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I have a question regarding work done / energy done.

Say a $70\mathrm{kg}$ man walked $100\mathrm{m}$ on a horizontal surface as a constant speed, is it correct to assume the energy done by this man would be:

$$\text{Energy} = \text{force}\cdot\text{distance}=70\cdot9.81\cdot100=68,670\text{ joules}$$

Where the force is just the weight of the man.

Then say he walks another $100\mathrm{m}$, but this time he ascends $10\mathrm{m}$ and descends $10\mathrm{m}$ during the walk. To calculate the energy would you then have to add in energy done against gravity, so:

$$\text{Total energy}=(\text{force}\cdot\text{distance})+(m\cdot g\cdot h)=(70\cdot9.81\cdot100)+(70\cdot9.81\cdot10)=86,670+6,867=75,537\text{ joules}$$

Qmechanic
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Kieran Holmes-Martin
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2 Answers2

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The work done by the force given by $$W=\mathbf{F}\cdot \mathbf{s}=Fs\cos\theta$$ where $s$ is the distance travel by the particle or object.

Now when you travel a horizontal distance, the work done by the gravitational force (weight) is zero as the displacement vector and force are perpendicular to each other i.e. $\theta =\pi /2\Rightarrow W=0$. Although if the person ascends in that case the angle $\theta =\pi $ that means $W=-Fs$. Similarly for the case of descending the angle $\theta=0$ implies $W=Fs=mgs$.

Now you can proceed!

Young Kindaichi
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  • In that case is the equation for the work done moving up an incline W = mgh?

    Also, what happens when he moves down the incline? Does he do the same amount of work as above moving the same distance down the incline?

    Thanks

     – Kieran Holmes-Martin Mar 13 '21 at 21:37
  • Use the general expression, I have given that is $W=Fs\cos\theta$ when $\theta $ is the angle between the force and displacement vector. For example, if the person goes down by the amount $h$ then the work done would be $W=mgh\cos 0=mgh$. If it goes down to some inclined plane $W=mgh\sin\theta$ where $\theta,h$ are inclined angle and the distance travelled. – Young Kindaichi Mar 13 '21 at 21:44
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Say a 70kg man walked 100m on a horizontal surface as a constant speed, is it correct to assume the energy done by this man would be:

Energy = force x distance = 70 x 9.81 x 100 = 68,670 joules

Where the force is just the weight of the man.

The man does no physics work. (He does, however, do physiological work by his muscles that burns calories, as discussed at the end). Physics work is defined as

$$W=\int \overrightarrow F.d\overrightarrow s=F ds\cos\theta$$

Where $\theta$ is the angle between the direction of the force and the direction of the displacement $ds$ of the man. Since the man's weight is downward and his displacement is horizontal, $\theta=90 ^o$, $\cos\theta = 0$ and $W=0$..

Then say he walks another 100m, but this time he ascends 10m and descends 10m during the walk. To calculate the energy would you then have to add in energy done against gravity, so:

Total energy = (force x distance) + (m x g x h) = (70 x 9.81 x 100) + (70 x 9.81 x 10) = 86,670 + 6,867 = 75,537 joules

When the man ascends 10 m he applies a force in the downward direction and, per Newton's 3rd law, the normal reaction force of the ground on the man is equal and in the upward direction. Since the displacement of the man is also in the upward direction $\theta = 0$, $\cos \theta =+1$ and the work done by the force is $W=+mgh$ = +6,867 J.

At the same time the man ascends doing positive work, gravity does an equal amount of negative work, taking the work done by the man and storing it as gravitational potential energy in the man/earth system.

When the man descends 10 m he loses the gravitational potential energy he acquired from his ascent. Now the normal upward force exerted on the man by the ground is in the direction opposite to his displacement and $\theta =180^o$, $\cos\theta = -1$, and $W=-mgh$ = -6,867 J. The work done by the force is now negative.

The net work done by the man for the ascent plus descent is therefore zero. Therefore, the total physics work done by the man for the entire 200 m is zero.

However, as I stated at the outset, even though the man may do no no physics work, internal physiological work is done by his muscles which results in burning calories. An example of doing absolutely no physics work yet doing physiological work is holding a heavy object in place. Richard Feynman, in his physics lectures, described it this way:

The fact that we have to generate effort to hold up a weight is simply due to to the design of striated muscle. What happens is when a nerve impulse reaches a muscle fiber, the fiber gives a little twitch and then relaxes, so that when we hold something up , enormous volleys of nerve impulses are coming in to the muscle, large numbers of twitches are maintaining the weight, while other fibers relax. When we hold a heavy weight we get tired, begin to shake, ...because the muscle is tired and not reacting fast enough.

Hope this helps.

Bob D
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  • Hi,

    Thank you for the response - I can understand that there is no 'work done'. But like you said, there is physiological work done by the muscles.

    Do you know how I would calculate that? Would it be the same as what I have stated above?

    Thanks

     – Kieran Holmes-Martin Mar 14 '21 at 09:36