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Magnitude at top of loop

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass m = 1.8 kg starts at rest on a rough inclined plane a height H = 8m above the ground. It slides down the plane, across a frictionless horizontal floor, and then around a frictionless loop the loop of radius R = 2.0 m. On the floor the speed of the block is observed to be 11 m/s.

    What is the magnitude of the normal force exerted on the block at the top of the loop?

    2. Relevant equations
    (Flipping the y axis)
    [tex]N + mg = \frac{mv^2}{R}[/tex]

    3. The attempt at a solution
    [tex]N = \frac{mv^2}{R} - mg[/tex]
    [tex]N = \frac{(1.8kg)(11m/s)^2}{2m} - (1.8kg)(9.8m/s^2) = 91.26N[/tex]

    The solution says it's 20.7N, I'm not sure where I'm messing up. Though, I feel that maybe I should be using PE=KE?
     
  2. jcsd
  3. Apr 7, 2015 #2
    I think I got it, I'm using the velocity at the bottom of the loop instead of at the top, will post the correct solution once I get it.
     
  4. Apr 8, 2015 #3
    For anyone else wondering here was the solution I came too.
    [tex]KE_i = PE_f + KE_f[/tex]
    [tex]\frac{1}{2}mv_i^2 = mgh + \frac{1}{2}mv_f^2[/tex]
    h = -2R
    [tex]v_i^2-4gR = v_f^2[/tex]
    [tex]N+mg=\frac{mv^2}{R}[/tex]
    [tex]N=\frac{m(v_i^2+4gR)}{R}-mg[/tex]
    m=1.8kg
    v_i=11[itex]\frac{m}{s}[/itex]
    g=9.8[itex]\frac{m}{s^2}[/itex]
    R=2m
    [tex]N=\frac{1.8((11)^2-4(9.8)(2))}{(2)}-(1.8)(9.8)[/tex]

    Phew, everything check out?
     
  5. Apr 8, 2015 #4

    haruspex

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    Looks right now.
     
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