# Energy conservation and string tension problem

• crybllrd
In summary, a string with a tension force of 55N pulls a 10kg block up a frictionless incline for a distance of 5m. The total external work done on the block-earth system is 275J. The change in gravitational potential energy is 300J and the change in kinetic energy is 275J. The work-energy theorem states that the net work done is equal to the change in kinetic energy. The other force doing work is the normal force, which does not contribute to the total work done. The system can also be viewed as the block moving across a horizontal surface, with the tension force increasing both the kinetic and potential energy of the system.

## Homework Statement

A string exerts a tension force of 55N on a 10kg block as it moves a distance of 5m up a frictionless incline. The string pulls parallel to the incline.

A) Find the total external work done on the block-earth system as the block moves 5m up the incline.

B)Find the change in gravitational potential energy for the block-earth system as the block moves 5m up the incline.

C)Find the change in kinetic energy for the block-earth system as the block moves 5m up the incline.

## Homework Equations

$$KE=\frac{1}{2}mv^{2}$$
$$PE_{g}=mgh$$
Our teacher tells us to round gravity up to 10m/s/s.

## The Attempt at a Solution

A)The external force should be the tension, so:
$$(55N)(5m) = 275J$$

B)I used trig (3-4-5 triangle) to find the height to go up 3m.
$$mgh_{f}-mgh_{i}=\Delta KE$$

$$(10)(10)(3)-(10)(10)(0)=300J$$

C)$$KE_{f}-KE_{i}=\DeltaKE$$
I am not sure what to do here because I don't know how to solve for velocity.

What does the work-energy theorem say about change in kinetic energy?

kuruman said:
What does the work-energy theorem say about change in kinetic energy?

$$W_{net}=KE_{f}-KE_{i}$$

If this is right, then I can set 275J equal to

$$KE_{f}-KE_{i}$$

$$\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}=275J$$

$$5v_{f}^{2}-5v_{i}^{2}=275J$$

But then I will have two variables $$v_{f}$$ and $$v_{i}$$

I think I am on the wrong track.

crybllrd said:
$$W_{net}=KE_{f}-KE_{i}$$
This is right.
If this is right, then I can set 275J equal to

$$KE_{f}-KE_{i}$$
No, you cannot. Tension is not the only force that does work. What other force is there?

$$\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}=275J$$

$$5v_{f}^{2}-5v_{i}^{2}=275J$$

But then I will have two variables $$v_{f}$$ and $$v_{i}$$
You don't care about the initial and final speeds. The problem is asking you to find the change in kinetic energy, not the initial and final speeds. If you find the correct Wnet, that is the change in kinetic energy.

kuruman said:
Tension is not the only force that does work.

It asks what the change for the block-earth system is. On a free body diagram for the block, I have three forces:

I have the Normal force (80N) by the ramp on the block,
the tension force (55N) ,
and a weight force (80N in y-dir, 60N in x-dir). (no friction)

Is the other force doing work the Normal force?

If so, then I am not sure how to calculate the work because the distance is perpendicular.

Yes there are three forces. As you say, the normal force does zero work. So what is the net work, i.e. the total work done by all three forces?

kuruman said:
So what is the net work, i.e. the total work done by all three forces?
This is where I am getting confused.
The problem is defining the system as block-earth, so if I am not mistaken, the weight force is the only force not doing work on the system.
That would leave Normal and Tension forces.
Normal is perpendicular, so equal to zero,
Tension is 275J,
so total work is [275+0] 275 Joules.
kuruman said:
Tension is not the only force that does work. What other force is there?
You said earlier I can't set the equations equal to 275J because there are other forces. What am I missing here?
Do I need to add in the weight force? I think it should be included in the block-earth system, not as external work.

The tension increases both the kinetic energy and the gravitational potential energy of the Earth-block system. If you ignore the potential energy change, it is as if the incline is not there and the block moves across a horizontal surface.

## What is energy conservation?

Energy conservation is the principle that energy can neither be created nor destroyed, but can only be converted from one form to another. In the context of a string tension problem, it means that the total amount of energy in the system remains constant, even as the string is pulled and stretched.

## Why is energy conservation important in a string tension problem?

Energy conservation is important in a string tension problem because it allows us to accurately predict the behavior of the string. By knowing that the total energy in the system remains constant, we can use this information to solve for the unknown variables, such as the tension in the string.

## What factors affect the energy conservation in a string tension problem?

The two main factors that affect energy conservation in a string tension problem are the tension in the string and the displacement of the string. As the tension increases, the energy stored in the string also increases. Similarly, as the string is stretched further, more energy is stored in the system.

## How can energy conservation be applied to real-life situations?

Energy conservation is a fundamental principle in physics and is applied to many real-life situations. For example, it is used to design structures such as bridges and buildings, to understand the behavior of materials under tension, and to optimize the performance of machines and devices.

## What are some common misconceptions about energy conservation in a string tension problem?

One common misconception is that the tension in the string is always equal to the weight of the object being suspended. In reality, the tension can vary depending on the displacement of the string and other factors. Another misconception is that energy is lost in the system, when in fact it is simply being converted from one form to another.