Object on table pulled by hanging object

[pnwguy17]

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?

 

[ehild]

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

 

[pnwguy17]

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?

 

[ehild]

Energy conservation is not valid. How much energy is lost because of friction?

ehild

 

[rwisz]

Your approach to solving this problem using energy methods is correct. However, there are a few errors in your calculations that have led to the incorrect answer.

1. In your initial energy calculation, you have used the wrong value for the potential energy of the hanging block. The correct value should be (0.42)(9.8)(0.7) = 2.91 joules.

2. In your final energy calculation, you have used the wrong value for the rotational kinetic energy of the pulley. The correct value should be (1/2)(0.35)(0.02^2+0.03^2)*(v/0.03)^2 = 0.0049v^2 joules.

3. You have also made a mistake in your final energy calculation for the hanging block. The correct value should be (1/2)(0.42)(v^2) = 0.21v^2 joules.

4. Finally, you have not taken into account the work done by friction in your calculations. The coefficient of friction is used to calculate the work done by friction, which should be subtracted from the initial energy to get the final energy.

Taking all of these into consideration, the correct final energy equation should be:

3.393 - (0.25)(0.85)(9.8)(0.7) = 0.761v^2

Solving for v, we get v = 1.59 m/s, which is the correct answer.