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.

Question

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!

Answer 1

made a mistake when I wrote Ef = Kf + Ugf

instead it's -->

Ef = Kf + Ugf + Usf

Answer 2

where Kf is = 0

Answer 3

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.

Answer 4

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?

Answer 5

and why isn't it Wg = Wspring - Wfriction

as friction is going to other way?

Answer 6

nevermind my first question, my mistake hehe

Answer 7

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

Answer 8

okay cool, thanks a lot Bob