Asked by Raf

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!
Answered by Raf
made a mistake when I wrote Ef = Kf + Ugf

instead it's -->

Ef = Kf + Ugf + Usf
Answered by Raf
where Kf is = 0
Answered by bobpursley
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(SinTheta-mu*CosTheta)

125x=98(.7-.3*.7)

x=98/125 *.7*.7= .392m

check my thinking.
Answered by Raf
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?
Answered by Raf
and why isn't it Wg = Wspring - Wfriction

as friction is going to other way?
Answered by Raf
nevermind my first question, my mistake hehe
Answered by bobpursley
your second quesion. Friction absorbs energy, as does the spring. They add to equal the energy input of the system.
Answered by Raf
okay cool, thanks a lot Bob
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