AP Physics Test: Energy Problem

In summary, the conversation discusses a physics test question about minimizing work against gravity and friction when moving crates up a ramp with constant velocity. The correct answer is found to be tan x = 1/µ, where x is the angle of the ramp and µ is the coefficient of kinetic friction. The reasoning behind this answer is explained using calculus. However, there is a debate about whether the answer should actually be tan x = 90°, as this would result in the shortest ramp length and minimum normal force. Eventually, it is determined that the answer of x = 90° is correct and the conversation ends with a discussion about technical difficulties with the LaTex generator.
  • #1
mooshasta
31
0
Recently my AP Physics class had a test on energy. The following is one of the questions verbatim from the test. I just can't work out the problem to any of the choices, let alone the "correct" choice. The problem is:

Movers want to set the ramp of their truck so that the work they do against gravity and friction is a minimum for crates moving up the ramp with constant velocity. µ is the coefficient of kinetic friction and x is the angle between the ramp and the ground. For the work to be a minimum, they must choose:

a. tan x = µ
b. tan x = -µ
c. tan x = -1/µ
d. tan x = 1/µ
e. tan x = 1 - µ


My reasoning is as follows:
Since the truck's height is unchanging, the work done against gravity should be constant regardless of the angle of the ramp; work agianst friction is the only variable work. To minimize the work against friction, we should try to minimize the normal force and the actual length of the ramp. The shortest ramp length AND the minimum normal force both occur at x = 90° (this isn't much of a ramp, but oh well). Wouldn't this be the correct answer? The "correct" answer is tan x = 1/µ, but I just don't see how this is possible.

Questions on these tests are often wrong, which is unfortunate.


Thanks in advance,
Andrew
 
Physics news on Phys.org
  • #2
The answer is [itex]tan\theta=\mu[/itex]
 
  • #3
Mmm... I think...

Hi Ho! :smile:

The correct answer is [tex]\tan(x)=\frac{1}{\mu}[/tex].

work agianst friction is the only variable work
You're correct.

To minimize the work against friction, we should try to minimize the normal force and the actual length of the ramp
Those are synonymous with varying the ramp's angle.

The shortest ramp length AND the minimum normal force both occur at x = 90°
Well, don't go straight to that point. Using calculus to guide us to a good common sense will prove that your reasoning above is not correct.

Please look at the attached picture.
Because the velocity of the crate is constant, there is no acceleration.
Thus,
[tex]Fx=-Fpush+Fg(\sin(x))+Ffriction=0[/tex]
[tex]Fy=N-Fg(\cos(x))=0[/tex]
Next, doing the [tex]Fy[/tex], [tex]N=Fg(\cos(x))[/tex]
Therefore, because [tex]Ffriction=\mu N[/tex], substituting it to the [tex]Fx[/tex] equation will result as, [tex]Fg(\sin(x))+\mu N=Fpush[/tex]
Remember that we have already had the value of [tex]N[/tex] as the result of doing the [tex]Fy[/tex].

Now we know that [tex]Fpush[/tex] depends on x as shown below.
[tex]Fpush(x)=Fg(\sin(x))+\mu Fg(\cos(x))[/tex]

Next, we use calculus to find its minimum value as follows.
[tex]\frac{d(Fpush)}{dx}=Fg(\frac{d(\sin(x))}{dx}+\mu \frac{d(\cos(x))}{dx})=0[/tex]
Then, [tex]\frac{d(Fpush)}{dx}=Fg(\cos(x)-\mu \sin(x))=0[/tex]
[tex]\cos(x)-\mu \sin(x)=0[/tex]
[tex]\cos(x)=\mu \sin(x)[/tex]
[tex]\tan(x)=\frac{1}{\mu}[/tex]

Hopefully, this solves your problem.

Your welcome :biggrin:
 

Attachments

  • picture.PNG
    picture.PNG
    1.7 KB · Views: 403
Last edited:
  • #4
I'm not sure I get it. :(

You've minimized the Fpush, which I guess minimizes the work.. but I can't see how the work done against the frictional force is a minimum here. Maybe if you could explain to me why 90° is incorrect, I would understand better.

Thank you
 
Last edited:
  • #5
Mentor, we need help!

Mmm... OK! :smile: Let's assume that [tex]\mu = \frac{1}{2}[/tex]. So,
[tex]Fpush( \arctan ( \frac{1}{ \frac{1}{2}}))=Fg( \sin ( \arctan ( \frac{1}{ \frac{1}{2}}))+( \frac{1}{2}) \cos ( \arctan ( \frac{1}{ \frac{1}{2}})))[/tex]
[tex]Fpush( \frac{1}{2})=1.12 Fg[/tex]

If we use your argument that x should be 90°,
[tex]Fpush(90°)=Fg( \sin (90°)+( \frac{1}{2}) \cos (90°)[/tex]
[tex]Fpush(90°)=Fg[/tex]

It turns out that when using a ramp, at a certain angle that is [tex]\tan (x) = \frac{1}{\mu}[/tex], the [tex]Fpush[/tex] will reach a maximum point.

Well, I found this out when I try all of the possible value of x using OpenOffice.org Calc.There, the minimum value of [tex]Fpush[/tex] happens when the [tex]x=0[/tex] (using [tex]\mu < 1[/tex] and the maximum value happens when [tex]x= \arctan ( \frac{1}{\mu})[/tex]

Well, we need a help from a mentor, guys!
Mentor, we need help, please! :biggrin:
 
Last edited:
  • #6
Hi Ho!

I've just discussed this problem with an assistant of school's physics laboratory and it turns out that your answer on choosing x=90° is correct. In fact, when you use a ramp, you'll get an angle that maximizes your Fpush.

Thank you.

PS: What's wrong with the LaTex generator? It can't generate the LaTex codes in my previous post correctly. I wonder.
 
  • #7
Thank you :) I would have replied to your other post but I was actually waiting for the latex to be generated this whole time, heh.




Thank you again :)
 

1. What is the AP Physics Test: Energy Problem?

The AP Physics Test: Energy Problem is a section of the AP Physics Exam that focuses on assessing students' understanding of energy concepts and their ability to apply them to problem-solving situations.

2. What topics are covered in the AP Physics Test: Energy Problem?

The AP Physics Test: Energy Problem covers topics such as work, kinetic energy, potential energy, conservation of energy, and power.

3. What are some common strategies for solving AP Physics Test: Energy Problems?

Some common strategies for solving AP Physics Test: Energy Problems include drawing diagrams, identifying known and unknown quantities, using energy equations, and checking for conservation of energy.

4. Are calculators allowed on the AP Physics Test: Energy Problem?

Yes, students are allowed to use calculators on the AP Physics Test: Energy Problem. However, they should be familiar with the calculator they will be using and know how to input equations and perform calculations efficiently.

5. How can I prepare for the AP Physics Test: Energy Problem?

To prepare for the AP Physics Test: Energy Problem, students should review key energy concepts, practice solving energy problems from past exams and practice tests, and seek help from teachers and tutors if needed. Additionally, practicing time management and test-taking strategies can also be beneficial.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
863
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
2
Replies
56
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top