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Energy, Work, and Power

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  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    "A 1480 kg car accelerates uniformly from a position of rest to a speed of 95km/h in 6 s?"

    1) calculate the cars acceleration
    2) determine how far the car traveled during this acceleration
    3) determine how much work was done by the car during the 6s
    4) identify what power was developed during this process


    2. Relevant equations
    v2 = v1 + at
    d = v1t + 1/2at²
    d = v2 - 1/2at²
    work = force / distance or w = f / d
    power = work / time or p = w / t
    f = m * g
    3. The attempt at a solution
    1) I rearranged v2 = v1 + at to a = v2 - v1 / t to find power
    so
    95km/h - 0km/h / 6s = 95km/h / 6s = 15.83m/s²
    Therefore the acceleration of the car should be 15.83m/s²

    2) This is where im stuck.
    I would assume I need to use the equation d = v1t + 1/2at²
    so
    d = 0 * 6 + 1/2 * 15.83 * (6)²
    from this i get 284.94m


    I usually check online for the answers after im done my work but none of the online sources im finding match my answers for question 2. The answer im finding online is 79.167. Would anybody be able to tell me if both of the answers for question 1 and 2 are correct?
    Any help is appreciated!
     
  2. jcsd
  3. Oct 7, 2015 #2

    phyzguy

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    You need to keep better track of your units. km/hr divided by s doesn't give m/s^2. Try converting the 95 km/hr into m/s.
     
  4. Oct 7, 2015 #3
    Ahh I completely missed that. so 95km/h becomes 26.39m/s

    1) so now 26.39m/s / 6s = 4.4 so acceleration is now 4.4m/s^2

    2) d = 1/2 * 4.4 * (6)^2
    d = 79.2m
     
  5. Oct 7, 2015 #4
    To determine the amount of work done by the car, how do I know which formula to use to calculate the force? There is f = m * g and then f = m * a
     
  6. Oct 8, 2015 #5

    haruspex

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    The question ought to state this, but I think you can assume it's a horizontal motion.
     
  7. Oct 8, 2015 #6
    Thank you for the reply. I ended up calculating both then checking which ones were correct online using various work, power, energy calculators to see if they matched the units of emasurement originally given by the equation. Turn outs f = m * a was correct!
     
  8. Oct 8, 2015 #7

    haruspex

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    That's not going to help you solve such problems in general. You need to understand why ma is correct here. Did you understand my reply?
     
  9. Oct 8, 2015 #8
    Yes I understand. Since the car is moving horizontally and not vertically I can assume that the equation should be f = m * a. If the car were motionless then I would need to use f = m * g
     
  10. Oct 8, 2015 #9

    haruspex

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    No, if the car were motionless then no work would be done, so you would not use either. When would you use g in finding work done?
     
  11. Oct 8, 2015 #10
    Im assuming you would use g only if the car had vertical displacement then?
     
  12. Oct 8, 2015 #11

    haruspex

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    Yes.
     
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