question in conservation of energy with fricition
Posted on 2007-07-29
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 +