Physics  Verification
posted by Raf on .
Hello, just wanted to verify is what I did is good. I posted this question yesterday and Bob helped me, so just wanted to be sure it's good.
The problem is:
A bloc of 10 kg is put at the top of an inclined plan of 45 degrees (to the left), attached to a spring which has a spring constant of 250 N/m. The coefficient of kinetic friction between the bloc and the surface of the inclined plan is of 0,300.
What is the maximum elongation of the spring?
First I went to find fc :
EFy = 0
n  mg*sinTHETA
n = mg*sinTHETA*mu
n = 10kg*9.8N/kg*sin45*0.300
n = 20.8 N
then I know that Us = 1/2kx^2
and that Ug = mgycosTHETA*x
so:
Ef = Kf + Ugf
Ef = 1/2kx^2 + mg*cosTHETA*x
Ef = 125x^2 + 29.4x
Ef = Ei + Wnc < My friction force
so
125x^2  29.4x = 0  20.8x
125 x^2 = 8.6x
125x = 8.6
x = 0.0688 meters or 6.8 cm.
does that make sense?
Thank you!

made a mistake when I wrote Ef = Kf + Ugf
instead it's >
Ef = Kf + Ugf + Usf 
where Kf is = 0

No it does not make sense.
Workdone by gravity= force down plane*distance= mg*sinTheta*x
Work done on spring: 1/2 k x^2
Work done on friction: mg*cosTheta*mu*x
so
mgSinTheta*x=1/2 kx^2+mgCosTheta*mu*x
divide by x, then group terms
1/2 kx=mg(SinThetamu*CosTheta)
125x=98(.7.3*.7)
x=98/125 *.7*.7= .392m
check my thinking. 
ohhh I see now, but when you say Work done on spring, you take the Energy formula which is 1/2 k x^2 , wouldn't the work on a spring be mu*N*x?

and why isn't it Wg = Wspring  Wfriction
as friction is going to other way? 
nevermind my first question, my mistake hehe

your second quesion. Friction absorbs energy, as does the spring. They add to equal the energy input of the system.

okay cool, thanks a lot Bob