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Homework Help: Energy prob

  1. Dec 13, 2004 #1
    A steel ball has a mass of 4 kg and rolls along a smooth level surface at 62 m/s.

    a) Find its kinetic energy ( i found this.. )

    b) At first, the ball was at rest on the surface. A constant force acted on it through a distance of 22 m to give it the speed of 62 m/s. What was the magnitude of the force?

    so heres my thought process:

    .5 mv^2 + mgh = .5mv^2 + mgh
    0 = .5mv^2+mgh
    -.5mv^2=mgh
    mg=- .5mv^2/h
    mg= - .5 x 4 x 62^2/22
    mg= -350 N


    is the above method correct?

    Q2) In the 1950's, an experimental train that had a mass of 2.50x10^4 was powered across a level track by a jet engine that produced a thrust of 5x10^5 N for adistance of 509 m.

    a) Find the work done on the train. ( found this to be 2.545E8)
    b) Find the change in kinetic energy. ( 2.545E8)
    c) Find the final kinetic energy of the train if it started from rest. ( how do i find this?? .5 mv^2 wont wrk since i dunno waht the velocity is )
    d) find the final speed of the train if there were no friction ( how do i find this?? )

    thanks much
     
    Last edited: Dec 13, 2004
  2. jcsd
  3. Dec 13, 2004 #2

    Pyrrhus

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    On your b) you should have used

    [tex] \sum_{i=1}^{n} W_{i} = \Delta K [/tex]
     
  4. Dec 13, 2004 #3
    so ur saying i would use W= delta K = fd?

    but i'd still get the same answer of 350 N...
     
  5. Dec 13, 2004 #4

    Pyrrhus

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    Yes, but i don't know where you got Potential Gravitational energy, the problem doesn't refers to a not leveled surface, plus there's inconsistency, because you got mg = -350N when that is weight, and the weight of the object is about 39.2 N.
     
  6. Dec 13, 2004 #5
    is the magnitude of force = 1.69 x 10^5 n??

    thanks
     
  7. Dec 13, 2004 #6
    c) that's easy - it just means that KEi = 0 .. when KEi = 0, KEf = Work!
    d) well then .5v^2 = work and then u can find v :tongue:
     
  8. Dec 14, 2004 #7
    so c= 2.545 x 10^8 right?
     
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