Kinetic Energy being larger than Mechanical Energy in a Hotwheels Lab

[MrSand]

Homework Statement

The lab setup is rather simple we have a vertical ramp which has a height of .33m, and our total distance is 1.33m. A meter for the ramp and a foot for the rest of the track below. We let the Hotwheels car go one round and recorded the time using a slow-mo video and a stopwatch. We have a time of .85s. A mass of .02575 kg converted from 25.75 grams and a velocity initial of 0. Whilst there is friction and air resistance we don't need to account for it. The angle of the ramp is 18 degrees. The main issue is that when I input my solved kinematic equations I keep getting a KE greater than our initial PE. This is very conflicting as we can't gain more than the allotted PE, there shouldn't be any extra energy but there is for some reason and I can't see it.

Relevant Equations

GPE: m x g x h KE: 1/2(m)*(v^2) Kinematic equations: Vf^2= Vo^2 + 2ax, Vf = Vi + at, x = Vi*t + 1/2a*t^2

- work

 

[Orodruin]

You cannot use $s = at^2/2$ as acceleration is not constant (acceleration is zero during the horizontal portion and you seem to use the time for ramp+horizontal).

A more direct approach to find the final v would be to measure the time taken for the horizontal piece of the track.

 

[haruspex]

h=0.33m? How did you compute that?
It is not clear to me what period the 0.85s represents.
From your use of it to calculate an acceleration, it seems to be the time to descend the ramp. Is that correct? Then you use that to find the velocity at the bottom of the ramp.
It would be a lot simpler to recognise that, with no losses, the velocity would be the same as if it fell straight down, $v^2=2as=2gh$.

 

[Orodruin]

From your use of it to calculate an acceleration, it seems to be the time to descend the ramp. Is that correct?

No. If you check the numbes used, they are using 1.33 m for the distance travelled. Based on the drawing of the track, this would seem to imply they have assumed acceleration constant for the entire ramp and horizontal part when computing the final v.

 

[Orodruin]

It would be a lot simpler to recognise that, with no losses, the velocity would be the same as if it fell straight down, v2=2as=2gh.

But this assumes conservation of mechanical energy, which is what they are trying to verify experimentally.

 

[Orodruin]

As to the angle, an angle of 18 degrees for a 1 m ramp gives a height of ca 31 cm. A 33 cm height for a 1 m ramp gives 19 degrees so it depends on which was actually measured in the experiment - height or angle. Although the difference is likely within the experimental error bars.

 

[haruspex]

But this assumes conservation of mechanical energy,

Not really. It is a matter of kinematics that for constant acceleration $v^2-u^2=2as$. Considering it only as downslope motion, it is just geometry that $h=s\sin(\theta)$ and $a=g\sin(\theta)$, whence $v_f^2=2gh$.

 

[Orodruin]

Not really. It is a matter of kinematics that for constant acceleration $v^2-u^2=2as$. Considering it only as downslope motion, it is just geometry that $h=s\sin(\theta)$ and $a=g\sin(\theta)$, whence $v_f^2=2gh$.

I mean, regardless of how you want to see it, the experiment is to verify that $v^2 = 2gh$. You cannot then just make the assumption that $v^2 = 2gh$.

 

[Orodruin]

$a=g\sin(\theta)$

which, by the way, also becomes equivalent to the conservation of mechanical energy in this situation. It is ultimately a statement of forces, but equates to conservation of energy through the work-energy theorem.

 

[haruspex]

the experiment is to verify that $v^2 = 2gh$.

I see nothing in post #1 or title that says this is an experiment to verify conservation of ME. I presume you base this on experience of such courses.
But what does it really mean? If you assume g is constant and that the velocity is constant on the flat then it can all be deduced from first principles. If you cannot use the constant acceleration kinematic equations then I see no way to analyse the data.

Seems to me it is an experiment to determine whether the student can conduct experiments and analyse the result. Which is fair enough.

 

[Orodruin]

I see nothing in post #1 or title that says this is an experiment to verify conservation of ME. I presume you base this on experience of such courses.
But what does it really mean? If you assume g is constant and that the velocity is constant on the flat then it can all be deduced from first principles. If you cannot use the constant acceleration kinematic equations then I see no way to analyse the data.

Seems to me it is an experiment to determine whether the student can conduct experiments and analyse the result. Which is fair enough.

So let me ask what you think is the point of the experiment. Computing mgh twice and concluding that you got the same result?

 

[haruspex]

So let me ask what you think is the point of the experiment. Computing mgh twice and concluding that you got the same result?

Maybe it verifies that speed is constant in the absence of a force in the direction of the velocity? Other than that, what I wrote in post #10.
As regards teaching physics, it would be more appropriate to ask what you would have to assume to deduce from kinematics that the lost GPE would equal the gained KE. E.g., g is constant, m is constant, and what I wrote just above about speed.

 

[Orodruin]

Maybe it verifies that speed is constant in the absence of a force in the direction of the velocity?

A more direct test of that would measure the velocity for different sections of the flat part. You don’t even need a ramp, you can give the car a velocity by hand.

If you cannot use the constant acceleration

That acceleration is constant does not mean it is equal to $g\sin\theta$.

It appears to me that what the OP is attempting is to compare kinetic energy to lost potential energy. This is precisely the work-energy theorem and if you use mgh to compute kinetic energy, well, then you have assumed that it holds.

OP should chime in and clarify what the actual experimental task is as described by the teacher. @MrSand

Edit: And regardless what the answer is, the basic mistake of the OP is assuming acceleration to be constant over the full 1.33 m track.

 

[MrSand]

A more direct test of that would measure the velocity for different sections of the flat part. You don’t even need a ramp, you can give the car a velocity by hand.


That acceleration is constant does not mean it is equal to $g\sin\theta$.

It appears to me that what the OP is attempting is to compare kinetic energy to lost potential energy. This is precisely the work-energy theorem and if you use mgh to compute kinetic energy, well, then you have assumed that it holds.

OP should chime in and clarify what the actual experimental task is as described by the teacher. @MrSand

Edit: And regardless what the answer is, the basic mistake of the OP is assuming acceleration to be constant over the full 1.33 m track.

We are supposed to account for energy loss and the lab is asking us to announce the factors that may have led to it. I spoke to my teacher and he kept it vague so I'm going with the assumption that we do need to account for friction

 

[MrSand]

this is the track

 

[Orodruin]

We are supposed to account for energy loss and the lab is asking us to announce the factors that may have led to it. I spoke to my teacher and he kept vague so I'm going with the assumption that we do need to account for friction

Just to be clear: So the task is to compare the loss in potential energy to the final kinetic energy?

You are “supposed” to find the kinetic energy to be lower and argue where energy may have gone?

Regardless, did you understand the points above regarding the errors in your approach?

 

[MrSand]

I'd say so.