Coefficient of Friction question for a cart going down a wooden ramp

[cutielollipop]

(mentor note: moved from Classical Physics forum hence no template)

Hello, I am having trouble with this question: Imagine in real life there was a coefficient of kinetic friction of 0.4 between the plastic wheels of the cart and the wooden ramp. If there is only friction on the flat part of the track, how far would the cart go before it stopped if it started from a
height of 4m? The mass of the cart is 10 kg. Here is what I have done and I would appreciate it if someone could verify my work.

Velocity using the simulation at the bottom of ramp: 8.854 m/s

Step 2: Calculating the force of friction on the flat part of the track:

Fk = µkFn

Fn = (5 kg)(9.8 m/s^2)

Fn = 49 N

Fk = (0.4)(49 N)

Fk = 19.6 N

Step 3: Calculating distance before the cart stops.

The work done by friction:

Wfriction = Ffriction x distance

The work done by friction is also equal to the loss in kinetic energy:

Wfriction = Ek_initial – Ek_final

Wfriction = ½ mv^2 – 0 (because Ek_final = 0)

Equating expressions:

Ffriction x d = ½ mv^2

Rearranging for “d”

d = 10.0 m

Therefore, the car would go 10 m before it stops on the flat part of the track.

 

[haruspex]

Are the wheels stuck?
Is the cart mass 5kg or 10kg?
Should the cart mass matter? The better style is to work purely algebraically, not plugging in numbers until the end. It has many advantages.

 

[cutielollipop]

The wheels are attached to the cart and the mass is 5 kg. Sorry for the typo.

 

[kuruman]

Yes, the wheels are attached to the cart. The issue is whether they are allowed to roll freely or not. There is also mention of the "flat part of the track." Is there a part that is not flat? What does the wooden ramp look like?

 

[cutielollipop]

The ramp looks like this. (but the height isn't 2.5 m. It's 4m)

 

[PeroK]

View attachment 329591
The ramp looks like this. (but the height isn't 2.5 m. It's 4m)

The cart rolls down the slope a vertical height of $4m$. When it reaches the flat section the brakes are applied, the wheels lock and the coefficient of kinetic friction is 0.4. How far does the cart slide?

If that's the question, then $10m$ is correct.

Can you see why $\mu_k d = h$? Where $d$ is the sliding distance and $h$ is the height?

 

[cutielollipop]

Yes I see that. Thank you.

 

[haruspex]

Yes I see that. Thank you.

Are you confirming that the wheels lock at the bottom of the ramp?
Please state the question exactly as given to you.

 

[Lana]

Imagine in real life there was a coefficient of kinetic friction of 0.4 between the
plastic wheels of the cart and the wooden ramp. If there is only friction on the flat
part of the track, how far would the cart go before it stopped if it started from a
height of 4m? Show your calculations

 

[haruspex]

Imagine in real life there was a coefficient of kinetic friction of 0.4 between the
plastic wheels of the cart and the wooden ramp. If there is only friction on the flat
part of the track, how far would the cart go before it stopped if it started from a
height of 4m? Show your calculations

Hmm… nothing there about the wheels becoming stuck. And nothing about axle friction or rolling resistance.

Taking the description literally, and assuming no rolling resistance, here is what would happen:
Since there is no friction on the slope, the wheels will slide without rotating there, so the cart will reach the horizontal at the speed you calculated.
On reaching the horizontal, the friction will cause the wheels to start to rotate, reducing the speed a little; but with no data on the wheels we cannot calculate that.
Thereafter, with no rolling resistance, the speed will remain constant until the cart climbs the far side. The wheels continue to rotate at the same speed although the cart now slows.
Having lost some KE to get the wheels rotating (which will include some lost to friction before rolling rate was achieved) the cart will not reach 4m before sliding back down.
Back on the flat, there will be more KE lost to friction while the direction of the wheel rotation gets reversed.
This process will continue, but without more detailed calculations I am unsure whether the total distance traversed is finite.

Clearly this is not what the problem setter had in mind. What was in that mind I cannot guess.

 

[PeroK]

Clearly this is not what the problem setter had in mind. What was in that mind I cannot guess.

I assume the problem setter, as is often the case, believes that kinetic friction applies to a rolling wheel.

 

[haruspex]

I assume the problem setter, as is often the case, believes that kinetic friction applies to a rolling wheel.

But 0.4?!

 

[jbriggs444]

But 0.4?!

The sticky wheel gets the grease.

 

[zan123_]

the wheels dont get stuck, there is not clarification for that in the question

 

[zan123_]

is the answer above correct? and is there any way someone can clarify why/how Work done is 1/2mv^2. It is not clear for me thank you.

 

[jbriggs444]

is the answer above correct? and is there any way someone can clarify why/how Work done is 1/2mv^2. It is not clear for me thank you.

You are talking about the answer that you was provided in your the original post?

The work done on the cart by the track is equal to the energy gained by the cart. That is the work energy theorem.

In this case, the cart loses the energy. Its initial energy, when it hits the level portion of the track, is given by $\frac{1}{2}mv^2$. Its final energy, when it comes to a stop, is zero. So the energy "gained" is $-\frac{1}{2}mv^2$. The work "done" must be equal to this.

Of course, the cart is losing energy so both the change in energy and the work done are negative.

One problem with this is that $v$ is not given in the problem. It could be calculated based on solving $\frac{1}{2}mv^2 = KE_\text{bottom} = PE_\text{top} = mgh$ for v in terms of h.

But why bother?!

You want kinetic energy. You have $g$ and $h$ along with the irrelevant $m$. Why bother calculating $v$ if you are only going to use it to calculate the energy that you already know.[/S]

 

[PeroK]

You are talking about the answer that you provided in your original post?

That's not the OP!

 

[zan123_]

You are talking about the answer that you provided in your original post?

The work done on the cart by the track is equal to the energy gained by the cart. That is the work energy theorem.

In this case, the cart loses the energy. Its initial energy, when it hits the level portion of the track, is given by $\frac{1}{2}mv^2$. Its final energy, when it comes to a stop, is zero. So the energy "gained" is $-\frac{1}{2}mv^2$. The work "done" must be equal to this.

Of course, the cart is losing energy so both the change in energy and the work done are negative.

One problem with this is that $v$ is not given in the problem. It could be calculated based on solving $\frac{1}{2}mv^2 = KE_\text{bottom} = PE_\text{top} = mgh$ for v in terms of h.

But why bother?!

You want kinetic energy. You have $g$ and $h$ along with the irrelevant $m$. Why bother calculating $v$ if you are only going to use it to calculate the energy that you already know.

thank you so much for your help! I was talking about the answer in the original post

 

[kuruman]

thank you so much for your help! I was talking about the answer in the original post

The answer of 10 m in the original post would be correct if a brake were applied to lock the wheels and prevent them from turning throughout the motion. Under this assumption, the wheels slide without friction while the cart is on the incline and with friction once they reach the flat part. The statement of the problem in post #9 mentions wheels but why bother if they are not allowed to roll?