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Finding net work done

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A woman pushes a 38kg box on a horizontal floor with a force F⃗ that is directed 30° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.53. If the initial speed of the box is 7.89m/s and the box accelerates at 1.63m/s^2 , what is the net work done on the box by all forces acting on it after it has traveled a distance of 2.51m ?
    The answer provided is 155.47J.

    2. Relevant equations
    Fk = ukN
    W= fdcos(x)
    W= fd
    W= Kf - Ki

    3. The attempt at a solution
    Method 1:
    Fcos(30) - Fk = 38 (1.63)
    Fcos(30) - 0.53N= 38 (1.63)

    Finding the Normal Force:
    N-mg-Fsin(30) =0
    N =38 (9.81) + Fsin(30)

    Subbing N back into the equation

    Fcos(30) - 0.53[38 (9.81) + Fsin(30)]= 38 (1.63)
    F = 38(1.63) + 0.53(9.81)(38) / cos(30) - 0.53sin(30)

    Work: Fcos(30) x 2.51 = 1541.18J

    Method 2:

    W= Kf - Ki
    = 0.5(38)(54.07) - 0.5(38)(62.25)
    = -155.47J

    Can somebody tell me why I got the wrong answer using the first method, yet when I use the second method I get the answer with a minus sign (which indicates negative work?) ?

    If that is the case, then shouldn't the answer then be -155.47J?

    Thank You!
  2. jcsd
  3. May 15, 2015 #2
    Hi dyel, Welcome to PF !:smile:
    This much is enough.
    You need the net work done.
    You already have the force value in horizontal direction.
    What is the distance. Find the horizontal work done.
    What is the force in upward direction?
  4. May 15, 2015 #3


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    This way is going to calculate the work done by the pusher. This is not the same as the net work done on the box.
    How do you get 54.07?
  5. May 15, 2015 #4
    Vf^2 = Vi^2 + 2as
  6. May 15, 2015 #5
    Why is W = fd wrong? I interpret the work on the box done by the mass multiplied by net acceleration and the distance traveled.
  7. May 15, 2015 #6
    W = fdcosθ
    Note θ = 0° for forward horizontal motion taking horizontal force component and it is 90° for vertical force component.
  8. May 15, 2015 #7


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    What if there is no "net acceleration" as is the case here?
  9. May 15, 2015 #8
    I got it. Looked up Wikipedia as well. You take the dot product of F ds, which is F d cos θ.
  10. May 15, 2015 #9
    Work is the dot product of two vector quantities, a force vector and a displacement vector. Because of this, work is always a scalar and therefore has magnitude and no direction. So whenever you calculate work done using the dot product of your force vector and your displacement vector, you want to use absolute value bars.

    You're given a lot of information, most of which you don't need. You're given your displacement, your mass and your acceleration. The horizontal component of the force being applied by the woman is found with one very simple calculation, because the acceleration we're given is the box's horizontal acceleration (the box does not go up or down, therefore there is no vertical component of acceleration as all the vertical components of force are balanced). We don't need to care about the friction force because we're already given the horizontal acceleration.

    Armed with this knowledge, we now know that we have a horizontal displacement (2.51m), the mass of the box (38kg) and the constant acceleration of the box (1.63m/s^2).
    We know that F = ma, and we know that work is the dot product of the force vector and the displacement vector. What about the angle, though? The dot product always multiplies the magnitudes of the vectors by the cosine of some angle theta.

    In this case, we have all the horizontal components we need. Therefore, theta is 0, and the cosine of 0 is 1. Thus

    W = ||F|| ||d|| = ||ma|| ||d||

    W = ||(38kg)(1.63m/s^2)|| ||2.51m||

    W = 155.4694kgm^2/s^2 = 155.4694N*m = 155.4694J

    W = 155.47J

    And voila, there's our answer!

    As you can see, physics is all about understanding the concepts and applying these concepts to a problem. If we can understand a problem, most of the time a solution will easily present itself. Instead of jumping straight into calculations, I thought about concepts I've learned, saw where I can apply these to this particular problem, and only then started to solve. Physics is the art of thinking, and mathematics is simply a tool with which we can describe our thoughts in an applicable way.

    Best of luck in the future!
  11. May 15, 2015 #10


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    The initial speed is nearly 8m/s, and the acceleration is positive, so I don't see how that formula will give you a value less than 60 for vf2. Did you subtract instead of adding?
  12. May 16, 2015 #11
    Vf^2 =Vi^2 + 2as
    = 7.89^2 + 2(1.63) (2.51)
    = 54.06
    Vf = 7.35 m/s
  13. May 16, 2015 #12


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    7.89^2 + 2(1.63) (2.51)=70.4347
    How can the speed go down from 7.89 to 7.35 while accelerating at +1.63?

    7.89^2 - 2(1.63) (2.51)=54.07
  14. May 16, 2015 #13
    I messed up on my end. Opps
  15. May 16, 2015 #14


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    So now you should get the right sign on the work as well as the right magnitude, yes?
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