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Speed problem

  1. Dec 17, 2006 #1
    1. The problem statement, all variables and given/known data
    I have to find the speed needed to push a cart on top of a hill and I only have the height as a number.


    2. Relevant equations
    A boy places his matchbox car at the top of the track he has set up with his brother. If the track starts atop the 55 cm toy box, descends to the floor and then up to the 75cm high bed where his brother waits for the car. How fast must the boy push his cart so it just makes it to the brother on the bed ?


    3. The attempt at a solution

    PE = m(10m/s).55m = 55JM ?:rolleyes:
     
  2. jcsd
  3. Dec 17, 2006 #2

    OlderDan

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    You have nothing about speed in your attempted solution. Where do you think it applies?
     
  4. Dec 17, 2006 #3
    I have an awnser sheet and it says 2m/s... I'm supposed to find what force pushed the cart above the second hill which I assume is kinetic energy but I have no mass wich I don't understand...
     
  5. Dec 17, 2006 #4

    OlderDan

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    Kinetic energy is involved in the problem. The initial speed gives the car initial kinetic energy. During the motion some energy is transformed fom potential to kinetic and then the kinetic is transformed to potential.

    The mass is a factor in both the kinetic and potential energies. Its value does not matter. Do the problem with the mass represented by M. In the end you will find the initial speed required is the same for any mass of the car.
     
    Last edited: Dec 17, 2006
  6. Dec 17, 2006 #5

    ShawnD

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    One trick of simple math problems is to immediately start filling things in and use arbitrary numbers if needed, as long as that variable is a part of every term in the equation. This method is particularly useful for ratios between things, where factoring can get complicated. Making up numbers only works if you can look at the equation and think "this can be factored, but i don't feel like doing that."

    For the problem given it's as simple as finding the energy difference between where the car starts and where it finishes, which is a difference of 20cm.
    (m)(g)(h) = (1/2)(m)(v^2)
    m can be factored out, g is gravity (9.something), h is 20cm (convert it to meters)
    Solve for v.
     
  7. Dec 17, 2006 #6
    Thanks a lot guys!
     
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