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What force will be needed to start the sled moving?

  1. Dec 11, 2003 #1
    i need help with figureing out this problem

    A sled of mass 57 kg is pulled along snow-covered, flat ground. The static friction coefficient is 0.30, and the sliding friction coefficient is 0.10.

    What force will be needed to start the sled moving?

    Once moving, what total force must be applied to the sled to accelerate it 4.6 m/s2?

    What force is needed to keep the sled moving at a constant velocity?

    I figured out the weight, but i need the formula to figure out the rest.
  2. jcsd
  3. Dec 11, 2003 #2
    To start the sled moving, you need to overcome static friction, the force of which is mu*F_n. F_n in this case is equal to the mass*gravity of the sled.

    So the force to start = mu*m*g = .3*57*9.81 = 167.751 N

    Now to accelerate it to 4.6 m/s^2, you need to opposte the force of sliding (kinetic) friction

    Net F_x = F_applied - F_friction = ma
    F_applied is what we're looking for, F_friction again is mu*m*g where mu is KINETIC friction this time (not static)

    So F_applied = mu_k*m*g+m*a (a is 4.6 m/s^2, the acceleration we want to get it to)
    F_applied = .1*57*9.81+57*4.6 = 318.117 N
  4. Dec 11, 2003 #3

    thanks. do you think you can help me w/ my last problem?
  5. Dec 11, 2003 #4
    edit: whoops, didn't see it

    To keep something at constant velocity the net Force must be zero.. so here we have F_applied - F_friction = 0

    or F_applied = F_friction = mu_k*m*g = .1*57*9.81 = 55.917
  6. Dec 11, 2003 #5

    A force of 45 N accelerates a 6.0-kg block at 7.0 m/s2 along a horizontal surface.

    How large is the frictional force and what is the coefficient of friction?
  7. Dec 11, 2003 #6
    Re: thanks!

    ok.. again, just set up your net f_x forces. We have the applied force (let's call it F_a) and the frictional force (F_f)

    Net F_x = F_a - F_f = m*a
    Solving for F_f, we have F_a - m*a = F_f

    The applied force is 45N, the mass is 6.0kg, and acceleration is 7.0 m/s^2.. plug them in, you get a frictional force of 3N

    Now, F_f = mu_k*m*g
    mu_k (coefficient of friction) = F_f/(m*g) = 3N/(6.0kg*9.81m/s^2) = 0.05096 ~ .051
  8. Dec 11, 2003 #7

    yeah i figured that one out later on, but thanks anyway
  9. Dec 11, 2003 #8
    Re: Re: thanks!

    thanks again! do you have AIM?
  10. Dec 11, 2003 #9
    Re: Re: Re: thanks!

    yep: simplydelta

    feel free to message me anytime
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