Kinetic Energy including a pulley

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Lydia22
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Homework Statement


Two blocks, of masses M = 2.0 kg and 2M are connected to a spring of spring constant k = 200 N/m that has one end fixed, as shown in the figure below. The horizontal surface and the pulley are frictionless, and the pulley has a moment of inertia I=0.005 kg.m^2 with radius R=0.050 m. The blocks are released from rest with the spring relaxed.i.e in its equilibrium position).

(a) What is the combined(total) kinetic energy of the two blocks when the hanging two-block-pulley system has fallen a distance of 0.090 m?
J

(b) What is the rotational kinetic energy of the pulley when the hanging blockhas fallen 0.090 m?
http://www.webassign.net/hrw/W0155-N.jpg]2. Homework Equations [/b]
Mass 1
Fnet(X)= [T[/1] -[F][/sp] =[M][/1][a][/x]= [T][/1] -KΔs

Pulley
[T][/net] =[T][/1]R -[T][/2]R= -IΩ
Hanging Mass
[F][/net]=[M][/2]-[T][/2]=[M][/2]

The Attempt at a Solution



Three equations Four unknowns ,I do not know how to proceed from this point.
 
Last edited:
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I was trying to get the acceleration the lead to velocity
 
If the velocity is final 'v'
K =0.5M[V][/2] + 0.5I[ω][/2] + 0.5{2M}[V][/2] + mgh
=0.5M[V][/2] + 0.5(0.5M[R][/2])[V][/2][R][/2] + 0.5{2M}[V][/2] - 2Mgh
 
But how do I know i don't need to? won't this be in terms of unknowns ?? #swimming in the pool of confused (EXAM in a few hours) :cry:
 
(try using the X2 button just above the Reply box :wink:)
Lydia22 said:
If the velocity is final 'v'
K =0.5M[V][/2] + 0.5I[ω][/2] + 0.5{2M}[V][/2] + mgh
=0.5M[V]2 + 0.5(0.5M[R]2)[V]2/[R]2 + 0.5{2M}[V]2 - 2Mgh

Yes, but

i] I is given in the question

ii] you've left out the springy part! :wink:
 
If the velocity is final 'v'
K =0.5M[V][/2] + 0.5I[ω][/2] + 0.5{2M}[V][/2] + mgh
=0.5M[V][/2] + 0.5(0.005[V][/2][0.050][/2] + 0.5{2M}[V][/2] - 2(9.8)(0.090) + 0.5(200)(ΔS)
 
Sorry i cannot use those fancy buttons.Thank you for the little light