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Crate pulled up rough incline

  1. Nov 6, 2016 #1
    1. The problem statement, all variables and given/known data
    A crate is pulled up a rough incline. The pulling force is parallel to the incline. The crate is pulled a distance of 5.79 m. The acceleration of gravity is 9.8 m/s^2
    theta=23.4 degrees
    mass= 8.91kg
    Force pulling on box= 142N
    coefficient of friction= 0.292
    Initial velocity of box= 1.48m/s

    A.What is the magnitude of the work is done by the gravitational force? Answer in units of J.
    B.How much work is done by the 142 N force? Answer in units of J.

    2. Relevant equations
    PE= mgh
    W=F*displacement*cos(theta)

    3. The attempt at a solution
    A. I tried using the mgh formula but height is not given. Then, I tried the second formula given but I do not know how to calculate the displacement.
    B. Used the second equation got W= 142N*5.79*cos(23.4). Ended up being incorrect.
    UPDATE: got the second one as the cos(theta) ended up being 1 as the angle between the displacement and the force was 0.


    Clearly lost. I do not know where to properly start so a hint at that would be helpful!
     
    Last edited: Nov 6, 2016
  2. jcsd
  3. Nov 6, 2016 #2

    hilbert2

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    Calculating the height from the distance is basic trigonometry. The pulling force of 142 N has the same direction as the displacement, so you don't need any trig functions in part B.
     
  4. Nov 6, 2016 #3
    So cos(23.4)=x/5.79. 5.79cos(23.4)=x. x=5.314
    mgh= 8.91*9.8*5.314= 463.99
    Does this mean the work done by gravity is 463.99?
     
  5. Nov 6, 2016 #4

    hilbert2

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    ^ If theta is the angle between the incline and the horizontal direction, then h = d sin(theta), where d = 5.79m
     
  6. Nov 6, 2016 #5
    Plugging in I get h= 2.299m
    mgh=8.91*9.8*2.299= 200.786
    Does this mean the work done by gravity is 200.786?
     
  7. Nov 6, 2016 #6

    hilbert2

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    ^ Looks like it's correct, but I suggest you try to include the units of the physical quantities in all calculations in the future. That way you can often spot mistakes by checking whether the result is dimensionally correct.
     
  8. Nov 6, 2016 #7
    Thanks, that answer was correct.
    One last question. I am looking for the change in kinetic energy. I tried 822.18J- 200.786J= 621.394J with the equation Change in kinetic energy= work by gravity+ work by applied force.
     
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