# Conservation of energy: Block pulled up frictionless plane

• nickclarson
In summary: Your Name]In summary, the problem involves a block of mass 3.16 kg being pulled up a frictionless inclined plane by a string with a tension of 45.8 N. After traveling a distance of 2.12 m, the block's speed is 3.37 m/s. The task is to find the work done by the tension force. The correct equation to use is W = KE_{f} - KE_{i}, where KE_{f} is the final kinetic energy and KE_{i} is the initial kinetic energy, which is zero since the block starts from rest. After substituting in the values, the correct answer for the work done by the tension force can be found.
nickclarson

## Homework Statement

A block of mass 3.16 kg, starting from rest, is pulled up a frictionless inclined plane that makes an angle θ = 27.1° with the horizontal by a string parallel to the plane. The tension in the string is 45.8 N. After traveling a distance 2.12 m, the speed of the block is 3.37 m/s. Find the work done by the tension force.

## Homework Equations

$$U=mgh$$
$$W = KE_{f} - KE_{i}$$

## The Attempt at a Solution

I came up with:

$$KE_{i} + U_{i} = KE_{f} + U_{f}$$
$$0 + 0 = KE_{f} - U_{f}$$

so since I am looking to find the final kinetic energy which is equal to work...

$$KE_{f} = -U_{f} = -mgh$$

but positive because W is positive.

I got the wrong answer though and I could be way off, could somebody point me on the right track?

Thanks,
Nick

Dear Nick,

Thank you for sharing your attempt at solving this problem. It seems like you are on the right track, but there are a few things that may need clarification.

First, it is important to note that the work done by the tension force in this scenario is actually equal to the change in the kinetic energy of the block. This is because the only force acting on the block is the tension force, which is parallel to the displacement of the block. Therefore, the work done is given by the equation W = KE_{f} - KE_{i}, as you mentioned.

However, in your attempt at a solution, you set up the equation KE_{i} + U_{i} = KE_{f} + U_{f}. This equation is actually correct, but it is not necessary to use it in this scenario since the initial potential energy (U_{i}) is equal to zero, and the final potential energy (U_{f}) is also equal to zero because the block ends at the same height as it starts. Therefore, the equation simplifies to KE_{i} = KE_{f}.

Next, you mentioned that you got the wrong answer, but you did not provide the answer you got. It would be helpful to know the actual numbers you calculated to determine where the error may be.

To solve this problem, you can use the equation W = KE_{f} - KE_{i} and substitute in the given values for the final kinetic energy and the initial kinetic energy (which is zero since the block starts from rest). The final kinetic energy can be calculated using the equation KE_{f} = 1/2 * m * v^2, where m is the mass of the block and v is the final velocity. After substituting in the values, you should get the correct answer for the work done by the tension force.

I hope this helps. Keep up the good work!

Your attempt at a solution is partially correct. You correctly used the conservation of energy equation, but you made a mistake in your calculation for the final potential energy.

The correct calculation for the final potential energy is Uf = mghf, where hf is the final height of the block. Since the block is pulled up the inclined plane, the height will change from 0 to h = 2.12sin(27.1°) = 0.99 m.

Therefore, Uf = (3.16 kg)(9.8 m/s^2)(0.99 m) = 30.99 J.

Using the conservation of energy equation, we can then calculate the work done by the tension force:

W = KEf - KEi = (1/2)(3.16 kg)(3.37 m/s)^2 - 0 = 16.94 J.

Since there is no friction, all the work done by the tension force is converted into kinetic energy. Therefore, the work done by the tension force is 16.94 J.

## 1. How does the conservation of energy apply to a block being pulled up a frictionless plane?

The conservation of energy states that energy cannot be created or destroyed, it can only be transformed from one form to another. In the case of a block being pulled up a frictionless plane, the potential energy of the block at the top of the plane is equal to the kinetic energy it gained as it was pulled up the plane. This means that the total energy of the system (block + Earth) remains constant.

## 2. What is meant by a frictionless plane?

A frictionless plane is a hypothetical scenario where there is no resistance or force opposing the motion of an object on a plane surface. This means that there is no friction between the object and the surface, allowing the object to move without losing any energy due to friction.

## 3. Why is the conservation of energy important in understanding the motion of a block on a frictionless plane?

The conservation of energy is important in understanding the motion of a block on a frictionless plane because it allows us to predict and analyze the behavior of the block without considering the effects of friction. This simplifies the problem and allows us to focus on the transfer of energy between potential and kinetic forms.

## 4. How does the height of the plane affect the potential energy of the block?

The height of the plane directly affects the potential energy of the block. The higher the block is on the plane, the greater its potential energy. This is because the potential energy of an object is directly proportional to its height above the ground.

## 5. Can the conservation of energy be applied to real-world scenarios?

Yes, the conservation of energy can be applied to real-world scenarios. While it may be difficult to find a truly frictionless surface, the concept of conservation of energy still holds true in real-world situations. For example, a roller coaster ride or a pendulum swing both demonstrate the conservation of energy, where potential energy is transformed into kinetic energy and back again.

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