How do we calculate the energy we used to do something?

[John Constantine]

TL;DR Summary

energy conversion

Usually, I like to take a physical approach to phenomena that occur in everyday life. But I feel difficult to solve problems because I don't have higher education

My question stems from this question (What's the difference between running up a hill and running up an inclined treadmill?), which is similar to this one, but a little different.

Assume the following circumstances.

Factors such as air resistance are not considered.

There are two identical people, A and B. A is riding on an execise bike, which is fixed to the ground, and B is riding on a road bike. Press the pedal at the same torque. If the torque required to turn the pedal at a constant angular velocity is the same in both cases, The pedals will rotate at a same angular velocity in both cases ( If A and B push pedals at same torque) ---> I think Since they have traveled the same distance with the same magnitude of torque, in both cases what torque did to the object would be the same.

If I compare how the energy consumed by A and B was converted

Calories(energy) consumed by A = Energy required to rotate the wheel

Calories(energy) consumed by B = Energy required to rotate the wheel + translational energy of bicycle and B ( Kinetic energy)

From a certain point of view, the case of B is likely to consume more energy due to translational kinetic energy than the case of A.

But the two apparently pedaled the same distance at the same torque. What's going on? What's the point I'm missing or wrong with?

 

[Mister T]

When you run up a hill you are raising your body to a higher altitude. That takes work.

On a treadmill you are not raising your body, in other words, you don't have to lift your body, instead the treadmill descends.

 

[nasu]

You don't need to do work to keep a specific kinetic energy. So, comparing the kinetic energies of the two situation at the steady state (constant velocity) is not relevant. The work you do is used to overcome the enegy dissipated during motion with constant velocity. I suppose that the exercise bike is designed so that you feel the same resistance as you would in a real situation and consequently, do the same work.
But the human body is a tricky machine. The efficiency of using the work may be different for the two machines (real bike versus exercise bike) even though you apply the same torque over the same distance. This does not guarantees that the amount of chemical energy your body uses in order to get the same amount of mechanical energy is the same. And this is true in general. Activities resulting in the same amount of mechanical energy does not necessarily require the same amount of "effort" from the physiologic point of view.

 

[PeroK]

When you run up a hill you are raising your body to a higher altitude. That takes work.

On a treadmill you are not raising your body, in other words, you don't have to lift your body, instead the treadmill descends.

If you are suggesting that it's fundamentally less work on an inclined treadmill, then that is not the case. It's fundamentally the same work in both cases, leaving aside secondary factors like air resistance.

 

[sophiecentaur]

Work is a slippery customer. Although the basic definition of Work is Force times Distance, you need to take the directions of both the force and the distance into account. Then, although you may be sweating your guts out, working yourself to a frazzle, the effective / useful work done could be actually zero.

In sport, you try to maximise the useful work for minimal effort (winning a long race etc.) but, in training, you will often be maximising the work done by the body but getting nowhere (rowing machine) and increasing your capacity.

Bottom line is that the work done by your system and the energy consumed (respiration) are two different things. And we all know how tiring it can be to just stand there, holding a heavy weight. PF gets a regular supply of the OP's type of question and there's actually no definitive answer.

 

[John Constantine]

If you are suggesting that it's fundamentally less work on an inclined treadmill, then that is not the case. It's fundamentally the same work in both cases, leaving aside secondary factors like air resistance.

I'm just wondering if the same resistance from the pedals, whether they're on a road bike or stationary bike, and if the speed of the pedal is the same, the energy that a cyclist consumes would be the same.

I think it will be similar to the situation below.

There's a person A in outer space. A pushes a 5kg object. And then A pushes a 10kg object. (The initial velocity of both object and A is zero.)

After pushing an object, both systems have kinetic energy.

If A pushes both object at the same distance with the same force, regardless of its mass, what the net force has done is equal to the increase in kinetic energy of the object.

Therefore, the total kinetic energy of the two systems will be the same, and the energy consumed will be the same because Ke all comes from A. That's what I think.

I think the existing question is no different from this, so I thought that the energy that people consume would be the same.

Is there anything wrong with this assumption?

 

[PeroK]

I'm just wondering if the same resistance from the pedals, whether they're on a road bike or stationary bike, and if the speed of the pedal is the same, the energy that a cyclist consumes would be the same.

I think it will be similar to the situation below.

There's a person A in outer space. A pushes a 5kg object. And then A pushes a 10kg object. (The initial velocity of both object and A is zero.)

After pushing an object, both systems have kinetic energy.

If A pushes both object at the same distance with the same force, regardless of its mass, what the net force has done is equal to the increase in kinetic energy of the object.

