Object on table pulled by hanging object

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The discussion revolves around a physics problem involving an object on a table connected to a hanging mass via a pulley. The initial calculations for final velocity, using energy conservation methods, yielded a result of 2.27 m/s, while the textbook states the correct final velocity is 1.59 m/s. Participants highlight that friction plays a crucial role in the energy dynamics, indicating that mechanical energy is not conserved due to the coefficient of friction of 0.25. The conversation emphasizes the need to account for energy lost due to friction in the calculations. Understanding how to incorporate friction into the energy equations is essential for solving the problem accurately.
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Homework Statement


An object of m=0.85kg is on a table. It is attached via pulley to a hanging object of m=0.42kg. The pulley is a hollow cylinder of mass 0.35 kg, with an inner radius of 0.02m and an outer radius of 0.03m, and it turns without friction. The coefficient of friction between the table and first object is 0.25.
The velocity when the object passes a certain reference point is 0.82m/s.

Using energy methods, I am supposed to find the final velocity after the block has moved 0.7m


Homework Equations


Kinetic Energy:
K=1/2(mv^2 )

Rotational kinetic energy:
Kr=1/2(Iω^2)

Potential energy:
U=mgh

Moment of inertia for a hollow cylinder:
I=1/2(M)r1^2+r2^2)

Also, v=r\omega

The Attempt at a Solution


Since energy is conserved through the whole process, the initial kinetic energy of the two blocks plus the initial rotational energy of the pulley plus the potential energy of the hanging block will equal the final kinetic energies of the two blocks plus the final rotational energy of the pulley.

Initial Energy:
(1/2)(0.42)(0.82)^2 + (1/2)(0.85)(0.82)^2 + (1/2)(1/2)(0.35)(0.02^2+0.03^2)*(0.82/00.03)^2 + (0.42)(9.8)(0.7) = (0.141204 + 0.28577 + 0.08498 + 2.8812 = 3.393 joules

Final energy:
(1/2)(0.42)(v^2) + (1/2)(0.85)(v^2) + (1/2)(1/2)(.35)(0.02^2+0.03^2)*(v/00.03)^2 =
0.21v^2 + 0.425v^2 + 0.126v^2 = 0.761v^2

3.393=0.761v^2
v=sqrt(3.393/0.761)= 2.27 m/s

However, my book says that final velocity is 1.59 m/s.

where have i gone wrong? And why isn't the coeffcient of friction used in the equations?
 
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pnwguy17 said:
1. The coefficient of friction between the table and first object is 0.25.


The Attempt at a Solution


Since energy is conserved through the whole process, ...


Read the problem: there is friction. The mechanical energy is not conserved.

ehild
 
That's true...but since none of the equations call for the friction force, let alone the coefficient of friction, how do I inocrporate it into the problem?
 
Energy conservation is not valid. How much energy is lost because of friction?

ehild
 
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Thread 'A cylinder connected to a hanging mass'
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