Can a Hot Wheels Car Travel 42 cm with Initial Kinetic Energy Alone?

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Homework Help Overview

The discussion revolves around a physics problem involving a Hot Wheels car and an attached mass, focusing on the prediction of the car's travel distance based solely on its initial kinetic energy. The scenario includes a mass of 200g for the car and 150g for the block, with an initial velocity of 1.7 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relevance of specific equations related to kinetic energy and forces, questioning the use of certain variables and the implications of the setup. There is a focus on understanding the relationship between the forces acting on the system and the predicted distance.

Discussion Status

The conversation is ongoing, with participants providing hints and questioning the assumptions made by the original poster. Some guidance has been offered regarding the application of the work-energy theorem, but there is no consensus on the approach to take or the validity of the initial prediction.

Contextual Notes

There is uncertainty regarding the relevance of certain equations and the interpretation of variables such as theta in the context of a flat surface. The original poster's attachment is pending approval, which may limit the clarity of the discussion.

suekz
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Homework Statement



A Hot Wheels car, whose mass is 200g is hauling a mass block of 150g along a flat table surface
with a string as shown in the graph below. Assume the car and the attached mass is moving with
an initial velocity of 1.7 m/s. A student predicts that the car will travel a maximum distance of
42 cm. Is this a reasonable prediction? Explain fully.


Same

Homework Equations



KE = 1/2mV2
[tex]\Sigma[/tex]F = mgcos[tex]\theta[/tex]s

The Attempt at a Solution



I know that the formulas listed above are needed to solve this. After plugging in:

m = .350, v = 1.7, I get KE = .505 J
But where do I go from here to get distance? What would be theta be in this instance?

I have this formula KEf = KF0 + [tex]\Sigma[/tex]F = mgcos[tex]\theta[/tex]s, but I'm not sure what to use for theta?

I've attached the image to this post

Thanks for any hints!
 

Attachments

Last edited:
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We can't see your attachment yet, because it is pending approval. So it's a bit hard to help you with your problem.

Theta typically refers to an angle. There is no angle in this problem, because, as it says in the description, the table surface is flat.

What was the the information source that gave you the idea that you needed to use those formulae? Have you considered that perhaps they just may not be relevant to this problem? Also, what is the letter s in your sum of forces equation? mgcos(theta) already has units of force. So, if s is distance, then that equation makes no sense.
 
The s stands for distance in the equation. This equation was what we discussed in class, so i assume it is what we should use, since it is the only one I've seen with all the variables:

velocity, distance, and mass.

The picture is something like this, where the 200g car is hauling that 150g block off the side of the table:

| (pulley)--------rope------[200g car]
| ________table_________________
|
r
o
p
e
|
|
|
[150g block]
 

Attachments

  • prelab.JPG
    prelab.JPG
    2.8 KB · Views: 433
Welcome to PF.

The block then - all m*g of it - is retarding the forward motion?

Initially the car and the block are moving, as a system at 1.7 m/s positive x?

That would seem to suggest that the retarding deceleration of the system is 3/7*g?
 
First of all can you see that

[tex]\sum{F} = mg\cos \theta s[/tex]

makes no sense?

By a dimensional analysis, your left hand side and your right hand side are not consistent. You have

[force] = [force] * [distance}

This makes no sense. The left hand side should be the work done i.e. [itex]W = (\sum{F})s[/itex]

In words, you are trying to invoke the work-energy theorem, which states that

Work done = net force*distance = change in kinetic energy = difference between final and inital KE

or in symbols:

[tex]W = (\sum{F})s = \Delta KE = KE_f - KE_i[/tex]

Second of all, can you see that the equations you have written are not necessarily relevant to the problem (esp. the factor of cosine theta, which has nothing to do with anything here!).

suekz said:
This equation was what we discussed in class, so i assume it is what we should use, since it is the only one I've seen with all the variables:

Well, you would be much better off approaching problems using your understanding of physics, rather than using blind matching of formulae to variables! The whole point of doing these labs is to apply your knowledge of physics, rather than to go through a mechanical exercise in applying formulae that is not accompanied by any conceptual understanding of what is going on.

You would be well-advised to look at LowlyPion's hint, which can be bit cryptic, unless if you realize that the word "that" in his third sentence refers specifically to the student's estimate of a distance of 42 cm.
 

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