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    Hi,

    Im stuck on a couple of M1 questions, they are all similar though, so if you could help me with one then i'd be sorted !

    A package of mass 10kg is released from rest on a rough slope inclined at 25 to the horizontal. After 2s the package has moved 4m down the slope. Find the coefficient of fricion between the package and the slope.

    Ok, I have done this, but the answer is meant to be 0.241

    a = 2 m/s/s
    R - 10gcos25=0
    R=88.82

    F=ma = 10 x 2 = 20
    mu = F/R = 0.225

    thanks xxx
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    R=10gcos25
    a=2ms^2
    F(1)=ma=10x2=20
    so F(2)=10gsin25-20 F(1) and F(2) act in opposite directions so minus one from the other to give resultant force
    F=Ru
    U=(10gsin25-20)/10gcos25=0.241
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    20 = Resultant force = Force done by gravity - frictional force
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    (Original post by franks)
    Hi,

    Im stuck on a couple of M1 questions, they are all similar though, so if you could help me with one then i'd be sorted !

    A package of mass 10kg is released from rest on a rough slope inclined at 25 to the horizontal. After 2s the package has moved 4m down the slope. Find the coefficient of fricion between the package and the slope.

    Ok, I have done this, but the answer is meant to be 0.241

    a = 2 m/s/s
    R - 10gcos25=0
    R=88.82

    F=ma = 10 x 2 = 20
    mu = F/R = 0.225


    thanks xxx
    you seem to have confused "F".
    that F is not friction. that is the resultant force down the slope.
    friction would be that F minus the force due to gravity down the slope ie:

    20 - 10gsin25 = Friction
    friction = -21.4N

    mu = 21.4/88.8 = 2.41
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    How silly am I... hehe, thanks a lot - I just forgot half of the bluming resolving business. Thanks a lot for your help
    xxx
 
 
 
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