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Kinetic Energy and Work and Power

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

    Question:

    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?

    Answer:

    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...

    Question:

    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.

    Answer:

    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

    Doc Al

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    Staff: Mentor

    Please don't double post! (You have the same question in the other Homework forum.)
    The force (and acceleration) are not constant. Find the work done by the force. (Integrate!)
    Where did the cos(theta) come from?
     
  4. Feb 18, 2004 #3
    okay after i integrated the work done by the force i got 2.13333j..now where do i go from here to find the kinetic energy?

    Also how can i found the max. kinetic engery of the block between x=0 and x=2m?


    and for the second question..isnt the formula w=mgdcos(theta)?\

    thanks
     
    Last edited: Feb 18, 2004
  5. Feb 18, 2004 #4

    Doc Al

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    Staff: Mentor

    The work done is the kinetic energy. That's the so-called "Work-Energy" theorem.
    You can find the KE as a function of x. Find the max of that function. (Hint: How did you find the KE above?)
    Sure, W = FDcosθ where θ is the angle between F and D. What angle did you use?
     
  6. Feb 18, 2004 #5
    ok thanks for the first part of the question..

    but i am still confused about the 2nd part of the quesotin..i still seem cant seem to get it right..this is how i tried to solve it


    this is how my triangle looks like....Hypoteneuse = 2m Height = .4158234m and length = 1.956295201m..and the angle between the length and hypotenuse to be 12 degreee and the box lays on the hypoteneuse while moving up at a constant velocity of 5.31 m/s

    W=Fdcos(theta)
    F=mg

    so W=mgdcos(theta)

    m=2kg g= 9.8m/s^2 d=2m and (theta)=??

    So W=(2)(9.8)(2)cos(??)=x

    Power=W/T

    and we know constant velocity .50 m/s and d=2 so t = 4 seconds

    so x/4 = y Watts

    is my work correct provided my theta is correct? If yes..then how do i find my theta?


    also..how would i find the force of the conveyor belt doing work on the box as the box moves horizontally?

    isnt that just W=FD ..and W=(2)(9.8)(2)=39.2

    and Power = W/T and time =4 so 39.2/4 = 9.8 Watts
     
  7. Feb 18, 2004 #6

    Doc Al

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    Staff: Mentor

    Sure. Theta is angle between the force F and the displacement D. What's the force? What's the displacement?
    What happened to the cosθ factor? You can only skip it if the force and displacement are in the same direction. (Which just means that θ = 0 and thus cosθ = 1.) Are they? What's the force and the displacement?
     
  8. Feb 18, 2004 #7
    thx!!
     
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