Weird mechanical advantage/energy transfer problem

[cxz]

Homework Statement

There is an engineer with a mass of 100kg standing on a platform raised 10m above ground. On the ground is a platform that holds a box with mass m. Connecting the two platforms is a machine of unknown design (possibly pulleys) that generates a mechanical advantage of 2. When the engineer steps on platform1, platform 2 is raised accordingly. The mechanical advantage of the machine cannot be cannoted.

Problem: Assume mass(box) m is twice the mass of the engineer and the engineer gives himself a push downwards to get moving. Including the push, the work done on the mass as platform 2 rises to its top height of 5m (maximum height) will be equal to:
(note: the answer must be given in terms of the original potential energy and final kinetic energy of the engineer)

Homework Equations

Conservation of mechanical energy of the entire system.

ΔKEsystem + ΔPEsystem = 0

Mechnical work done on the box by the engineer:
Work = Fd = ΔKEbox + ΔPEbox

The Attempt at a Solution

"The net force on both masses is zero so velocity from the push remains constant. The work done on the mass is

m2*g*h2 + 1/2*m2*v2 ^2

2*m1 = m2, 2*v2 = v1, and 2*h2 = h1 (v=h/t where t is equal for both platforms).

Substituting those variables in, the equation for the work done on the box becomes m1*g*h1 + 1/2 * (1/2*m1*v1^2), or the original potential energy of the engineer + 1/2 of the final kinetic energy of the engineer."

That's the solution from the book, and it makes perfect sense to me. What I'm confused about is this:

The initial energy of the engineer is his potential energy at a height of 10m + the kinetic energy from the push. This expression is equal to the total energy present in the system initially.

When the engineer reaches the bottom, his potential energy is 0 and his kinetic energy is 1/2*m1*v1^2, where v1 is the final velocity of the engineer.

Now let's call the engineer's original velocity v0. The total energy lost by the engineer is: m1*g*h1 + (1/2*m1*v1^2 - 1/2*m1*v0^2).

Equating that expression to the expression for the work done on the mass (the solution), we have (1/2*m1*v1^2 - 1/2*m1*v0^2) = 1/2*(1/2*m1*v1^2), which can be interpreted to mean that half of the kinetic energy from the original "push" was lost as the engineer descended down the height of the machine.

This is where my confusion lies. This concept is not intuitive to me at all. I can see how the math works out from the conservation angle, but I can't picture how the kinetic energy is transfered, or why exactly 1/2 of it is transfered. Maybe there's a flaw in my logic. Can someone nudge me in the right direction? Thank you very much.

 

[anathema]

Thank you for bringing up this interesting problem! I can understand your confusion and I am happy to provide some clarification.

Firstly, let's consider the conservation of mechanical energy in this system. As you correctly stated, the total energy of the system at the beginning is equal to the total energy at the end. This means that the initial potential energy of the engineer (at a height of 10m) must be equal to the final kinetic energy of both the engineer and the box (at a height of 5m).

Now, let's break down the motion of the engineer and the box. When the engineer gives himself a push downwards, he is essentially transferring some of his potential energy to kinetic energy. This initial push gives him a velocity v0, which remains constant throughout his descent. As the engineer reaches the bottom, his potential energy is 0 and his kinetic energy is 1/2*m1*v0^2.

On the other hand, the box is being raised by the machine, which is essentially doing work on the box. This work is equal to the change in potential energy of the box. As the box reaches the top, its potential energy is equal to m2*g*h2.

Now, let's consider the engineer and the box at the same height of 5m. At this point, the engineer's kinetic energy is still 1/2*m1*v0^2, but the box's potential energy is now equal to m2*g*h2. This means that the initial potential energy of the engineer (at 10m) must have been equal to the final kinetic energy of both the engineer and the box (at 5m). Therefore, the remaining potential energy of the engineer is equal to 1/2*m1*v0^2, which is half of the initial potential energy.

I hope this explanation helps to clarify the concept of energy transfer in this system. It is important to remember that energy is conserved and can be transferred between different forms, such as potential and kinetic energy. Keep up the good work in your studies and don't hesitate to ask for further clarification if needed.Scientist

 

[m2003]

I can provide a response to this content by explaining the concept of energy transfer and mechanical advantage in this problem.

Firstly, let's break down the problem into smaller parts. We have two platforms connected by a machine with unknown design. The engineer standing on the first platform is providing a push downwards to get the system moving. This push is a form of energy transfer, in this case, from the engineer's body to the system. This energy transfer causes the second platform to rise, and the box on it to move.

Now, let's focus on the machine that connects the two platforms. The machine has a mechanical advantage of 2, which means that for every unit of force applied by the engineer, the machine can generate 2 units of force. This allows the engineer to lift a heavier box than they would be able to without the machine. However, the trade-off is that the distance the second platform moves is half the distance the engineer moves. This is where the concept of mechanical advantage comes into play. The machine is able to increase the force, but at the cost of distance.

Now, let's look at the energy transfer during this process. The engineer starts with a certain amount of potential energy due to their mass and height on the first platform. When they push down, they are converting this potential energy into kinetic energy, which is then transferred to the system. The system, in turn, converts this kinetic energy into potential energy as it rises. This is where the 1/2 factor comes into play. Since the distance the second platform moves is half the distance the engineer moves, the potential energy gained by the system is also half the potential energy lost by the engineer. This is why the solution in the book has 1/2 in front of the final kinetic energy term.

I hope this explanation helps in understanding the concept of energy transfer and mechanical advantage in this problem. Remember, in science, it is important to break down complex problems into smaller, more manageable parts to understand them better.