How Is Velocity Calculated in a Pulley System with Friction?

[Peter P.]

Homework Statement

Determine the velocity of the 60- block A if the two blocks are released from rest and the 40- block B moves 0.6 up the in cline. The coefficient of kinetic friction between both blocks and the inclined planes is = 0.05.

Here is the given diagram:

The Attempt at a Solution

Since there is a pulley system, I began with the length equation so that i can determine the displacement of A with respect to B.

Δs_A = -0.5Δs_B
Once everything is plugged in, Δs_A = -0.3 (or just 0.3m down along its plane).
Using this, i calculated the change in height as it falls from the top to the next position. Δh = 0.2598076211 m. Then i calculated the normal force exerted by the plane and the force of friction acting on the block A (F_f = 14.715 N (directed towards the pulley system). So here's the part that i don't get. I am not sure how to tie everything together. I know that:
Kinetic Energy (T), Work and potential Energy (U)
T₁ + U_{1 -> 2} = T₂
0.5m_Av₁² + U_{1 -> 2} = 0.5m_Av₂²

From the free body diagram of the isolated mass A, i know that there is a normal force, the weight of the block, and a friction force acting on it. So from this information, I believe that there should be gravitation potential energy (V_G = mgΔh), and i feel like the friction force needs to be included as well. So i would integrate the force i calculated, with respect to s (∫F_f ds). Then i feel like it should be integrated from 0 -> 0.3. Is this correct so far?

Thanks in advance for any help.

 

[PeterO]

Homework Statement

Determine the velocity of the 60- block A if the two blocks are released from rest and the 40- block B moves 0.6 up the in cline. The coefficient of kinetic friction between both blocks and the inclined planes is = 0.05.

Here is the given diagram:

The Attempt at a Solution

Since there is a pulley system, I began with the length equation so that i can determine the displacement of A with respect to B.

Δs_A = -0.5Δs_B
Once everything is plugged in, Δs_A = -0.3 (or just 0.3m down along its plane).
Using this, i calculated the change in height as it falls from the top to the next position. Δh = 0.2598076211 m. Then i calculated the normal force exerted by the plane and the force of friction acting on the block A (F_f = 14.715 N (directed towards the pulley system). So here's the part that i don't get. I am not sure how to tie everything together. I know that:
Kinetic Energy (T), Work and potential Energy (U)
T₁ + U_{1 -> 2} = T₂
0.5m_Av₁² + U_{1 -> 2} = 0.5m_Av₂²

From the free body diagram of the isolated mass A, i know that there is a normal force, the weight of the block, and a friction force acting on it. So from this information, I believe that there should be gravitation potential energy (V_G = mgΔh), and i feel like the friction force needs to be included as well. So i would integrate the force i calculated, with respect to s (∫F_f ds). Then i feel like it should be integrated from 0 -> 0.3. Is this correct so far?

Thanks in advance for any help.

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

Firstly I have never seen Kinetic Energy given the symbol T ? That is usually reserved for Tension.

Secondly, when you looked at just body A, I didn't see any mention of the Tension force in the string??

EDIT: Thirdly, I wouldn't be integrating anything to solve this problem.

 

[Peter P.]

Firstly I have never seen Kinetic Energy given the symbol T ? That is usually reserved for Tension.

Secondly, when you looked at just body A, I didn't see any mention of the Tension force in the string??

EDIT: Thirdly, I wouldn't be integrating anything to solve this problem.

Using T for the kinetic energy is the convention used by my professor and i agree that it was odd when i first saw it as well. Second, i completely forgot about the tension, which only confuses me ever so more in how i need to incorporate it into the equation. I am thinking it depends on m_B.

 

[PeterO]

Using T for the kinetic energy is the convention used by my professor and i agree that it was odd when i first saw it as well. Second, i completely forgot about the tension, which only confuses me ever so more in how i need to incorporate it into the equation. I am thinking it depends on m_B.

Since you thread title is "Princilpe of Work and Energy" - I would be limiting my throghts to those "global" ideas rather than the nitty gritty of each mass.

You have already deduced the 2:1 ratio between movements of each mass.

I would work on:

Mass A goes down - loses Grav. Potential Energy
Mass B goes up - gains Grav. Potential Energy
Mass A slides its distance against its friction [ μN where N stands for Normal Reaction Force] losing some energy.
Mass B slides its distance against its friction, losing some more energy.

The potential energy gains&losses present a start for the final Kinetic Energy of the masses - but some of that is lost to friction on each block.
What is left should enable you to find v using (1/2)mv², making sure you use the entire mass, as both will be moving since they are tied together.