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Deceleration on a slope

  • Thread starter geejodi
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  • #1
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Homework Statement


A car traveling at 22.0 m/s runs out of gas while traveling up a 25.0 degree slope. How far up the hill will it coast before starting to roll back down?


Homework Equations


Vf^2 = V0^2 + 2ax
x = V0t + (1/2)at^2 (maybe?)


The Attempt at a Solution


I am using an online homework system called Mastering Physics. I did this problem and got an answer, which was marked wrong. Here's what I did:

Used Vf^2 = V0^2 + 2ax
where x = vertical position up the slope.
Vf = 0
V0 = 22sin25
a = -9.8

I solved for x, and got x=4.41 m. This is the vertical distance, or one leg of the right triangle across from 25 degrees. I used the law of sines to find the hypotenuse, and got 10.43 meters for the hypotenuse, which would be the distance that the car moved before it stopped.

This was marked wrong. Please explain why my thinking is wrong, and the right way to do the problem!! Thank you!!

~G
 

Answers and Replies

  • #2
Doc Al
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Used Vf^2 = V0^2 + 2ax
where x = vertical position up the slope.
Vf = 0
V0 = 22sin25
a = -9.8
The acceleration is not 9.8 m/s^2; that would be true for a body in freefall, not something rolling down an incline.

Measure the position, speed, and acceleration parallel to the incline. Then that kinematic formula will make sense.
 
  • #3
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Lets forget horizontal velocity altogether for a minute. How fast is the car moving against gravity:

that would be 22*sin(25)=9.30m/s. From here a few branch points as to soln.

using the eqn you posted: Vf^2=Vo^2+2ax
I get 9.3^2/(2*9.8)="x"=4.41m

So I agree with the answer, the question is what is asked, by up the hill? then it would be tan 25=4.41/x =9.45m
 
  • #4
ranger
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It seems like you are using the wrong acceleration (a = -9.8). Is the car in free fall?

EDIT: Doc Al beat me to it.
 
  • #5
Doc Al
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Lets forget horizontal velocity altogether for a minute. How fast is the car moving against gravity:

that would be 22*sin(25)=9.30m/s. From here a few branch points as to soln.

using the eqn you posted: Vf^2=Vo^2+2ax
I get 9.3^2/(2*9.8)="x"=4.41m

So I agree with the answer, the question is what is asked, by up the hill? then it would be tan 25=4.41/x =9.45m
You are assuming, like geejodi did, that the acceleration is 9.8 m/s^2 down. Not so.
 
  • #6
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Ok... but how do I find the acceleration if I don't know the distance it traveled?
 
  • #7
Doc Al
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Ok... but how do I find the acceleration if I don't know the distance it traveled?
By analyzing the forces acting on the car parallel to the incline, and using Newton's 2nd law.
 
  • #8
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I disagree, there are two ways to approach the problem, rotate the coordinate system or break it down in to a pure Vo(y) which is subject to -9.8m/s^2 by resolving the vecocity into usual x,y vectors.

And then solving for the vert displacement, go back and get the up the hill displacement. I considered both, this seemed easier to explain.
 
  • #9
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I disagree, there are two ways to approach the problem, rotate the coordinate system or break it down in to a pure Vo(y) which is subject to -9.8m/s^2 by resolving the vecocity into usual x,y vectors.

And then solving for the vert displacement, go back and get the up the hill displacement. I considered both, this seemed easier to explain.
That is what I did, and it was marked wrong.
 
  • #10
Doc Al
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I disagree, there are two ways to approach the problem, rotate the coordinate system or break it down in to a pure Vo(y) which is subject to -9.8m/s^2 by resolving the vecocity into usual x,y vectors.
You are welcome to rotate the coordinate system, but that won't help. The acceleration of the car is not 9.8 m/s^2 down. (The incline exerts a force on the car that modifies the acceleration.)
 
  • #11
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By analyzing the forces acting on the car parallel to the incline, and using Newton's 2nd law.
Mass is not given, and we haven't even done force yet. I know about force caues I did it last year but I really don't think this problem should involve force.
 
  • #12
Doc Al
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Mass is not given, and we haven't even done force yet. I know about force caues I did it last year but I really don't think this problem should involve force.
Perhaps you've covered the fact that the acceleration down an incline (in the absence of friction) is the component of the freefall acceleration along the incline.
 
  • #13
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Perhaps you've covered the fact that the acceleration down an incline (in the absence of friction) is the component of the freefall acceleration along the incline.
Please oh please, I appreciate your help, but I really wish you would stop being so cryptic. I breezed through my other physics problems and this one has had me going for an hour. I am exhausted and want to sleep. Please explain the concept behind what you are trying to say, clearly? You don't have to give me the answer, just a concrete way to do it.... please...
 
  • #14
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Ok, no thanks to you, I got it. I replaced -9.8 with -9.8sin25. Next time don't be so cryptic!!! :(
 
  • #15
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In the absence of friction, I say figure out high high could it fly? Answer is simple, upward initial vel less time it takes for gravity to stop it. Coud be 1 degree or 90. same answer. I just don't get this normal force entering into the eqn? Are you suggesting that it has no "weight" while traveling up the incline since the Earth is pushing back? In these cases, I like to flip to limits, purely upwards, no normal force, purely horizontal, mg. Lost so lets gets this man in the proper direction, and me too, if we can.
 
  • #16
Doc Al
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In the absence of friction, I say figure out high high could it fly? Answer is simple, upward initial vel less time it takes for gravity to stop it. Coud be 1 degree or 90. same answer.
You could just look at the vertical motion, if you like. Take the vertical component of initial velocity combined with the vertical component of the object's acceleration. The acceleration of the object is g sin(theta) down the incline. The acceleration is not g! (It is not in free fall.) The vertical component of acceleration is therefore g [sin(theta)]^2. Use that and you will get the right answer. (To me, it's more complicated.)

I just don't get this normal force entering into the eqn? Are you suggesting that it has no "weight" while traveling up the incline since the Earth is pushing back?
The acceleration of an object is due to the net force on it. In this case two forces act: the normal force and gravity.

Note to geejodi: I thought I was being clear when I said "the acceleration down an incline (in the absence of friction) is the component of the freefall acceleration along the incline". I want you to know where the sin(theta) comes from.
 
  • #17
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Thanks, I see your point. What I was really advocating now that I think about my own logic more carefully was setting it up as a potential energy question, i.e separating out the V(y-init) and equating it to the height H at which it would be converted to potential energy.
 

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