-1

I have a biomechanics question regarding how much work is done (or energy consumed) by a person moving at a constant speed from point $A$ to $B$.

  1. In the first scenario, $A$ to $B$ is $100 \; \text{m}$ along flat ground. Assume the weight of the man is $70 \; \text{kg}$.

For some reason I get work done to be $0$, because $W = f \cdot d = 0$ since the person is moving at a constant speed. But of course this isn't the case, since the person must have expended energy moving from $A$ to $B$. Is the work done $W = f \cdot d = 70 \; \text{kg} \cdot 9.81 \; \text{m/s}^2 \cdot 100 \; m = 68,760 \; \text{J}$.

  1. In the second scenario, $A$ to $B$ is $100 \; \text{ m}$ but there is an incline, by $50 \; \text{m}$ of horizontal travel you would have risen $10 \; \text{m}$, and then back down $10 \; \text{m}$ to $B$, making the tangent ratio of the angle $\tan(\theta) = \frac{1}{5}$.

Again, when I do the calculation, $W = 0$, due to the up and down movement against gravity cancelling each other out. But, this isn't the case since the person has done work (used energy) in going from $A$ to $B$.

How would I go about calculating the work done?

Dale
  • 99,825
Kieran Holmes-Martin
  • 15

4 Answers4

1

How would I go about calculating the work done?

You have correctly calculated the work done in both cases. The work done is zero in both cases. An ideal perfectly efficient machine could accomplish the motion without any net expenditure of energy.

I have a biomechanics question regarding how much work is done (or energy consumed) by a person

You have placed “(or energy consumed)” in parentheses indicating that you think these are two equivalent concepts. They are not. The ratio of the work done to the energy consumed is known as the efficiency. They are only equivalent in the special case that the efficiency is 100%.

Biological systems are nowhere near 100% efficient. In particular, for this example the efficiency is 0%. There is no way to obtain the energy consumed (unknown) from the work done (0).

Dale
  • 99,825
0

In scenario 1) it looks as though you've used the formula for Potential energy

$P.E = mgh$

$g$=9.81, but $h$ is a change in height, so for scenario 1) $h$ should be $0$.

In scenario 2) $h=10$.

So the person will need to do work walking up the hill. Probably best to ignore the downward part of the journey, the person has to expend energy to walk up the hill, but doesn't regain it on the downward journey, (technically they might have to use more energy to stop themselves speeding up on the downward journey, but you can probably ignore that).

If you are using $W=Fd$, the $F$ is zero if the person is moving horizontally but equal to the weight $mg$ if moving vertically.

John Hunter
  • 13,700
  • Hi,

    This isn't really answering my question, for scenario 1 I have used W = F x d to calculate the energy the person has used. And the force is the weight of the 70 kg man. So W = 70 x 9.81 x 100

     – Kieran Holmes-Martin Mar 14 '21 at 11:05
  • In $W=Fd$ the force must be in the same direction as the distance travelled. If the peron is travelling horizontally, we must use a horizontal force, but there isn't one. The person has no resistance to motion and can travel with no work. If you are applying the same formula vertically, then yes the $F$ is the weight of the man as that force is directed parallel to the 10m rise in height. All the best with it – John Hunter Mar 14 '21 at 11:09
  • Thanks John - but again, this doesn't answer my question. The person has spent energy moving from A to B (some number of joules), how do I calculate this? – Kieran Holmes-Martin Mar 14 '21 at 11:11
  • Hi Kieran, it's $0$ for part 1) for part 2) it's 70x9.81x10. Try this, if you walk 100m around your house horizontally it doesn't seem to use much energy. If you tried 10m up stairs, perhaps 4 times up and down a 2.5m staircase, it'll take more work – John Hunter Mar 14 '21 at 11:15
  • I still don't understand, by your reasoning I would be able to walk forever in a horizontal direction because I am not expending any energy. This is not the case, even if I walked 1 meter in any direction, I would have used energy. How do I calculate the energy used to move? – Kieran Holmes-Martin Mar 14 '21 at 11:18
  • Sorry wrong place, it's here again: It seems as though the question is designed to help students get used to P.E.=mgh or $W=Fd$, it mentioned constant speed. It's an idealised question to help people practice the concepts. In reality you are right a 10 mile walk isn't easy, but if you ignore the stopping and starting e.g. on a bike on a smooth road, 10 miles horizontally is actually quite easy. If you tried a walk or bike ride that did 10 miles uphill, up and down mountains, then that would really be exhausting. – John Hunter Mar 14 '21 at 11:24
  • @KieranHolmes-Martin the energy expended by a person in moving horizontally is primarily spent in applying friction force on the ground (brushing the feet against the ground) leading to the equal and opposite reaction friction force on the person. Friction is a non-conservative force and the negative of the work done equals the energy expended by the person. This energy is primarily converted into heat energy which is absorbed by the feet (which is why they get hot if the person walks too far or while applying high friction force) and the ground. – kbakshi314 Mar 14 '21 at 11:25
  • The question is regarding find the optimal route between A and B with the criterion of reducing the net cost of energy. But at the moment I don't know how to calculate the energy used getting from A to B because all the equations seem to say there is no energy used. Which just can't be the case. – Kieran Holmes-Martin Mar 14 '21 at 11:32
  • @JohnHunter I agree that the goal of the question is to practice computing the work done by gravity. However, I think the spirit of the question is to assume negative work done by gravity implies energy is spent by the moving person and positive work done by gravity implies that energy is gained by them. The former scenario is illustrated easily since climbing a mountain is energy draining while the latter is exemplified by a person climbing down a mountain (sliding down slide) who gains speed at the bottom of the mountain (bottom of slide) by virtue of work done by gravity to accelerate them. – kbakshi314 Mar 14 '21 at 11:33
  • @KieranHolmes-Martin, the simplistic model of the motion of the person being used is introducing the apparent contradiction (for instance ascending and descending mount Everest would consume no energy as per this model). However, it is a good start if combined with the energy expended in applying friction as explained in the comment to the answer I posted earlier. – kbakshi314 Mar 14 '21 at 11:35
  • got to go after this comment, all the best with it: The up and down hill does need energy because then $W=Fd$ is 70x9.81x10 and ignore the downhill part. The optimal route is the flat one, that in theory uses no energy (although does use a bit in reality) – John Hunter Mar 14 '21 at 11:37
  • @JohnHunter I agree with your comment if the goal is to compute the real energy expended by the person in climbing up and down a hill. However, if the goal is to optimize the real energy, the calculation of a work done along a path is more involved since it needs to include the work done in applying friction for locomotion across the path. Therefore, it is non-trivial to compute the work done on a path and therefore not evident that the flat path is the optimal one. – kbakshi314 Mar 14 '21 at 11:42
0

