question in conservation of energy with fricition

Posted on 2007-07-29
Last Modified: 2010-07-27
i have a question in friction. In general the friction i thought is is fk*DX*cos180 which is always (-)
in this example :

a freight company uses a compressed spring to shoot 2 kg packages up to a 1m high frictionless ramp. this spring constant k= 500 N/m and the spring is compressed 30cm = 0,3 m
Someone spills soda on the ramp .That creates a 0.5 m sticky spot with a coefficient of kinetic energy 0.3
will the package make it to the truck ?

i have seen the solution but there is one part i dont get.
My solution is
I found that fk*DX = 2.94 (so does the solution)
1/2K*DX = 22.5 (we agree as well)
1/2 mv^2 +mgh+fkdxcos(180)-1/2K*DX^2 = 0  here is the trick now
v final =0
thus h = (22.5 + 2.94)/mg

the solution does h=(22.5-2.94)/mg = 0.998 (so it barely misses)

since at the initial equation of conservation of energy , all the examples i have studied so far they treat fk*DX with a negative sign  , how to we have a positive sign here , thus we subtract at the very end ?
thus mgh =

=> mgh = 1/2 *500*0.3^2 +
Question by:c_hockland
    LVL 22

    Expert Comment

    What is D?
    What is X?
    What is f?
    How long is the ramp?  
    At what angle is the ramp inclined?
    LVL 84

    Accepted Solution

    The energy dissipated is positive.
    LVL 22

    Expert Comment

    Well, I'm glad at least somebody understood. :)

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