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Use Work (net) to find coefficient of static friction and speed

  1. Feb 25, 2014 #1
    1. a 5.0 kg block is pushed up a wall by a 90N force. Assume that the block starts from rest and that it has an acceleration of 0.5m/s^2.

    a. find the coefficient of kinetic friction between the block and the wall.
    b. Find the work done by each force acting on the block.
    c. Find the speed of the block after it has moved 3.0m up the wall.

    I broke the 90N force into x and y components, but it made it difficult to see what work done by other forces was negative or zero, so I then made the 90N force point in the positive x direction, so that Ncos(55) = zero, Nsin(55) = m*g*sin(55) = 40.1N, and the work done by friction and weight equal zero in the -x (cos(55)) direction. and work done by weight = -m*g*sin(55) = -40.1, but other than knowing the work done by friction is = -μk*N*sin(55), I don't know how to solve for μk.

    Also I think it's weird that work done by the normal force cancels out work done by the weight force - they are in opposite directions so it makes sense, but I thought I should be able to add up all the forces involved and get zero - is that not true because there is acceleration?

    Should I be using the acceleration given in the problem instead of the acceleration due to gravity?

    I think that the total work is equal to work of friction, plus normal, plus weight, plus the 90N force (some of these are negative of course), and that that number is = 0.5*m*v^2, and that will be how I solve for c. Basically I'm really stuck on a. - finding the coefficient of kinetic friction. Once I have that, I can calculate the work done by friction and complete part b., and use that part to solve for c - only I don't know how to plug in the 3.0m - I know it will have something to do with the acceleration given but I'm not sure what.

    So to sum up my questions - how do I find the coefficient of kinetic friction?
    What does the acceleration being equal to 0.5 mean for how I solve/think about this problem, and does it replace g?
    How am I supposed to use the information in part c (distance traveled) to solve for the velocity at that point?

    And of course, are my thoughts above laid out correctly or am I trying to use the wrong expressions to solve this problem?

    Thank you.
     
  2. jcsd
  3. Feb 25, 2014 #2

    BvU

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    How did you come to ignore the template? Would very much like to know where the 55 comes from ?
    Please give a statement of the problem and a listing of variables and given/known data.
    Apparently you are not using any expressions at all, at least they aren't to be found under 2. relevant equations. They are relevant also for potential helpers to check you are mobilizing the right stuff.
     
  4. Feb 25, 2014 #3

    BvU

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    says who ? Who says they are in opposite directions and who says that they cancel?

    Read what you wrote before posting. Lots of potential helpers have PhDs but they are not clearvoyant.
    Did you make a drawing ? What IS the positive x direction ?
     
  5. Feb 25, 2014 #4

    Doc Al

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    Please describe the problem more clearly. Presumably, that 90 N force makes some angle with the vertical wall.

    Identify all the forces acting on the block and apply Newton's 2nd law to start you off.
     
  6. Feb 25, 2014 #5
    Yes - sorry forgot to include information from the drawing - the force of 90N is pushing the block at an angle of 55 degrees from the horizontal. So towards the wall and up - I set up my axis for the problem to make that force just be the positive x direction.

    Forces acting on the wall are friction = uk*m*g*sin(55) (will be negative), weight = m*g*sin(55) also negative, and normal = m*g*sin(55) - positive.

    The block is accelerating so I think that since I know m*a and I know the forces it should be:

    m*a = F - (N+W+ukN) where N = m*g*sin(55) and so on from above. But when I try to solve for Uk I get impossible numbers.

    (m*a - F + N +W)/m*g*sin(55) = uk, but it's greater than one. So, no. Also I'm supposed to be using Work/energy things but I have no idea how to relate those to this because of friction - Ui+Kf = Uf+Ki (potential and kinetic energies).
     
  7. Feb 25, 2014 #6

    Doc Al

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    OK.
    Are you saying that you took your x axis to be in the direction of that 90 N force? Why in the world would you do that?

    Stick to horizontal and vertical components.

    What's the normal force between block and wall?

    ?

    ?

    I'd say start over using normal horizontal and vertical axes.
     
  8. Feb 25, 2014 #7
    YES! I think my TA hates me - he suggested I set up the axes that way. So I just redid it using a "normal" axis so that weight, friction, and the normal force don't have to be broken up into components.

    Here's what I have:

    Work by normal force = Work done by x-component of external force (Fcos(55)).

    F=m*a=Fsin(55) - (W+f)

    -f = m*a - Fsin(55) +W

    -22.2 = (5*0.5) - 90sin(55) + (5*9.8)

    f = uk*N
    22 = uk*Fcos(55)
    uk = 22.2/90cos(55)

    uk = 0.43

    That seems like a reasonable answer. I think what really confused me on this problem was that we are supposed to be using only work and energy principles on this homework. I wouldn't know how to use the work/energy equations to find the coefficient of friction. I may or may not get credit for this problem, but at least from here I can experiment with ways to find the coefficient using the work/energy equations - if possible. Since there's no time parameter I don't think I could find velocity - otherwise I could use the K = .5*m*v^2 to find it.

    Part b. I will redo using the "normal-person" axis, which should be much easier, and part C is clearly a work problem, with two distinct positions and velocities (from rest to some velocity). Thank you!
     
  9. Feb 25, 2014 #8

    haruspex

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    Since you are given nothing that limits the duration of the process until part c, part b is insoluble.
     
  10. Feb 25, 2014 #9
    Good to know - I just put everything * s (displacement).
     
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