Therefore, the total kinetic energy of the two systems will be the same, and the energy consumed will be the same because Ke all comes from A. That's what I think.

Work done is Force x Distance through which the force acts. When pushing an object horizontally that work translates into KE of the object. Outer space has nothing to do with it!

I think the existing question is no different from this, so I thought that the energy that people consume would be the same.

Is there anything wrong with this assumption?

It's clear that if you are pushing the pedals with the same force around the same diameter circle in the same way, then the construction of the bike doesn't affect how much work you are doing.

 

[sophiecentaur]

Work is a slippery customer. Although the basic definition of Work is Force times Distance, you need to take the directions of both the force and the distance into account.

Moreover, Energy is Frame Dependent; the answer depends on who's measuring it.

There's a person A in outer space. A pushes a 5kg object. And then A pushes a 10kg object. (The initial velocity of both object and A is zero.)

After pushing an object, both systems have kinetic energy.

So that is assuming the reference frame measures distances and velocities relative to the initial point / velocity / time when the experiment starts. If A is doing the pushing, momentum has to be conserved (total is zero at the start). The directions of each push will affect the result. Momentum is conserved but KE will change due to the Energy from the person.

So doing the thing in outer space doesn't actually make things much easier. It's probably better to do it on an Earth with 'infinite mass' and on a very slippery table. You can then 'ignore' what happens to the Earth when the objects are thrown because it will acquire almost zero KE.

 

[nasu]

Calories(energy) consumed by A = Energy required to rotate the wheel

It may help if you realize that this equality is not true. Neither the one following it in the OP. The left hand side is non zero even if the right hand side is. You can just sit in the bicycle (any kind) or just sit on a chair and while there is no kinetic energy on the right hand side, there are a few hundred joules "consumed" every second by A (the body). The fact that the right hand side may be the same in two different situations does not implies that the left hand sides will be equal. They may or may not.

 

[jbriggs444]

Moreover, Energy is Frame Dependent; the answer depends on who's measuring it.

The answer certainly can depend on the reference frame that is adopted. The wise analyst will bear this in mind. But it can depend on what question is asked. Sometimes the answer does not depend on the reference frame.

If we ask "how much work is done by a bicycle rider", "how much work is done by a motor" or "how much energy is absorbed by the brakes", we can get an invariant answer. It does not matter what reference frame we use. We just have to restrict our attention to a sub-system which is subject to no net external force [and which does end up changing its rotational energy]

A somewhat complicated example has been given already. A cyclist pedalling a bike up a hill versus a cyclist pedalling on a stationary exercise bike. The scenarios are identical except for a change in reference frame. The work done by the cyclist is identical regardless of what reference frame we choose.

We can shift to a simpler example to make the point more clear. Consider an ideal massless cylinder and piston that is held in a fixed horizontal orientation. There are two interfaces to the outside world. One where the left hand end of the cylinder is pinned to something in the environment and one where the right hand end of the piston is pinned to something else in the environment.

The cylinder expands, by a displacement of $\Delta s$ exerting some force $+\vec{F}$ on the left and an equal and opposite force $-\vec{F}$ to the right. Maybe it is a hydraulic cylinder into which we are pumping fluid. Maybe it is filled with gunpower which has been ignited. Maybe it is filled with compressed air or a spring. Maybe there a fellow perched inside with a torque wrench and a jack screw.

The claim is that no matter what reference frame is adopted, moving or accelerating in any manner to the left or right (or up and down or rotating, for that matter), the sum of the work done by the left end of the cylinder on the environment plus the work done by the right end of the piston on the environment will be identical and will be given by $\vec{F} \cdot \Delta s$.

The algebraic trick that makes this work is the fact that the displacements of the two ends along the axis of the cylinder are related. No matter how you jiggle the frame of reference to increase the one displacement, you'll be decreasing the other.

 

[sophiecentaur]

It may help if you realize that this equality is not true. Neither the one following it in the OP. The left hand side is non zero even if the right hand side is. You can just sit in the bicycle (any kind) or just sit on a chair and while there is no kinetic energy on the right hand side, there are a few hundred joules "consumed" every second by A (the body). The fact that the right hand side may be the same in two different situations does not implies that the left hand sides will be equal. They may or may not.

It should only be true if the process were 100% efficient, with no losses and no Energy going elsewhere. The OP's error is a very common one but it's very excusable because of the nonsense that's talked by many faux Scientists and people involved with the Physical Training industry.

The problem is that Work is Energy and Energy is not conserved. Otoh, Momentum IS conserved so the gun and the bullet end up with equal and opposite momenta after it's fired. Force times Time is equal but not Force times Distance is not the same both sides.