Newtonian Mechanics – Work and Energy in an Elevator

energynewtonian-mechanicswork

I was reading some notes that offer the standard definition of energy,
which is the ability to do work. Consider an elevator that is going
up. That elevator is doing work because it is exerting force in the
same direction as the displacement. My question is: how does this
work relate to energy? As the elevator goes up, its total potential
energy, $i.e.$ $m\times g\times h$ is increasing. However, is this
energy separate from the energy required to do the work in
the first place? Os is the correct reasoning that as the elevator
is rising up, it is generating exactly enough potential energy to
exert force to go back down? I am not a physics student and I find
energy super difficult to grasp. Any help is much appreciated.

Best Answer

Since you are new to physics, in order to help answer your questions the following is a brief primer on work and energy.

  1. The relationship between work and energy is that work is one of the mechanisms by which energy is transferred from one thing to another. In the case of work, the transfer of energy is due to force times displacement. (The other main mechanism is heat. Heat is energy transfer due solely to temperature difference).

  2. If the direction of the force is the same as the direction of the displacement, the work done by the force is called positive work because it transfers energy to the object being displaced. This is the case with the work done on the elevator. A motor does work on the elevator (and its contents) by lifting it up with a cable against the force of gravity. The motor transfers energy to the elevator and its contents.

  3. If the direction of the force is opposite to the direction of the displacement, the work done by the force is called negative work. When negative work is done by the force it takes energy away from the object being displaced. In this case when the elevator rises gravity does negative work on it since its force is opposite to the displacement of the elevator. Gravity takes energy away from the elevator and its contents and stores it as gravitational potential energy of the earth-elevator system. More on this below. Gravity is what is called a conservative force. That means the work done by gravity only depends on the initial and final positions of the object, independent of the path between the positions.

As the elevator goes up, its total potential energy, $i.e.$ $m\times g\times h$ is increasing. However, is this energy separate from the energy required to do the work in the first place?

It is important to understand that it is not the potential energy of the elevator that is increasing. It's the potential energy of the combination of the elevator and the earth (earth-elevator system) that is increasing. That's because potential energy is a system property. Although you will commonly see people refer to the potential energy of objects, that is technically incorrect. The object without the earth and the earth without the object experiences no change in gravitational potential energy due to the movement of either.

That said, the increase in potential energy is not seperate from the energy required to do the work in the first place. Thats because the origin of the energy that gravity took away from the elevator to store as gravitational potential energy is the energy transferred to the elevator by the motor. But that does not necessarily account for all the energy transferred to the elevator by the motor. Part of that energy may be the change in kinetic energy of the elevator. To explain:

A. As the elevator passes a floor a height $h$ on the way up gravity takes energy of $mgh$ away from the elevator to store as gravitational potential energy. But since the elevator is still moving up, it has kinetic energy of $\frac{1}{2}mv^2$. That kinetic energy equals the difference between the positive work done by the motor on the elevator, minus the energy taken away by gravity (gravitational potential energy).

B. Just before the elevator reaches its intended floor at say height $h'$, it decelerates to come to a stop. At that point its kinetic energy is zero just like it was at the start of the trip. The total change in KE is thus zero. This means that the net work done on the elevator, $+mgh'$ by the motor and $-mgh'$ by gravity, equals zero.

This last point illustrates the important work-energy theorem, which states that the net work done on an object equals its change in kinetic energy.

Hope this helps.

Related Question