Potential energy dissipation

[ritwik06]

Homework Statement

Imagine a right angled triangle ABC. AC is the hypotenuse.

The actual problem is:
An inlined plane is kept along AC (5m metre long). An object A is kept at the base of it. A force is applied on it parallel to AC. AB is 3m long. Account for the difference between the rise in potential energy (the object reaches A) and the actual work down by the force.

Homework Equations

Nil

The Attempt at a Solution

I think that there is also a horizontal displacement which accounts for the difference. My book says its entirely pure friction. But i don't agree. please help.

 

[anathema]

Thank you for bringing up this interesting problem. I would approach this problem by first defining some assumptions and simplifications. We will assume that the object A has a mass m and is initially at rest at the base of the inclined plane. The force applied parallel to AC is denoted as F.

Now, let's consider the energy changes in this system. The object A will experience a change in potential energy as it moves from the base of the inclined plane to point A. The potential energy at the base is zero, so the potential energy at point A is simply mgh, where h is the height of point A above the ground. This change in potential energy is equal to the work done by the force F, which is given by W = Fd, where d is the distance moved by the object along the inclined plane.

Here is where the difference between the rise in potential energy and the actual work done comes in. The work done by the force F is only in the vertical direction, while the displacement of the object A is both vertical and horizontal. This means that there is a horizontal displacement that does not contribute to the change in potential energy, but does contribute to the work done.

In order to account for this difference, we need to consider the components of the force F. The vertical component of the force, Fv, is responsible for the change in potential energy, while the horizontal component, Fh, does not contribute to the change in potential energy. The work done by the force F is therefore given by W = Fv * d, where d is the distance along the inclined plane. This means that the difference between the rise in potential energy and the actual work done is due to the horizontal component of the force, which is indeed friction.

I hope this explanation helps to clarify the problem for you. Keep up the critical thinking and questioning in your scientific studies!

 

[m2003]

I would approach this problem by considering the concept of potential energy dissipation. Potential energy is the stored energy an object has based on its position or configuration. In this case, the object A has potential energy due to its position on the inclined plane. When a force is applied parallel to the hypotenuse AC, the object A will experience a change in its potential energy as it moves up the plane. However, the actual work done by the force may not be equal to the change in potential energy.

This difference can be attributed to potential energy dissipation, which is the loss of potential energy due to factors such as friction and other forms of energy conversion. In this case, the force applied on the object A may not be perfectly parallel to the plane, resulting in some of the force being used to overcome friction and other resistive forces. This leads to a decrease in the overall potential energy of the object A, even though it has reached its intended position at the top of the plane.

Therefore, the difference between the rise in potential energy and the actual work done by the force can be explained by potential energy dissipation. This concept is important to consider in real-world scenarios, as it can affect the efficiency and accuracy of calculations and experiments. As a scientist, it is important to take into account all factors that may contribute to changes in potential energy, and to continually evaluate and refine our understanding of these processes.