Rotational Dynamics involving a pulley, spring, and block

[JennV]

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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! :smile:

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

[JennV]

Hi JennV! :smile:

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


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? :smile:

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