A 1.5kg block starts to slide up a 25 degrees incline with an initial speed of 3m/s.It stopes after sliding 0.4m and slides back down.Assuming the friction force impeding its motion to be constant.
1).How large is the friction force?
2)What is the blocks speed as it reaches the bottom?

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asked by Pierre
  1. I really do not care about the mass so will just call it m
    on the way up
    normal force up on block = m g cos 25
    so friction force down slope = mu m g cos 25
    work done by friction = force * distance
    = mu m g cos 25 * .4
    work done against gravity = m g * .4 sin 25
    total stopping work done during ascent = .4 (mu m g cos 25+m g sin 25)
    that comes out of the kinetic energy at the start
    (1/2) m v^2 = .5 m (9) = 4.5 m
    4.5 m = .4 (mu m g cos 25 + m g sin 25)
    m cancels out as we know
    4.5 = .4 g (mu cos 25 + sin 25)
    solve for mu
    then your friction force is mu (1.5* 9.81) cos 25

    use that same force going down but up now
    the total energy lost to friction is the friction force going up times .8 meters
    (1/2) m v^2 at the end = (1/2) mv^2 at start - .8 * friction force

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    posted by Damon

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