Object on table pulled by hanging object

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Homework Help Overview

The problem involves an object on a table connected via a pulley to a hanging mass. The setup includes a hollow cylindrical pulley and considers the effects of friction between the table and the object. The goal is to find the final velocity of the object after it has moved a certain distance, using energy methods.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of energy in the context of friction and question how to incorporate the coefficient of friction into their calculations. There is an exploration of the implications of friction on energy conservation.

Discussion Status

Some participants have pointed out that energy conservation may not apply due to the presence of friction, leading to questions about how to account for energy loss. The discussion is ongoing, with participants seeking clarification on the role of friction in the problem.

Contextual Notes

The problem statement includes specific values for mass, friction coefficient, and initial conditions, but there is an indication that the friction force has not been adequately addressed in the energy equations presented.

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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[tex]\omega[/tex]

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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