Kinetic Energy including a pulley

In summary, the hanging two-block-pulley system has a combined kinetic energy of -9.8 Joules when it has fallen a distance of 0.090 meters.
  • #1
Lydia22
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0

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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  • #2
Hi Lydia! :smile:

Why forces? :confused:

Just call the speed "v", and calculate all the energies. :wink:
 
  • #3
I was trying to get the acceleration the lead to velocity
 
  • #4
i know! :smile:

you don't need to
 
  • #5
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
 
  • #6
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:
 
  • #7
(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:
 
  • #8
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)
 
  • #9
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
 

FAQ: Kinetic Energy including a pulley

What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is defined as the work needed to accelerate an object from rest to its current velocity. Kinetic energy is a scalar quantity and is measured in joules (J).

How does a pulley affect kinetic energy?

A pulley can change the direction of motion and the amount of force needed to move an object, but it does not affect the kinetic energy of the object. This is because kinetic energy depends on the mass and velocity of the object, which do not change when a pulley is used.

Is kinetic energy conserved in a pulley system?

Yes, kinetic energy is conserved in a pulley system. This means that the total amount of kinetic energy before and after the use of a pulley remains the same. However, the distribution of kinetic energy may change depending on the arrangement of the pulley system.

How is kinetic energy calculated in a pulley system?

The formula for calculating kinetic energy in a pulley system is the same as in any other system: KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity. However, in a pulley system, the velocity may need to be calculated using the principles of rotational motion.

Can kinetic energy be converted into potential energy in a pulley system?

Yes, kinetic energy can be converted into potential energy in a pulley system. This can happen when the object attached to the pulley is lifted to a higher position, increasing its potential energy. However, the total energy (kinetic + potential) remains constant in a closed pulley system.

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