1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work done by constant force

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    2.40 *102 N force is pulling 85.0 kg refrigerator across a horizontal surface. The force acts at an angle of 20.0 degrees above the surface. The coefficient of kinetic friction is 0.200 and the fridge moves a distance of 8.00 m. Find the the work done by the kinetic frictional force and the pulling force.


    2. Relevant equations
    Equation 1
    W = F(cos(angle))*s
    s = displacement

    Equation 2
    Fs = Fn * coefficient of kinetic friction



    3. The attempt at a solution
    So I'm just working to get the work done by the kinetic frictional force first. My answer doesn't match the book's though.
    Fn - mg + 240sin20= m*ay
    Fn - mg + 240sin20 = 0
    Fn = mg - 240sin20

    Put Fn = mg into equation 2.
    Fs = Fn * coefficient of kinetic friction
    Fs = (mg - 240sin20) * coefficient of kinetic friction

    Put into equation 1
    W = F(cos(angle))*s
    W = (mg - 240sin20) * coefficient of kinetic friction(cos(angle))*s

    Solve
    W = (85kg*9.8m/s2 - 240sin20) * 0.200(cos(20))*8m
    W = 1130 J
    Book says -1.2*103 J
     
  2. jcsd
  3. Feb 10, 2010 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    what is the angle beween the friction force and the displacement?
    correct for the proper angle between the friction force and displacement, and your answer for the work done by friction will be correct (watch plus /minus signs, however). Then also find the work done by the applied pulling force (part 2 of the question).
     
  4. Feb 10, 2010 #3
    Ah, okay. Then since the the frictional force would be opposite the displacement the angle would be 180? That makes it
    W = (85kg*9.8m/s2 - 240sin20) * 0.200(cos(180))*8m
    W = -1200 J
     
  5. Feb 10, 2010 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes..now on to part 2...if you would, please.
     
  6. Feb 10, 2010 #5
    So, basically just solve since we have all the values already...
    W = 240N(cos(20))*8m
    W = 1800J
     
  7. Feb 10, 2010 #6

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    :cool:Now you're making it look easy.........:smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Work done by constant force
Loading...