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Car pulled by constant force

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A 900-kg car is pulled by a constant force of 1800N. Assume friction on the road surface is negligible.

    a) What is the acceleration of the car?
    b) If the car starts from rest, what is its speed after 10s?
    c) What is the average velocity of the car during the 10s?
    d) What is the distance traveled in the 10s?
    e) What is the work done by the pull force in the 10s?
    f) What is the kinetic energy of the car at the end of the 10s?
    g) If the call falls off a cliff of height 200cm at the end of the 10s, what is its kinetic energy right before it lands?

    2. Relevant equations

    vf = vo + at
    x = vot + 1/2at2
    vf2 = vo2 + 2ax
    F = ma

    3. The attempt at a solution

    a) F=ma
    1800N= (900kg)(a)
    a= 2 m/s2
    b) vf = vo + at
    vf = (2 m/s2)(10 s)
    vf = 20 m/s
    c) I am not sure what the differnce is between questions b and c
    d) x = vot + 1/2at2
    x = 0 + 1/2(2 m/s2)(10s)2
    x = 100 m
    e) 1800N times 100m = 18000 J but I am not sure of this one
    f) I need help with this one
    g) I need help with this one too =(

    I truly appreciate any help you can offer!
     
  2. jcsd
  3. Aug 25, 2009 #2

    tiny-tim

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    Hi starhallie! :smile:
    c) the velocity starts at 0 and finishes at 20, so the average is … ? :smile:
    f) KE (kinetic energy) is defined as (1/2)mv2
    g) use KE + PE = constant :wink:
     
  4. Aug 25, 2009 #3

    kuruman

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    (b) Asks you to find the instantaneous velocity at t = 10 s and you found it. (c) asks you to find the average velocity from t = 0 to t = 10 s. Because it is an average, it has to be a number between 0 m/s and 20 m/s.

    For (f) and (g) you need to write down more equations, one for kinetic energy and one for conservation of mechanical energy.
     
  5. Aug 25, 2009 #4
    c) 10 m/s
    f) KE= 1/2(900kg)(20 m/s)2
    KE= 100,000 kgm2/s2
    g) vcar= 200m/10s = 20 m/s
    PE= 1/2(900kg)(20 m/s)2
    PE= 100,000 kgm2/s2

    I hope that's at least a start? :shy:
     
  6. Aug 25, 2009 #5

    kuruman

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    Good, except for (g). What is the expression for potential energy? It can't be the same as kinetic energy.
     
  7. Aug 25, 2009 #6
    PE = mgh = (900kg)(9.8 m/s2)(200m)
    PE = 1,764,000 kgm2/s2

    I think? :uhh:
     
  8. Aug 26, 2009 #7

    ideasrule

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    That's right, if the cliff is 200 m high and not 200 cm (which would be one puny cliff).
     
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