# Homework Help: How to use the coefficient of kinetic friction?

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1. Apr 4, 2017

### Hannah

1. The problem statement, all variables and given/known data
A child is playing with their Hot Wheel cars. They have set up a track which is initially horizontal but then ascends to a second horizontal section which is 50 cm higher than the initial track. The track is friction-less until the car reaches the upper section of the track where some spilled soft drink exerts a frictional force on the car. They launch the 100 gram cars on the lower horizontal section using a spring loading launcher which requires the spring to be compressed by 2 cm. (Use conservation of energy and work to solve this problem.) (25 marks) a. If the car is travelling at 1 m/s when it initially reaches the upper section of the track what is the spring constant of the launcher?
b. If the coefficient of kinetic friction on the upper portion of the track is 0.4, how far does the car travel before it stops?
c. Draw energy bar charts for:
i. The initial situation (the car is in the spring loading launcher with the spring compressed),
ii. When the car arrives at the start of the upper horizontal section,
iii. When the car comes to a final stop.
d. What work does the child need to do to compress the spring?
e. How much work does the frictional force do?

2. Relevant equations
Fsc,x= -k(x-xo)
Fec,x=-mg
Fk,12=(u,k)(Fn,12)

3. The attempt at a solution
So I think I got part a)So since the track is frictionless we can assume the intial velocity before going up the inline= final velocity at the top of incline so there is no acceleration. The only force that the car has to overcome is the force of gravity. So i used Fec,x= mg=(100g)(1kg/1000g)(9.8m/s^2)=0.98N. This would also be the force that the spring has to exert on the car.
So then i used Fsc,x=-k(x-xo)
0.98N=-k(-0.02m)
k=49

b) for part b I am completely lost. I know I will probably have to use the equation Fk,12=(u,k)(Fn,12)
but I don't know what the normal force is? Could someone explain that to me and how to figure that out please.

c)i)this would be all potential energy
ii)There would be kinetic energy, maybe some potential energy? and some thermal energy
iii) This would all be converted to thermal energy

d and e should be easy after I figure out the rest.

2. Apr 4, 2017

### Mastermind01

What is a normal force?

3. Apr 4, 2017

### Hannah

That is what i am wondering I found in my textbook that Fn,12 = Fc, 12 so basically the normal force is the contact force of surface one on surface 2

4. Apr 4, 2017

### Mastermind01

Nope, the normal force is the perpendicular component of the contact force. Friction is also part of the contact force. So then what are the perpendicular forces on the car?

5. Apr 4, 2017

### Hannah

It would just be the spilled drink on the car.

6. Apr 4, 2017

### Mastermind01

No , I'm talking about forces on the car in the upper part.

7. Apr 4, 2017

8. Apr 4, 2017

### Hannah

the force of gravity?

9. Apr 4, 2017

### Mastermind01

That is correct, now we have two forces in the perpendicular direction , the normal and gravity. Now does the car have acceleration in the perpendicular direction when it's in the upper part?

10. Apr 4, 2017

### Hannah

Ya i was starting to rethink that part I think since the spring accelerates the car from 0m/s to 1m/s Fsc,x=mg+ma

11. Apr 4, 2017

### Hannah

No just in the horizontal direction it has negative acceleration.

12. Apr 4, 2017

### Mastermind01

Nope, you are close but wrong. Your question mentions that there is a horizontal lower section before elevation to an upper section. So, for the horizontal section it should be $kx = ma$ but we don't know the length of the horizontal section. So how else can we do it?

13. Apr 4, 2017

### Mastermind01

Yep, so now what should be the normal force? ( Hint: Newton's second law)

14. Apr 4, 2017

### Hannah

So is it zero? I was thinking that before but I didn't think it made sense because then kinetic force would be 0.

15. Apr 4, 2017

### Mastermind01

No by Newton's second law the net force on a body is equal to the product of it's mass and acceleration (here it's zero). The key word here is net.

16. Apr 4, 2017

### Hannah

well if we know the final velocity at the top we could figure out the initial at the bottom by vf^2=vi^2 +2ad
so Vi=3.286 but to get the acceleration at the bottom we would need either time or distance .

17. Apr 4, 2017

### Mastermind01

Yes. So what other approach can you use? (It's given in your question)

18. Apr 4, 2017

### Hannah

okay so normal force would be -mg to counteract the force of gravity on the object = -(0.1kg)(9.8m/s^2)=0.98

19. Apr 4, 2017

### Mastermind01

You got it!

20. Apr 4, 2017

### Hannah

con
so it says conservation of energy but im still confused on how that would me find acceleration? sorry I'm so bad at this lol

21. Apr 4, 2017

### Mastermind01

Using conservation of energy you don't need to find acceleration. What is the law of conservation of energy?

22. Apr 4, 2017

### Hannah

well in an isolated system energy is conserved, i found this equation online but i have never used it in class before
1/2mvi^2+mgh,i+1/2kxi^2=1/2mvf^2+mghf+1/2kxf^2+EHf

23. Apr 4, 2017

### Mastermind01

Then how come their's a question on your homework?

Anyway, do you know about kinetic and potential energy?

24. Apr 4, 2017

### Hannah

yes, i think we just may have used something else for potential energy without the spring constant which we just learned not too long ago

25. Apr 4, 2017

### Mastermind01

Well then you should read up about spring potential energy from your textbook or online resources (MITOCW or KA) and then ask for help if you have some difficulty.