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2 questions,

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data
    The coefficent of static friction between the floor of a truck and a box resting on it is 0.39. The truck is traveling at 87.8 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?

    2. Relevant equations

    3. The attempt at a solution
    I really am not sure how to work this out. I attempted using formula Vf^=V0^2 + 2adx but I don't think that's right. Help please!
  2. jcsd
  3. Feb 3, 2008 #2
    Sorry, my other question is this:
    M1 and M2 are two masses connected as shown. The pulley is light and frictionless. Find the mass M1, given that M2 (7.00kg) accelerates downwards at 3.43m/s2, that theta is 20.0o, and that muk is 0.390.

    I'm pretty sure I set it up correctly:
    for block 2 i said, T-mg=ma --> T=ma+mg
    for block 1, T-fk-mgsin(theta)=ma AND N-mgcos(theta)=0 (no acceleration in vertical direction)

    also, i know that m2>m1 for m2 to accelerate downwards

    basically i plugged everything in but the answer is still wrong..
    Last edited: Feb 3, 2008
  4. Feb 3, 2008 #3
    I'm also having trouble with this problem. Here is what I tried (which didn't work):
    First my numbers are different. Us=0.25 and v=81.4 km/hr.

    First I converted 81.4 km/hr to 22.61 m/s.

    Fs = Us*N => ma = Us*mg => a = Us*g => a = (0.25)(-9.81) = -2.4525 m/s^2
    v^2 = v0^2 + 2a(delta x) => 0 = 22.61 + 2(-2.4525)(delta x) => delta x = 4.61 m

    This is the wrong answer according to LONCAPA, so I don't really know what to do now.
  5. Feb 3, 2008 #4
    You forgot to square the initial velocity in the second equation.
    Last edited: Feb 3, 2008
  6. Feb 3, 2008 #5
    Oh simple mistakes... they kill me.

    I have the right answer now... 104.2 m
  7. Feb 4, 2008 #6
    I'm also having trouble with this one if anyone has any advice to offer.
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