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Mechanics net tractive force question

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Ascending a 1 in 10 hill, a 1500 kg car slows down from 100 km/h to 50 km/h in a distance of 400 m. Calculate the net tractive force F between the tyres and the road assuming F remains constant and wind resistance can be neglected. Is the driver braking during this period?

    I am told the answer is 379N but have no idea how to go about it
    any help would be greatly appreciated
     
    Last edited by a moderator: Apr 2, 2013
  2. jcsd
  3. Apr 2, 2013 #2

    rude man

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    Compare kinetic energy at the start and after the 400m with gain in potential energy. Then look at the energy lost by the friction.
     
  4. Apr 2, 2013 #3
    would i convert the velocities into m/s?
    Do I find the differences between the KE and compare that with the PE?

    Would you be able to demonstrate it?
     
  5. Apr 2, 2013 #4

    rude man

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    k.e. after = k.e. before - gain in p.e. - friction energy loss.
    k.e. = 1/2 m v2 in general
    friction energy loss = friction force x distance covered

    You are looking for the friction force.

    Always use SI units or you'll get the wrong answer:
    meter, kilogram, Newton, meters/second, g = 9.81 m/sec2 etc.

    Don't know what you mean by "demonstrate it".
     
  6. Apr 2, 2013 #5
    but how to you calculate gain in PE? PE1 - PE2?
    I know PE = mgh but what would the value for h be?

    I meant worked example by demonstrate it :)
     
  7. Apr 2, 2013 #6

    berkeman

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    We do NOT do your schoolwork for you here. Re-read the PF Rules if you have any doubt about that (under Site Info at the top of the page).

    Your problem refers to climbing 400m up a 1-in-10 hill. That is all the info you need to calculate the rise in height to give you the change in PE. Please show some effort in this thread, or it will be deleted.
     
  8. Apr 2, 2013 #7
    ok sorry i didnot understand it first - i think i was confused by a 1 in 10 hill - what is it?
     
  9. Apr 2, 2013 #8

    rude man

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    1m up for every 10m horizontal
     
  10. Apr 3, 2013 #9

    Would the value of h be 400m?
    from my working i can't seem to get the answer
    I converted 100km/h to 27.8m/s and 50km/h to 13.9m/s
    for KE before i get: 1/2 * m * V before^2 = 1/2 * 1500 * (27.8)^2 = 579630
    for KE after i get: 1/2 * m * V after^2 = 1/2 * 1500 * (13.9)^2 = 144907.5
    for the PE(gain) i get PE=mgh = 1500*9.81*400 = 5886000

    using the formula quoted, k.e. after = k.e. before - gain in p.e. - friction energy loss.
    i re-arranged to get friction energy loss = -k.e. after +k.e. before - gain in p.e.
    from this i get:friction energy loss = -144907.5 + 579630 - 5886000
    = -5451277.5

    working backwards, using the formula quoted,: friction energy loss = friction force x distance covered, i calculated that the value i need for friction energy loss = 379*400 = 151600.

    the only thing i can think is that the height isn't the same as the distance 400m :S
     
  11. Apr 3, 2013 #10

    rude man

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    pe = mgh where h = height. Look at h again. Your last sentence above is a good hint! Surely you know the difference between distance and height ...

    Also, the problem asks for the tractive (friction) force, not the friction energy loss.
     
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