Principle of Work and Energy

[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.

 

[asteriatic]

The principle of work and energy states that the total work done on a system is equal to the change in its kinetic energy. In this problem, the work done by the forces acting on the system (gravity, normal force, and friction) will result in a change in the kinetic energy of block A.

To solve for the velocity of block A, you can use the work-energy theorem:

W = ΔKE

Where W is the total work done on the system and ΔKE is the change in kinetic energy.

First, calculate the work done by the friction force on block A as it moves down the incline. This can be done by multiplying the force of friction by the displacement, which is 0.3m in this case.

Wfriction = Ffriction * Δs = 14.715 N * 0.3 m = 4.4145 J

Next, calculate the work done by the normal force, which is directed perpendicular to the displacement and therefore does no work.

Wnormal = 0 J

Finally, calculate the work done by gravity, which can be calculated by multiplying the weight of block A by its change in height (Δh = 0.2598076211 m).

Wgravity = m*g*Δh = 60 kg * 9.8 m/s^2 * 0.2598076211 m = 151.887 J

Now, we can plug these values into the work-energy theorem:

W = ΔKE

151.887 J + 0 J + 4.4145 J = 0.5 * 60 kg * v^2 - 0.5 * 60 kg * 0^2

156.3015 J = 30 kg * v^2

v^2 = 156.3015 J / 30 kg = 5.21005 m^2/s^2

v = √(5.21005 m^2/s^2) = 2.282 m/s

Therefore, the velocity of block A is 2.282 m/s.

Note: The coefficient of kinetic friction is not needed in this calculation, as it was already taken into account in the calculation of the friction force. Also, the potential energy of the block is not needed in this calculation since it is not changing during the motion.