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Quick Applied Maths question

  • 22-06-2007 11:29am
    #1
    7woohoo7
    Closed Accounts Posts: 8 ✭


    I know there's another thread at the moment but there's not much time before the exam now so I thought I'd have a better chance of getting my question answered in a new thread (sorry!).

    In 2005 Q10(b):
    (iii) Find the work done by the pull on the rope when the mass has been
    raised by 15 m.

    Why do you say Work done Change in K.E. and leave out the gain in Pot.E.?


    In 2003 Q4(b):
    After 1 second the string is cut. Determine whether or not the block will reach
    the pulley.     (The plane is inclined at 60 degrees)

    Why does the solution give the acceleration as = -(gsin60+g/8)
    where does g/8 come from? Is it something to do with the frictional force?

    Please help.


Comments

  • #2
    lemur option
    Closed Accounts Posts: 75 ✭✭lemur option


    ok I know this is a little late, but in case you're wondering

    1) you only factor in one as "change in KE =gain in PE" to factor in both would either cancel or double the work in that action. depending on how you add them.

    2) the accelleration is +g/8 because as you said there is a frictional force acting against the block R=g/2 "mu"=1/4 therefore the force is "mu"R= g/8.

    hope you did ok in the test today. and from now on you should probably only post in main threads.


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