How Does Friction Affect the Speed of a Descending Block in a Pulley System?

[sadakaa]

A rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the first block (8 kg) and the table is $$\mu$$=.250. The blocks are released from rest. USE THE ENERGY METHODS TO CALCULATE THE SPEED OF BLOCK 2 (6 kg) after it has descended 1.5 meters.

Basically there's a diagram: Block #1 rests on the table and is connected by a rope that goes through a pulley to block #2, which is hanging off the table.

E = K + U
W = FDcos$$\Theta$$ = $$\Delta$$E
$$\Sigma$$F = F(Tens) - F(fric)

Block #1:
E-initial = PE = (6)(9.8)(1.5)
E-final = KE = 1/2 (6) v^2

Block #2:
E-initial = PE = ?
E-final = KE = ?

F = F(tens) - (8)(9.8)(.25)


Whats the next step??

 

[jbsteele]

The next step is to calculate the energy change for both blocks. For Block #1:E-initial = PE = (8)(9.8)(1.5)E-final = KE = 1/2 (8) v^2For Block #2:E-initial = PE = (6)(9.8)(1.5)E-final = KE = 1/2 (6) v^2Now that we have the initial and final energies for both blocks, we can calculate the work done by the tension force:W = Fdcos\theta = \DeltaE W = (F(tens) - F(fric))(1.5) W = [F(tens) - (8)(9.8)(.25)](1.5)We can now use the work-energy theorem to calculate the speed of block 2 after it has descended 1.5 meters:W = \DeltaE [F(tens) - (8)(9.8)(.25)](1.5) = \DeltaE [F(tens) - (8)(9.8)(.25)](1.5) = (6)(9.8)(1.5) - 1/2 (6) v^2 [F(tens) - (8)(9.8)(.25)](1.5) = (6)(9.8)(1.5) - 3v^2 Now we can solve for v:[F(tens) - (8)(9.8)(.25)](1.5) - (6)(9.8)(1.5) = -3v^2 [F(tens) - (8)(9.8)(.25)](1.5) + (6)(9.8)(1.5) = 3v^2 3v^2 = [F(tens) - (8)(9.8)(.25)](1.5) + (6)(9.8)(1.5) v = \sqrt{\frac{[F(tens) - (8)(9.8)(.25)](1.

 

[baba89]

The next step would be to use the energy conservation equation, E-initial = E-final, to solve for the final kinetic energy of block #2. This can be done by setting the initial potential energy of block #2 equal to the final kinetic energy of block #1, as the rope and pulley system will transfer the potential energy of block #1 to the kinetic energy of block #2. The equation would look like this:

(6)(9.8)(1.5) = 1/2 (6)v^2

Solving for v, we get v = 4.43 m/s. This is the speed of block #2 after it has descended 1.5 meters.

To check this answer, we can also use the work-energy theorem to calculate the work done by the tension force on block #2. The work done by the tension force is equal to the change in kinetic energy of block #2. Using the equation W = Fdcosθ, we can calculate the work done by the tension force as:

W = (6)(9.8)(1.5)cos90 = 88.2 J

This is also equal to the change in kinetic energy of block #2, which we calculated to be 88.2 J. This confirms that our answer for the speed of block #2 is correct.