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Work done by a friction force, block moving up a ramp-question

  1. Apr 17, 2007 #1
    Work done by a friction force, block moving up a ramp--question

    1. The problem statement, all variables and given/known data

    100kg block moves up on a rough surface, at an incline of 30 degrees, a constant force P=800 N is applied horizontally moving the block a distance of 3 M up the ramp in a time interval of 2 seconds, v1=0.8 m/s, v2=2.2 m/s, what is the work done by the friction force

    2. Relevant equations

    d-displacement, W-work, F-force, m-mass, g-gravity, v1-initial velocity, v2-terminal velocity
    eq1) W=F*cos(30)*d
    eq2) F(friction)=(coefficient of kinetic friction)*mg
    eq3) W(friction)=F(friction)*d*cos(theta)
    eq4) 1/2m(v2)^2-1/2m(v1)^2=F(friction)*d*cos(theta)

    3. The attempt at a solution
    Change in KE of the 8block is equal to the work done on block by friction force so, plugging in for eq4 i get 210, it should be negative because its in the opposite direction-right? Is my set up correct? Thanks for your time!
     
  2. jcsd
  3. Apr 17, 2007 #2

    Astronuc

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    Staff: Mentor

    There is a force accelerating the block, which means changing the velocity, and friction is not part of that. The force doing the accelerating is the applied force minus the friction force.

    The total force includes the force accelerating the block AND the friction force which opposes the movement of the block.

    The work done by friction is a function of the friction force applied over the distance.
     
  4. Apr 17, 2007 #3
    (astronuc, good to see you here, simultaneity at work here)

    It is all good, though eqn 4 may not be the best choice for solving and I didn't check the results, but I suspect may be in error cause it neglects the slowing that would be present due to work done by/ or against gravity. Either way 4 is best when there is no change in Potential Energy. I like 3 here.

    PS: this may be handy and signs can be awfully confusing in work.
    Work sign rule is : if component of force is in the direction of displacement, then work is positive, otherwise negative.
    Positive work means transfer of energy "to" the particle.
    Negative work means transfer of energy "from" the particle.
     
  5. Apr 17, 2007 #4
    Wow! that was fast! Thank you!

    Is the applied force P=800 N? For the accelerating force do I use the F=ma? Do I use F=m ((v2)^2-(v1)^2)/(2d)? so I get 730 for my friction force?? So confused!
     
  6. Apr 17, 2007 #5
    Thank you for that tip, denverdoc! im trying to figure out a way to use eq3, but I dont have the friction force, thats what im solving for, but I have velocities and a mass, which is why I chose to use eq4
     
  7. Apr 17, 2007 #6
    You might want to look at equation 2 again -- I think it should be [tex]F_f = \mu_kF_N[/tex]
     
  8. Apr 17, 2007 #7
    Indeed, it is. And in this case critical to soln as the normal force on a level surface= mg and is less on an an incline.
     
  9. Apr 17, 2007 #8
    Thats good reasoning, but only if you account for the difference in potential energies, after all even in a vacuum, a ball will still slow when thrown upwards,
     
  10. Apr 17, 2007 #9
    So, taking a different approach--I need to find all forces acting on the block, so if force P=800 in a horizontal direction (incline is 30), then its x and y components are 800N*cos(30), and 800Nsin(30). ANd if its weight force is mg=980N, then its x and y components are 980N*cos(30) and 980*sin(30), and those would both be negative because they are in the direction opposite of the movement, and my friction force is going to be -980sin(30)=-490 N, does that seem right so far?
     
  11. Apr 17, 2007 #10
    Sooo True!! THanks for helping me out! I really appreciate it!
     
  12. Apr 17, 2007 #11
    Hey I'm a rocket guy and know all too well what goes up usually comes down, and occasionally in many more pieces, esp without the help of a parachute. I put a lot of work into it, and nowhere in the work-energy theorum does that get accounted for. Universe you owe me!
     
  13. Apr 17, 2007 #12
    I found the friction force to be -133 N, to find work, I multiply the force by the distance, do I need to account for the angle of the incline?
     
  14. Apr 17, 2007 #13
    Not unless there are other energetic terms to account for.

    Pe1+KE1=PE2+KE2+frictonal work

    edtorial: and VIPoint, some forces like friction burn energy, while others do not, big difference as those that are conservative do not(electric fields, gravity,etc) store or release energy, The conservative forces are like a freeway, while the frictional forces more like a tollway.
     
    Last edited: Apr 17, 2007
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