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How high does a block on a ramp go, with friction?

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data

    A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30∘ angle. The block's initial speed is 10m/s. The coefficient of kinetic friction of wood on wood is μk=.200.

    What vertical height does the block reach above its starting point?

    The answer is 3.79m, but I didn't arrive at it using my method.

    Looking it up on google, I saw people talking about conservation of energy to answer it, but we aren't at that point in this class yet. We're working on friction.


    2. Relevant equations

    See below.


    3. The attempt at a solution
    m = 2 kg
    mu_k = 0.200
    v_0 = 10 m/s
    theta = 30 degrees

    First I found the net x and y forces:
    F_net_x = f_k + F_G*sin(30 degrees)
    F_net_y = F_N + F_G*cos(30 degrees)

    Normal force = F_G*cos(30 degrees)*(-1) = 16.97
    f_k = 0.200*16.97 = 3.39

    So acceleration for the block is -1.7 (a = F/m, negative because going in the -x direction.)


    Then I used kinematic equations:

    0 = 10 - 1.7t => t = 5.88s

    s = 0 + (10 m/s)(5.88 s) + (1/2)(-1.7)(5.88)^2
    = 28.59

    That was the hypotenuse of the triangle. Using trig to solve I came up with the y axis leg = 14.3. Not the right solution.

    Where did I go wrong?

    Thank you
     
  2. jcsd
  3. Sep 21, 2013 #2

    gneill

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    Staff: Mentor

    The acceleration that you calculated for the block looks to be too small. Can you expand on your work there?
     
  4. Sep 21, 2013 #3
    I calculated acceleration like this:

    F_N (normal force) = F_G cos(30 degrees) = 16.97.

    f_k (kinetic friction) = mu_k * F_N (normal force)
    f_k = 0.200*16.97 = 3.39N
    a = F / m = 3.39/2kg = -1.17 (negative because moving in the direction opposite the slide)
     
  5. Sep 21, 2013 #4

    gneill

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    Staff: Mentor

    Ah. Friction isn't the only down-slope force that's working against the block. What other force is acting on the mass?
     
  6. Sep 21, 2013 #5
    Gravity is, but I thought I accounted for that above? I was confused when trying to obtain f_k, because I cannot set F_net_x or F_net_y to 0, because there is acceleration in both directions due to gravity. So neither of the F_net components are 0. So maybe this is why I got a wrong f_k?

    I set f_k to be mu*F_N. F_N was the negative of gravity of the y component of F_G...
     
  7. Sep 21, 2013 #6

    gneill

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    Staff: Mentor

    Okay, the idea is to find the acceleration that is occurring in the direction of motion. Since the block is constrained to moving along the slope of the ramp you need to consider all the forces that are acting parallel to that surface. There are two of interest here. One is the friction force, which you've found nicely. The other is the component of the force due to gravity acting on the block's mass.
     
  8. Sep 21, 2013 #7
    Ok, now I think I might be understanding... I will try it out
     
  9. Sep 21, 2013 #8
    Now I get the x component total force
    = F_G_x + f_k
    = (2*9.8*sin(30 degrees)) + 3.394 = 13.194

    Solve for acceleration = F/m = 6.6

    Kinematic equation for time: 0 = 10 - 6.6t => t = 1.5s
    Kinematic equation for distance: s = 0 + (10)(1.5)+(1/2)(6.6)(1.5)^2 = 22.425
    Solved the triangle, came up with 11.2 for the height side... still not right. :(
     
  10. Sep 21, 2013 #9
    The acceleration is opposite to the direction of motion. Take this into account into the second kinematic equation. What is the sign of the acceleration?
     
  11. Sep 21, 2013 #10
    Thanks
     
  12. Sep 21, 2013 #11
    When I am solving for F_net_x, I messed up by doing this: F_net_x = 3.39 + 2*9.8* (-sin(30 degrees))

    Now I see I should have left it positive sin.

    But, why? It's in the third quadrant; thus negative. This would result in the net force being negative, as expected. With a positive sin, we say after "I guess the force is negative so I'll throw a negative sign on there from nowhere." Right? Do I always just leave the trig functions positive on these problems, regardless of quadrant?

    With positive sin F_net_x = 13.19. With -sin, -6.41. Where else would the negative sign for the net x force come from? Negative friction?
     
  13. Sep 21, 2013 #12

    gneill

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    Staff: Mentor

    Well, strictly speaking, the friction force "3.39" should be negative, too. That's because friction always opposes the direction of motion and you've tacitly assumed the direction of motion to be in a positive x-direction up the slope of the ramp. This implies your choice of coordinate axes (which you didn't do explicitly at the outset). If you don't declare your coordinates at the outset (always the best plan!) then you'll have to deal with the directions of things as you go along and write your equations after considering how things are related. (This is something I often do myself, but in my defense I draw a LOT of diagrams to guide my assignment of directions and signs to things as I write the equations!)

    The details are best sorted out by drawing a diagram, and in particular a Free Body Diagram depicting the forces that are acting. This will help to tame the sign issues so you will know which forces are acting in which directions, and so you can write your equations with the appropriate signs for each term.

    In this problem both the friction and gravitational forces should be acting down-slope, which by your tacit choice of coordinate systems would imply a negative sign for both. The acceleration will also turn out to be negative (opposing the initial direction of motion).
     
  14. Sep 22, 2013 #13
    Outstanding explanation, thank you.
     
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