Kinetic Energy including a pulley

AI Thread Summary
The discussion revolves around calculating the total kinetic energy of a two-block-pulley system and the rotational kinetic energy of the pulley after the blocks have fallen a distance of 0.090 m. Participants express confusion over how to relate forces, accelerations, and energies in their calculations, emphasizing the need to consider both translational and rotational kinetic energy. Key equations are provided, but there is uncertainty about the correct application of the spring constant and the inclusion of potential energy changes. The conversation highlights the importance of understanding energy conservation principles in solving the problem. Overall, the thread serves as a collaborative effort to clarify the approach to this physics problem.
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.
 
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Hi Lydia! :smile:

Why forces? :confused:

Just call the speed "v", and calculate all the energies. :wink:
 
I was trying to get the acceleration the lead to velocity
 
i know! :smile:

you don't need to
 
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)
 
a bit difficult to read :redface:

i think it's ok, except that the spring KE should be 1/2 k h2 :wink:
 
  • #10
Sorry i cannot use those fancy buttons.Thank you for the little light
 
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