Engineering science

2. A wooden block slides directly down an inclined plane, at a constant velocity of 6 m/s.
a. How large is the coefficient of kinetic friction if the plane makes an angle of 25 degrees with the horizontal?
b. If the angle of incline is changed to 10º, how long far will the block slide before coming to a stop?

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asked by Bombwe
  1. What is tangentTheta?
    forcedownplane=frictionupplane
    mg*sinTheta=mu*mg*cosTheta
    sintheta/costheat=mu
    tangent theta=mu

    b. net force down=ma
    mg*sinTheta-mu*mgCosTheta=ma
    solve for acceleration a. Mu is tangent of 25.

  2. normal force = m g cos 25

    friction force = mu m g cos 25

    = force down slope = m g sin 25

    so
    sin 25/cos 25 = tan 25 = mu
    = .466
    ===================
    at 10 degrees
    friction force = .466 m g cos 10
    = .459 m g
    work done by friction
    = .459 m g d
    = loss of kinetic energy
    = (1/2)m(36) = 18 m
    so
    18 = .459 g d = .459*9.81 * d
    d = 4 meters

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    posted by Damon
  3. oh and for time avergae speed = 6/2 = 3 m/s
    so
    t = 4/3 seconds

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    posted by Damon

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