Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Simple energy/work

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data
    A 890 N crate rests on the floor.
    How much work is required to move it at constant speed 5.6 m along the floor against a friction force of 180 N?
    How much work is required to move it at constant speed 5.6 m vertically?

    2. Relevant equations
    KE = (1/2)mv^2


    3. The attempt at a solution
    (1/2)(890/9.8) = x?
     
  2. jcsd
  3. Jan 27, 2010 #2
    You don't know the velocity so kinetic energy won't help you. Any other equations you can think of?
     
  4. Jan 27, 2010 #3
    Work force done = FaDcos(theta)
    Where:
    Fa = applied force
    D = displacement
    Cos(theta= im not sure would it be pie/2
     
  5. Jan 27, 2010 #4
    cos(pi/2) = 0, here your work is being done in the same direction as the motion so.

    That equation gets you half way there, need one more equation.
     
  6. Jan 27, 2010 #5
    I think cos(pi/2) = 1.?
    Mhm is the other equation:
    Work done by friction = FfD
    Where Ff= friction force
    D = displacement?
     
  7. Jan 27, 2010 #6
    Actually I wasn't quite right all you need is the equation you just wrote sorry bout that. So just use the equation you just wrote, (W done by friction) = (F of friction)*d
    Then just plug and chug.
     
  8. Jan 27, 2010 #7
    Thanks I was able to answer part 1.
    How about:
    How much work is required to move it at constant speed 5.6 m vertically?

    So here there is a gravitational component?
     
  9. Jan 27, 2010 #8
    W=f*d, you know the distance. Can you calculate the force due to gravity?
     
  10. Jan 27, 2010 #9
    The force due to gravity is just mg, (860)
    W = 890*5.6?
     
  11. Jan 27, 2010 #10
    there ya go easy right? In general if you have an equation with n variables you need n-1 equations to solve for the one variable you're looking for.
     
  12. Jan 27, 2010 #11
    Thanks, and thanks for the tip!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook