Kinetic Energy including a pulley

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Homework Help Overview

The discussion revolves around a physics problem involving kinetic energy in a system with two blocks connected to a spring and a pulley. The scenario includes frictionless surfaces and a specific moment of inertia for the pulley, with the blocks released from rest.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between forces, energy, and kinematics in the context of the problem. There is discussion about calculating kinetic energy using various forms, including translational and rotational components. Some participants express confusion about the necessity of certain calculations, particularly regarding acceleration and velocity.

Discussion Status

The conversation is ongoing, with participants offering different perspectives on how to approach the problem. Some guidance has been provided regarding the use of energy equations, but there is no clear consensus on the best method to proceed. Participants are actively questioning assumptions and clarifying details of the problem setup.

Contextual Notes

Participants note the presence of multiple unknowns and the challenge of integrating the spring's potential energy into their calculations. There is also mention of an impending exam, which adds to the urgency and stress of the discussion.

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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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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