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Posted by on Thursday, October 25, 2012 at 10:29pm.

A 4.170 kg block of wood rests on a steel desk. The coefficient of static friction between the block and the desk is μs = 0.505 and the coefficient of kinetic friction is μk = 0.255. At time t = 0, a force F = 12.7 N is applied horizontally to the block. State the force of friction applied to the block by the table at the following times:
t=0 and t>0

  • physics - , Thursday, October 25, 2012 at 10:41pm

    Friction = uN

    N=mg so

    F=umg

    F at t(0) = (.505)(4.17 kg)(9.8 m/s^2)

    F at t>0 = (.255)(4.17 kg)(9.8 m/s^2)

  • physics - , Friday, September 30, 2016 at 7:44pm

    idk

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