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Kinetic Energy and friciton

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    An 85.8 N box of clothes is pulled 24.5 m up
    a 21.1◦ ramp by a force of 118 N that points
    along the ramp.
    The acceleration of gravity is 9.81 m/s2 .
    If the coefficient of kinetic friction between
    the box and ramp is 0.26, calculate the change
    in the box’s kinetic energy. Answer in units
    of J.
    2. Relevant equations
    KE=1/2mv^2 ,Vf^2=Vi^2+2(a)(d),
    PE= m g h
    3. The attempt at a solution
    I tried to do a method told to me by a friend who said to use the formula
    F d - m g d ( sin Θ - µ cos Θ) = 1/2 m v² but when i tried it it turned it to be wrong at least the way i tried it maybe it was a calculation error. Please help with concept and formula
     
  2. jcsd
  3. Mar 2, 2009 #2

    Delphi51

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    That is a mean looking formula that might very well be the answer!
    But you really should know how to figure it out from scratch.
    Sketch the ramp with the force of gravity vector straight down from the box.
    Resolve that vector into a component along the ramp and a normal force into the ramp.
    Use trigonometry to find the two components.
    Figure out the force of friction.
    Then you can find the net force up the ramp and the work it does. It is that work that increases the KE.
     
  4. Mar 2, 2009 #3
    I tired that, It is telling me that the answer is wrong for the problem as the answer i got 5362.063648J. If u can can u let me know if it was a calculation error
     
  5. Mar 2, 2009 #4

    Delphi51

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    I got a very different answer - between 1000 and 2000. I'm not supposed to give answers here because it discourages people from doing their own work. Perhaps you could share some of your intermediate results. What did you get for the friction force? Parallel force?
     
  6. Mar 2, 2009 #5
    for force of friction i got 20.81232745 and normal force as 80.04741329. i jus added up Ff + Fn+F after that i just multiplied Fnet by the distance and got 4342.259603
     
  7. Mar 2, 2009 #6

    Delphi51

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    Ah, you need to take the direction of the forces into account when adding them!
    Some are up the ramp and some are down the ramp.
     
  8. Mar 2, 2009 #7
    so if i jus take into acount the force of friction is 20.812327146and subtract it from the force applied which is 118 i get 97.18767285. Then i just multiply it by the distance (24.5m) and i get 2381.097985 .is that rite or still kinda of
     
  9. Mar 2, 2009 #8

    Delphi51

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    Oops, I missed noticing that you have Fn in there instead of the parallel force.
    Fn is perpendicular to the ramp, so don't count it. But do count the component of the weight that is down the ramp. Add the 118N up the ramp. THREE forces to total. Yes, multiply the result by 24.5.
     
  10. Mar 2, 2009 #9
    so the force of friction calculation and everything is right. But is the perpendicular force positive or negative. I know that i subtract 118-Ff but do i add the perpendicular force or subtract it. Im confuesd on that part
     
  11. Mar 2, 2009 #10
    if i add the perpendicular force the awnser is gonna be close to 4000.00N and u said u got a ans between 1000 -2000 . If i subtract the perpendicular force i get 419.9430147 which below ur score that said was tht u got. SO i am confused now
     
  12. Mar 2, 2009 #11

    Delphi51

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    Have you got a diagram? It should show two arrows going down the ramp and only the 118 N arrow going up the ramp. One of the downs is the force of friction. The other is the parallel force - weight*sin(21.1). The perpendicular or normal force is NOT included here because it is perpendicular to the direction of motion. Only the forces in the direction of motion are counted when we consider the motion.
     
  13. Mar 2, 2009 #12
    did u get 1624.348687 as the change in KE
     
  14. Mar 2, 2009 #13

    Delphi51

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    That's it! Better round to 1620 or 1624 because you only had 3 digit accuracy in the given numbers.
     
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