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

In summary: As for that cosine theta, you can see that it is nowhere to be found in the problem. So why did you try to apply it? :) In summary, the problem involves a Hot Wheels car with a mass of 200g hauling a 150g mass block on a flat table surface with a string. The initial velocity of the car and attached mass is 1.7 m/s. The student predicts that the car will travel a maximum distance of 42cm. However, this prediction may not be reasonable and further analysis is needed. Some suggested equations, such as KE = 1/2mv^2 and ΣF = mgcosθ, may not be relevant to the problem and a better understanding of the physics
  • #1
suekz
2
0

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!
 

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  • #2
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.
 
  • #3
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]
 

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  • #4
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?
 
  • #5
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.
 

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

What is the Work and Energy Theorem?

The Work and Energy Theorem is a fundamental concept in physics that states that the work done on an object by a force is equal to the change in the object's kinetic energy.

How is the Work and Energy Theorem derived?

The Work and Energy Theorem is derived from the principles of conservation of energy and Newton's laws of motion. By analyzing the forces acting on an object and the resulting change in its energy, the theorem can be mathematically proven.

What is the significance of the Work and Energy Theorem?

The Work and Energy Theorem is important because it allows us to understand and predict the motion of objects in various situations. It also helps us to quantify the relationship between work and energy, which is crucial in many fields of science and engineering.

Can the Work and Energy Theorem be applied to all types of energy?

Yes, the Work and Energy Theorem can be applied to all forms of energy, including kinetic, potential, thermal, and even non-mechanical forms such as electromagnetic and nuclear energy. As long as there is a force acting on an object and causing a change in its energy, the theorem can be applied.

Are there any limitations to the Work and Energy Theorem?

The Work and Energy Theorem is based on ideal conditions and does not take into account factors such as friction, air resistance, and other forms of energy loss. In real-world situations, these factors may affect the accuracy of the theorem's predictions.

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