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Homework Help: Work and energy - 90% done, just need little clarification

  1. Oct 2, 2008 #1
    A 2.0kg box slides down a long, frictionless incline of angle 30 degrees. It starts from rest at time t = 0 at the top of the incline at height of 20m above ground.

    A) What is the potential Energy of the box relative to the ground at t = 0?
    U = MGH where y0 is at the bottom of the incline
    U = .39 KJ

    B)Use Newton's 2nd law to find the distance the box travels between 0 and 1s and its speed at t = 1s.
    F = MA
    A = 4.9 m/s^2
    D = 2.5m after 1s.

    C)Find the potential and kinetic energy of the box at t = 1s.
    KE = 1/2MV^2 = 24J

    Here is one I don't get since it traveled 2.5m in 1s shouldn't it be
    MGH where H = 20m - 2.5? I get U = .34KJ but the answer says .37KJ
    what is wrong?

    D)Find the kinetic energy and the speed of the box just as it reaches the ground at the bottom of the incline.

    I was thinking of using work theorem where total work done = change in kinetic energy

    The only force acting on the box is normal force and force of gravity,
    since normal force is perpendicular to the displacement = cos 90 = 0 work
    and force of gravity would be (m)(g)cos-120 * displacement
    What would be the displacement? 20m? from the height?

    Any help is appreciated.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 2, 2008 #2
    For question (B), you need to know the component of g, because its on the incline, so [tex]a=g sin\Theta[/tex].
  4. Oct 2, 2008 #3
    I already did B, I need help partially on C and D.
  5. Oct 2, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    2.5m is the distance traveled along the incline. What's the change in height?

    For part D: Using the total height and the angle, figure out the total distance along the incline.

    (Consider using just plain old conservation of energy. Better yet, use both methods and compare!)
  6. Oct 2, 2008 #5
    Oh thank you for pointing that minor detail out. WOW I was so delusional I completely forgot trigonometry.

    The change in height in part c is
    2.5*sin30 = h
    h = 1.25m
    U = MG(H-h) = .37KJ

    Part D:
    Sin 30 = 20 / D
    D = 40m

    Net Work= M*G*cos-120*40 = 392 = .39KJ
    sqrt = 19.8 rounded = 20 m/s

    Oh wow, thank you again Doc. =]
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