Difficulty in deciding when to apply work energy theorem

  • #1
physicsissohard
19
1
Homework Statement
Two blocks A and B of the same mass connected with a spring are placed on a rough inclined plane, which makes an angle $\theta$ with horizontal. What minimum velocity should be given to A up the incline so that B just moves
Relevant Equations
its
This is how I tried to do it. The force required to move B up the incline is $kx$ where x is elongation and k is spring constant. we know that spring force is greater than $mg(sin\theta+\mu cos\theta)$. And we can use work-energy theorem to figure out velocity.
$0.5*k*x^2=0.5*mv^2$ where $0.5*k*x^2$ is work done by spring force. and when you count all the chickens $v$ turns out to be $\sqrt{km}(gsin\theta+\mu gcos\theta)$. Which apparently is the wrong answer. And the correct answer apparently is $$\sqrt{(3m)/k}(gsin\theta+\mu gcos\theta)$$. I have no idea what I did wrong. Can somebody help? is there something wrong with the WOrk energy theorem, or what?
[![enter image description here][1]][1] [1]: https://i.stack.imgur.com/81yAK.png
 
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  • #2
The NET work is the one equal to the change in kinetic energy. This is what the work energy theorem "says".
 
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  • #3
physicsissohard said:
Homework Statement: Two blocks A and B of the same mass connected with a spring are placed on a rough inclined plane, which makes an angle $\theta$ with horizontal. What minimum velocity should be given to A up the incline so that B just moves
Relevant Equations: its

This is how I tried to do it. The force required to move B up the incline is $kx$ where x is elongation and k is spring constant. we know that spring force is greater than $mg(sin\theta+\mu cos\theta)$. And we can use work-energy theorem to figure out velocity.
$0.5*k*x^2=0.5*mv^2$ where $0.5*k*x^2$ is work done by spring force. and when you count all the chickens $v$ turns out to be $\sqrt{km}(gsin\theta+\mu gcos\theta)$. Which apparently is the wrong answer. And the correct answer apparently is $$\sqrt{(3m)/k}(gsin\theta+\mu gcos\theta)$$. I have no idea what I did wrong. Can somebody help? is there something wrong with the WOrk energy theorem, or what?
[![enter image description here][1]][1] [1]: https://i.stack.imgur.com/81yAK.png
You need two hash signs for your inline Latex.

Is the spring the only force on block A?
 
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Related to Difficulty in deciding when to apply work energy theorem

1. When should I use the work-energy theorem instead of kinematic equations?

The work-energy theorem is particularly useful when dealing with problems where forces and energy changes are more straightforward to analyze than motion parameters like acceleration, velocity, and time. Use it when you have information about the forces acting on an object and need to find the change in kinetic energy, or when dealing with non-constant forces.

2. How do I know if the work done by forces is easy to calculate?

The work done by forces is easier to calculate if the forces are constant or if you can easily integrate the force over the distance. If the force varies in a complex way or acts over a complicated path, it might be more challenging to use the work-energy theorem directly.

3. Can the work-energy theorem be applied to rotational motion?

Yes, the work-energy theorem can be extended to rotational motion. In this case, you would consider the rotational work done by torques and the change in rotational kinetic energy. The principles remain the same but are applied to angular quantities.

4. What are the limitations of the work-energy theorem?

The work-energy theorem does not provide information about the time it takes for the energy change to occur. It also requires knowledge of the forces acting on the object and assumes that energy is conserved within the system (i.e., no energy is lost to friction or other non-conservative forces unless they are explicitly accounted for).

5. How do I account for non-conservative forces when using the work-energy theorem?

When non-conservative forces (like friction) are present, you need to include the work done by these forces in your calculations. This work often appears as a loss of mechanical energy (e.g., thermal energy due to friction). The work-energy theorem can still be applied, but you must add the work done by non-conservative forces to the equation to account for the energy transformation.

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