll work it through that way: "Steady speed" means what in terms of "A?"

5. Jan 4, 2015

### Q7heng

Acceleration=0 at steady speed, but the trouble I'm having is converting everything to Newtons, since spring constant is Newtons/meter, and kinetic friction coefficient is Force of Kinetic Friction/Normal Force. The problem didn't give any units related to those calculations, maybe there is a way to convert it but I have yet figured it out/learned it.

6. Jan 4, 2015

### Bystander

This looks perfectly useful. What's the normal force?
Nothing wrong with this.
You aren't required to generate a numerical answer for every problem on the planet. Sometimes it's just a matter of coming up with a symbolic expression.

7. Jan 4, 2015

### haruspex

All good so far. Now, what is the relationship between F, Fk and Fspring?

8. Jan 5, 2015

### Q7heng

F=Fk+Fspring right?
If that is so then:
F=MA, A=0, and F=0
So 0=Fk+Fspring
0=-kx+μkN
kx=μkN
k/μk=N/x, since we are trying to find the ratio between the spring constant and the kinetic friction coefficient
Since N, normal force, is 36*9.8N right now, and x, distance, is 3.4cm or 0.034m right now, then it is 36*9.8/0.034=10,376.47, which is the ratio of spring constant to kinetic friction, is that correct?

9. Jan 5, 2015

### haruspex

Yes, that all looks right. But you should include the units in the answer.

10. Jan 5, 2015

Thanks!!!