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Hwk. Problem: Work, Energy, Power

  1. Apr 4, 2005 #1
    A runaway truck with failed brakes is moving downgrade at 146 km/h just before the driver steers, the truck travels up a frictionless emergency escape ramp with an inclination of 15°. The truck's mass is 5000 kg.


    (a) What minimum length L must the ramp have if the truck is to stop (momentarily) along it? (Assume the truck is a particle, and justufy that assumption.)


    I know that the minimum length stays the same if the truck's mass is decreased. and the minimum length decreases if the truck's speed is decreased.

    I am clueless on what equation to use to find the minimum length L. Can someone please help me out.
     
  2. jcsd
  3. Apr 4, 2005 #2

    Doc Al

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    Staff: Mentor

    Think conservation of energy. As the truck moves up the ramp, its kinetic energy is transformed into gravitational potential energy. So... how high does the truck rise? Then use a bit of trig to find the length of the ramp.
     
  4. Apr 4, 2005 #3
    To find the height, I used the equation:

    Vf^2-Vi^2 divided by 2g. Is the vfinal=40.6m/s and vinitial=0?

    Am I using the right equation?
     
  5. Apr 4, 2005 #4
    You have the values of vfinal and vinitial mixed up, and you have them mixed up in the equation. Technically not correct, but the mistakes cancel out to give the correct height.
     
  6. Apr 4, 2005 #5
    The mass is not necessary here.You have two ways for the result,1) Because it stopes, the square of the initial speed must be equal with double of L*acceleration which is g or sin15. From here you can find L.2)The phenomen is in the gravitational field - conservative , so the total energy is the same (equal) at the begining and at final.If we consider on start is the level zero so the potential energy is 0,and the body has only Kinetic=msquarev:2.It is= with the final where because stoped ,has not kinetic but has potential Wp= mgh where h is l*sin15. So if you write on a paper this eqaution you will obtain the same result like 1).
     
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