How to use the coefficient of kinetic friction?

[Hannah]

Homework Statement

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 traveling 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?

Homework Equations

Fsc,x= -k(x-xo)
Fec,x=-mg
Fk,12=(u,k)(Fn,12)

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.

 

[Mastermind01]

What is a normal force?

 

[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

 

[Mastermind01]

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

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?

 

[Hannah]

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?

It would just be the spilled drink on the car.

 

[Mastermind01]

It would just be the spilled drink on the car.

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

 

[Mastermind01]

Also I think the answer to your first part is wrong.

 

[Hannah]

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

the force of gravity?

 

[Mastermind01]

the force of gravity?

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?

 

[Hannah]

Also I think the answer to your first part is wrong.

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

 

[Hannah]

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?

No just in the horizontal direction it has negative acceleration.

 

[Mastermind01]

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

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?

 

[Mastermind01]

No just in the horizontal direction it has negative acceleration.

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

 

[Hannah]

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

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

 

[Mastermind01]

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

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.

 

[Hannah]

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?

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 .

 

[Mastermind01]

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 .

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

 

[Hannah]

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.

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

 

[Mastermind01]

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

You got it!

 

[Hannah]

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

con

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

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

 

[Mastermind01]

con

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

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

 

[Hannah]

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

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

 

[Mastermind01]

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

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

Anyway, do you know about kinetic and potential energy?

 

[Hannah]

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

Anyway, do you know about kinetic and potential energy?

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

 

[Mastermind01]

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

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.

 

[Hannah]

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.

1/2(0.1)(0)^2 +mg(0)+1/2k(-2)^2=1/2(0.1)(3.286)^2+mg(0)+1/2(k)(0)+Ehf
is that correct so far would the forces of gravity be 0 since there is no height and would the thermal energy be 0

 

[Mastermind01]

1/2(0.1)(0)^2 +mg(0)+1/2k(-2)^2=1/2(0.1)(3.286)^2+mg(0)+1/2(k)(0)+Ehf
is that correct so far would the forces of gravity be 0 since there is no height and would the thermal energy be 0

Yes the thermal energy would be zero but it's not correct so far. The equation for conservation of energy in this case is K.E_i+ P.E_i = K.E_f+ P.E_f. What's your final case?

 

[Hannah]

1/2(0.1)(0)^2+1/2k(-0.02m)^2+1/2(0.1)(3.286)^2+1/2k(0)^2
wouldn't in the final case the spring be back at equilibrium so x=0 and the velocity after is 3.286 which i determined earlier

 

[Hannah]

1/2(0.1)(0)^2+1/2k(-0.02m)^2+1/2(0.1)(3.286)^2+1/2k(0)^2
wouldn't in the final case the spring be back at equilibrium so x=0 and the velocity after is 3.286 which i determined earlier

there is supposed to be an equal sign in the middle there instead of +, right after (-0.02)^2

 

[Mastermind01]

1/2(0.1)(0)^2+1/2k(-0.02m)^2+1/2(0.1)(3.286)^2+1/2k(0)^2
wouldn't in the final case the spring be back at equilibrium so x=0 and the velocity after is 3.286 which i determined earlier

The problem could be made much simpler by taking the final case as just on top of the elevated portion. What is the energy equation now? (You now need to take gravity into consideration)

EDIT: Actually no, what you did is correct. Do you get the value of k now?

 

[Hannah]

The problem could be made much simpler by taking the final case as just on top of the elevated portion. What is the energy equation now? (You now need to take gravity into consideration)

EDIT: Actually no, what you did is correct. Do you get the value of k now?

0+1/2k(0.02)^2+mg(0)=1/2(0.1)(3.286)^2+1/2(m)(0)^2+(0.1)(-9.8)(0.5)
0.0002k=0.53989-0.49
k=249.45
seems pretty large

 

[Hannah]

0+1/2k(0.02)^2+mg(0)=1/2(0.1)(3.286)^2+1/2(m)(0)^2+(0.1)(-9.8)(0.5)
0.0002k=0.53989-0.49
k=249.45
seems pretty large

if i do it my original way i get k =2699.45 which is even bigger

 

[Mastermind01]

Did you do your calculation correctly and follow your equation in #28 ? Then your answer should be correct

 

[Hannah]

Did you do your calculation correctly and follow your equation in #28 ? Then your answer should be correct

okay thank you very much for all your help ! :)

 

[haruspex]

0+1/2k(0.02)^2+mg(0)=1/2(0.1)(3.286)^2+1/2(m)(0)^2+(0.1)(-9.8)(0.5)

Why are you subtracting mgh at the end there?
I believe what MM was suggesting in post #30 was to write
Initial spring PE = final KE + gained gravitational PE

if i do it my original way i get k =2699.45 which is even bigger

But it does have the advantage of being correct.