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Homework Help: Find the work done on a box on an inclined plane

  1. Mar 2, 2010 #1
    1. The problem statement, all variables and given/known data

    a constant force Fa of magnitude 85.0 N is applied to a 4.0 kg shoe box at angle φ = 53.0°, causing the box to move up a frictionless ramp at constant speed. How much work is done on the box by Fa when the box has moved through vertical distance h = 0.45 m?

    heres the picture to explain where the phi comes from http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/pict_7_32.gif
    2. Relevant equations
    W=F dot D
    or W = FDcostheta


    3. The attempt at a solution
    i know that if an object is going at constant speed it's net force is zero
    i tried solving for D which gave me .45/sintheta
    then i got y comp of the weight force = 4gsinthet
    the x comp of the wieght cut the phi angle in two = 53 - theta
    and using Fa and angle (53-thet) i got the force that balance the 4g sin = 85sin(53 - thet)

    then i used 4g sinthet = 85sin(53 - thet) to find theta because the net force is zero
    i then put the theta i got into the D equation above.
    I then found the x and y components of D and the x and y compnents of F using the angle( 53-thet) and did the dot product of that which came out to be 59.51455753 J
    and it's wrong....
    the answer is 17.658 J...supposedly
    can someone tell me what i'm doing wrong? am i using the wrong angle or am i just going abot the problem wrong from the beginning? help please
     
  2. jcsd
  3. Mar 3, 2010 #2

    rl.bhat

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    Homework Helper

    How much work is done on the box by Fa when the box has moved through vertical distance h = 0.45 m?
    When the work is done on a body, its kinetic energy increases. But in the problem it is stated that the box moves with uniform velocity. How is that?
    Next the work done on the box = rise in potential energy of the box = m*g*h
    If you substitute the values you get the answer.
     
    Last edited: Mar 3, 2010
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