Rotational Dynamics involving a pulley, spring, and block

[JennV]

Homework Statement

A block of mass m = 9.00 kg is situated on an incline of angle θ = 51.0^{\circ} with coefficient of kinetic friction \mu_k = 0.200, and is connected to a light spring of spring constant k = 40.0 N/m via a light rope passing over a massive pulley as shown. The block is released from rest when the spring is unstretched. The uniform solid cylindrical pulley is mounted on a frictionless axle and has the same mass as the block and a radius R = 0.550 m. The light rope which connects the spring to the block slides over the pulley without slipping.

Find the acceleration of the block at the instant it has slid a distance of x = 0.425 m parallel to the incline. (Take the positive direction to be down the incline.)

Diagram:
http://img72.imageshack.us/img72/8755/ch11qrotdyn.jpg

Homework Equations

mgsin(theta)
uk*mgcos(theta)
ma
0.5*Kx^2
Solid uniform cylinder moment of inertia = 0.5*MR^2

The Attempt at a Solution

I drew free body diagrams for each objects. & derived a equation to solve for acceleration:

m(block)gsin(theta)-ukm(block)gcos(theta)-0.5*kx^2 = 0.5m(pulley)a + m(block)a

But the acceleration that I got was really high, so I believe that it is wrong.

My question is: does the tension of the string between the spring and pulley proportional to the extension of the spring which is 0.5kx^2 ?

 

[tiny-tim]

Hi JennV!

(have a mu: µ and a theta: θ and try using the X² and X₂ icons just above the Reply box )

m(block)gsin(theta)-ukm(block)gcos(theta)-0.5*kx^2 = 0.5m(pulley)a + m(block)a

nooo …

i] the force from the spring is kx, not kx²/2 (that's the PE)

ii] the pulley isn't moving, it's only rotating, so you can't use the mass …

you either need the "effective mass" (if you know how to work that out),

or you need to call the tension either side of the pulley T₁ and T₂, and do two separate equations, F = ma for the block, and τ =Iα for the pulley …

what do you get? ​

 

[JennV]

Hi JennV!

(have a mu: µ and a theta: θ and try using the X² and X₂ icons just above the Reply box )


nooo …

i] the force from the spring is kx, not kx²/2 (that's the PE)

ii] the pulley isn't moving, it's only rotating, so you can't use the mass …

you either need the "effective mass" (if you know how to work that out),

or you need to call the tension either side of the pulley T₁ and T₂, and do two separate equations, F = ma for the block, and τ =Iα for the pulley …

what do you get? ​

Thank you so much tiny-tim for replying! But I actually already have derived the answer myself and it was 3.00! YAY! =D