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question in conservation of energy with fricition

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 +
0
c_hockland
Asked:
c_hockland
  • 2
1 Solution
 
NovaDenizenCommented:
What is D?
What is X?
What is f?
How long is the ramp?  
At what angle is the ramp inclined?
0
 
ozoCommented:
The energy dissipated is positive.
0
 
NovaDenizenCommented:
Well, I'm glad at least somebody understood. :)
0

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