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Working out Energy lost to friction after the car has travelled a certain distance

  • Thread starter physiman
  • Start date
  • #1
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First, a big hello to the Physics Forums community :)

Ok, now to business:

A car with mass 600 kg accelerates uniformly from rest down a steady hill inclined at 15 degrees to the horizontal[...]
- Work out the energy lost to friction after the car has travelled 100m
- The average frictional force acting on the car as it free-wheeled down the hill




For the first 3 parts I had to work out the speed of the car after 100m in 20s, the KE after the car has travelled 100m and the GPE the car has lost after tavelling 100m down that slope. I got 10 ms^-1; 30 000 J and 152340.9 J respectively.



So I tried using the Work done formula to out the energy lost due to friction:

W=Fd

For W I used the GPE the car has lost in PE, so 152340.9 J and got the value of F to be 1523.4 N. Is that the right method, I'm unsure because I think that I have to include the Kinetic Energy as well but the difference between KE and GPE looks too big for me.

Or do I have to use the F=ma formula? By that I mean work out a from the values I have, 0.5 m s^-2, and then I get 300N as the value for F but that doesn't look right to me as it has nothing to do with the energy loss.
well, I guess I can got to the next question only after I solve this one.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
993
13


Was the time given in the problem?
 
  • #3
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Yes, sorry forgot to mention that.
The time given was 20s.
 
  • #4
993
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Is the engine off?
 
  • #5
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That wasn't mentioned, it says it 'free-wheeled' down the hill.

Would that mean the engine was off?
 
  • #6
993
13


I think so.

BTW try to check your working for the GPE.
Are you taking 9.8 for g?
 
  • #7
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9.81 to be precise ;)
 
  • #8
993
13


What is your working for the GPE?
 
  • #9
993
13


Sorry - Your GPE is ok .
i was using 9.8 instead of 9.81/
 
  • #10
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I used mgh.

But I worked out the height from the ground, because GPE depends on it.

So I just used 100sin15 and got about 25.9 m for it.
 
  • #11
993
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I got F = 1223.41N
 
  • #12
13
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Did you use W=Fd?

Because my answer (1523.4 N) doesn't look far away from your one.
 
  • #13
993
13


Yes I used W = Fd.

Note that when you got 300 for the force, that was the RESULTANT or NET force.
You CAN use Fnet = ma and you still get 1523N for the frictional force.
 
  • #14
13
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How can I get to the Friction force from working out the resultant force?

Oh yh, and about the energy lost to friction bit:

Do I just work out the Net energy from before motion and after the car has travelled 100m?

i.e.:

Initial KE: 0 J
Initial GPE: mgh = 152340.8899...J

Total Energy before moving: 152340.9 J

After travelling 100m:

KE = 1/2mv^2 = 30000 J
GPE = mgh = 0 J (assuming that it has reached the horizontal)

Therefore making the energy lost to friction:

152340.8899... - 30000 = 122340.8899... J
 
  • #15
993
13


E lost to friction is ok
re friction from net force:
How many forces act along the track?
 
  • #16
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Not mentioned on the Question but the only forces I can think of is Weight of the car (perpendicular to the track, therefore 600(9.81)cos15 ), the frictional force and there should be a Normal reaction too.
 
  • #17
993
13


I mean forces ALONG THE TRACK. 600(9.81)cos15 ) is the component of the weight perpendicualr to the track and the normal reaction is also perpendicular to track.
Along the track there is the friction and the ...
 
  • #18
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oh alright, sorry.

tbh apart from the friction I can't think of any force, things like air resistance should be negligible
 
  • #19
993
13


Correct. Air resistance is being ignored but there is a component of the weight downward along the track opposing the friction upwards along the track. Right?
 
  • #20
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Ok, that must be the Horizontal force then, so 600(9.81)sin15 instead of using cosine?
 
  • #21
993
13


exactly!
Now you can find the friction form the NET force along the track.
 
  • #22
13
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Oh cool, I did get ~1523.41 N and from the 1223.41 worked out for friction it would support F=ma.

Thanks a lot, you were a real help. If there was a reputation system I would show my appreciation :)
 
  • #23
993
13


You are always welcome to this forum.
 

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