1. Limited time only! Sign up for a free 30min personal 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!

Finding velocity and work involving friction Work Energy

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data
    There are three boxes stacked on top of each other. Each of the boxes has a mass of 250 kg. There is a chord attached to the bottom box applying a force of 7,000N. The coefficient of static and kinetic friction are: μs = .950 and μk = .900. What is the work net if the boxes are pulled 5 meters? What is the change (change) in velocity of the boxes if they're moved 5 meters?



    2. Relevant equations
    Work = mass*acceleration*distance
    Work = F*d
    friction = μ * Normal Force



    3. The attempt at a solution
    Total mass of the boxes is 750kg. I'm guessing I have to find the Work done by friction and subtract it from the work done by the force.

    The amount of work done to the box is 7,000N*5m which is 35,000 J.

    The work done by kinetic friction is μk*Normal Force
    = .900*(750kg*cos(0°))
    = 675 N * 5m
    = 3,375 J of work against the boxes as they slide 5 meters.

    The work done by static friction is μs*Normal Force
    = .950*(750kg*cos(0°))
    = 712.5 N
    = 712.5N* <--------------So far is this right? What do I do next from here if this is right since static friction doesn't cover any distance?
     
  2. jcsd
  3. Mar 14, 2012 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    yes
    that's the work done on the boxes by the applied force, yes
    you forgot to multiply the total mass by 9.8 to get the normal force
    Does static friction between the floor and boxes apply here if the boxes are moving?
     
  4. Mar 14, 2012 #3
    From what I have researched they do not. The static friction can be discarded since static friction moves over a negligible distance. Here's my new attempt to solve for the work done on the boxes as they are pulled 5m.

    There are two forces acting upon the boxes. The rope and friction, friction is the resistive force.

    The work done by the rope is as follows:
    Wr = F*d
    = 7,000N*5m
    = 35,000 J

    The work done by friction is as follows:
    Wfriction = μk*N
    = .900*750kg*9.8m/s^2*cos(0°)*5m
    = 33,108.75 J
    μs is disregarded because it has no effect on the work because the distance = 0 in its equation.

    So, since friction is resistive force it is subtracted: 35,000 J - 33,108.75 J = 1,891.25 J ^Final answer. Is this correct?^

    No how would I begin to calculate the change in velocity of the boxes if they moved 5 meters?

    I can assume that initial velocity is 0.

    So, Wnet = .5(m)(v)^2 - .5(m)(vinitial)^2
    1,891.25 J = .5(750kg)(v^2) - .5(750kg)(0)^2
    1,891.25 = 375*v^2 - 0
    5.0433333 = v^2
    v = 2.25 m/s

    Does that seem right?
     
  5. Mar 14, 2012 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your explanation is wrong. Static friction is present only when things are static i.e. when the two surfaces are not sliding relative to each other. In the situation where the two surfaces are sliding relative to each other, kinetic friction is the type of friction that is present.
     
  6. Mar 14, 2012 #5

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, this is the total (or net) work done, but round it off to 1900 J (2 significant figures)
    looks good!
     
  7. Mar 14, 2012 #6
    I do not understand. In my scenario, I am attempting to calculate work. I am dealing strictly with distance. It definitely creates a resisting force. Is that not irrelevant when dealing with work, though, since distance would equal 0?
     
  8. Mar 14, 2012 #7

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't understand what you're asking. All I did was explain: 1. under what conditions static friction is present and 2. under what conditions kinetic friction is present. I didn't say anything about your particular scenario.
     
  9. Mar 14, 2012 #8
    "The static friction can be discarded when solving for work since static friction moves over a negligible distance."

    Is this now correct statement now?
     
  10. Mar 14, 2012 #9

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I just doesn't make any sense at all to say "the static friction moves", because the static friction is a force. An object can move under the influence of a force. A force itself cannot "move." It's not meaningful to say that.

    What actually happens is that since the force that the person is applying (using the cord) is larger than what static friction can supply, static friction is overcome immediately, and the boxes begin moving. Since the boxes are now moving, the only form of friction that is present is kinetic friction. One can speak of the work done by kinetic friction because this force is present while the displacement is occurring.
     
  11. Mar 14, 2012 #10
    I understand it now. Thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding velocity and work involving friction Work Energy
Loading...