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Homework Help: Kinetic Energy/work/power

  1. Feb 18, 2004 #1
    I cant seem to get the right answer to these questions


    A 1.8 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F(x)=(2.4-x^2)i N, where x is in meters and the initial position of the block is x=0

    (a) what is the kinetic energy of the block as it basses through x=2.0m?


    This is how i tried to solve it, since Vi(initial)=0 and Vf(final)= unknown, we can use the formula Vf^2=Vi+2a(X2-X1). But..Force = (2.4-x^2) and at x=2 it is (2.4-4)=-1.6....and f=ma so -1.6/1.8 will equal the acceleration which is -.89...but how the f is the acceleration negative when a horizontal force is applied to the box on a frictionless surface? i dont know where i went wrong with my calculatoins...

    Also here is another question that i am having trouble with...


    Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50m/s. At a certain location the conveyor belt moves for 2.0 m u an incline that makes an angle of 12degrees with the horizontal. Assume that the 2.0kg box rides on the belt without slipping

    (a) At what rate is the force of the conveyor belt doing work on the box as the box moves up the 12degree incline.


    this is how i tried to solve it...

    since m=2kg, a=0, theta= 12degree, and distance = 2m, and Vc(velocity at constant)=0.50...we have all the constants we need

    So first Work=mgdcos(theta) = (2)(9.8)(2)(cos12)=38.343

    and X2-X1 (change in distance)=(Vc)(time) sooo....2=38.343t and t = 4 seconds...

    and Power = Work/change in time ...sooo 38.343/4 = 9.586 which is not the answer? help what am i doing wrong guys! thanks
  2. jcsd
  3. Feb 18, 2004 #2
    You didn't go wrong anywhere, draw a graph of the force function. It is a negative parabola. Since F=(2.4-x^2), at x= sqrt(2.4) = 1.55m, the force will start pushing back in the negative direction.

    [tex]W = \int_{x_i}^{x_f} F(x) dx[/tex]
    [tex]W = \Delta KE[/tex]
    Last edited: Feb 18, 2004
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