Work done by a force on a material particle is defined as the path integral of the inner product of the force with the displacement. Therefore, the physical quantity computed in this context is the work done by a force on a material particle. In general, energy is expended to perform work although the work-energy theorem does not in general lead to mechanical energy conservation of a system of particles if non-conservative forces act on some of the particles of the system. We will assume that all forces are conservative in this discussion to simplify the analysis since this assumption means that the negative of the work done equals the energy consumed.

In the problem posed by the OP, the travel from point $A$ to $B$ occurs horizontally at a constant speed so that the net horizontal force on the person moving vanishes and can therefore not do any work. The spirit of the problem is therefore to compute the work done by the vertical force of gravity on the center of mass of the person which is given by $W = f \cdot d \cdot \pm 1$ where $f = 70 \; kg \cdot 9.81 m/s^2$, where $d$ is the distance moved in the vertically and the final multiplication is with $+1$ in the case that the displacement is along that of gravity and $-1$ if that is not the case.

In the first scenario, $W = f \cdot d = 0$ since $d = 0$. In the second scenario, let $d = 10 \; m$ so that $W = - f \cdot d + f \cdot d = 0$.

kbakshi314
  • 2,382
  • 1
    I understand that due to physics there is no work done. But it is not what I am asking - the physical human has spent energy moving from A to B. If you walked 100 miles horizontally you will know about it, you haven't performed no work.
    How do I calculate the energy spent by the person moving from A to B?     – Kieran Holmes-Martin Mar 14 '21 at 11:08
  • @KieranHolmes-Martin, you make a good point. The person of weight due to gravity $mg$ will spend energy moving the required muscles to apply a friction force $\mathcal{f}$ against the ground which will in turn cause the equal and opposite reaction force of friction to propel the person forward. Assuming a kinetic friction coefficient of $0 < \mu_k$ and neglecting the vertical oscillations which often accompany walking, the work done in traveling $d$ distance horizontally will be $- \mathcal{f} \cdot d = - \mu_k m g d$. – kbakshi314 Mar 14 '21 at 11:16
  • @JohnHunter, I agree. However, in a slightly better model of the biophysical energy of the system considered, the comment posted above should provide a good preliminary method to compute the work done by forces acting on the person during horizonal movement. The negative of the combination of the work done by the horizontal and vertical forces acting on the person should then be the energy expended by the person to achieve the motion. – kbakshi314 Mar 14 '21 at 11:27
0

I already answered a similar question under a different title: How much more net energy do I use walking up hill?. This will supplement that answer.

I believe you can find formulas on line to calculate exercise calories. Then you might be able to do a reality check. Your 68,760 J = 16.4 Kcal (food calories).

In any case, I believe you have left out a very important factor, namely, the rate at which the work is done, since you didn't specify the man's speed. Does the man walk or does he run. It certainly matters in terms of calories burned.

Hope this helps. .

Bob D
  • 71,527