5

In my textbook, there is written that:

Done work = spent energy

Suppose I am riding a motorcycle. I start my journey from a place and then going to various places I come back to the exact point(I never stopped in my journey). So my displacement is $0$, and the work done by me is also $0$.

But in my journey I have combusted fuel and spent energy and that energy is not $0$. So is my textbook wrong or I am wrong?

Captain Man
  • 105
Theoretical
  • 1,370
  • see related https://physics.stackexchange.com/questions/92758/is-any-work-done-if-i-walk-in-a-circle – pentane Feb 12 '18 at 07:25
  • 4
    You spent a lot of energy heating the engine of the bike, and making sound, and setting air molecules in motion. Thus you and your bike have done a lot of work on the environment, giving it energy. – Arthur Feb 12 '18 at 10:32
  • Because F, and the line over which you move are both vectors, when the path changes direction, or the Force changes (in direction or magnitude), you must perform the calculation separately for each segment where they are constant, and add up the individual pieces to get the correct answer. And the multiplication of the two vectors is what is called a dot product (https://en.wikipedia.org/wiki/Dot_product). so pushing on a moving object 90 degrees to it's direction of motion, does zero work. – Charles Bretana Feb 12 '18 at 11:44
  • related: https://physics.stackexchange.com/questions/360007/why-are-work-and-energy-considered-different-in-physics-when-the-units-are-the-s several answers to which discuss the care with which the language of work and energy should be used (something missing in some of the answers here). – dmckee --- ex-moderator kitten Feb 12 '18 at 14:36

3 Answers3

10

The definition of $W=Fx$ only works in some cases. The actual definition of work is $$W=\int \vec F\cdot d\vec x=\int F\cos\theta\ dx$$ Here $d\vec x$ is a tiny displacement vector and $\theta$ the angle between the force and displacement. You can approximate this formula by taking small steps and summing the total work. $$W\approx Fx_1+Fx_2\dots$$

In your case $F$ would be some friction force working in the opposite direction to $dx$, but because you only consider one displacement (beginning to end) your approximation of the work done is very bad. In fact it is zero. If you subdivide the path in smaller segments you will get work that isn't zero. As an example you could calculate the work done by a biker riding along a perfect square with sides 1. Considering each side seperately the work done on the biker amounts to $W=4\cdot F_{friction}$ instead of zero.

Energy always goes somewhere, it is never lost. Sometimes it is hard to tell where it goes. If you are pushing against a wall without moving it seems like the energy is lost because you aren't moving, but the proteins inside your muscles are moving and converting chemical energy to heat.

AccidentalTaylorExpansion
  • 17,187
  • Exactly right. The formula W = Fx is only correct when applied over a straight line path where force is constant, or on a small enough segment of a path that changes in force and direction can be ignored by averaging force and direction – Charles Bretana Feb 12 '18 at 11:35
  • @user3502079 How did you do that? I don't know very much about dot multiplication. – Theoretical Feb 13 '18 at 09:04
  • @AsifIqubal The definition of dot multiplication of two vectors is $\vec a\cdot b=ab \cos\theta=$(length of a)(length of b)cos(theta), where $\theta$ is the angle between $\vec a$ and $\vec b$. Just remember that if the vectors point in the same direction it becomes a*b and if they don't align the dot product will be less than a*b. – AccidentalTaylorExpansion Feb 16 '18 at 14:03
  • @user3502079 Suppose I am doing the same thing as in question but I'm doing it in space where friction ot gravitational force is negligible. Then how would you explain? – Theoretical Feb 17 '18 at 08:58
  • @AsifIqubal If a force is conservative (depends only on position, like gravity) the total work when going around a loop is always zero. This is not because the displacement is zero. Having no forces like in space also has this property. This means you can move around a loop without doing any work. If you fire your engine (ignoring friction again) all the work output will directly increase kinetic energy. – AccidentalTaylorExpansion Feb 17 '18 at 16:06
  • @user3502079 I didn't understand your engine part. And another question, So is moon not doing any work revolving around sun? Is she not spending any energy going this? – Theoretical Feb 19 '18 at 15:41
7

A better wording is:

Done work = useful energy spent

The thing is that it is easy to waste energy. I can easily stand and push on a wall for a whole day and waste my effort while spending large amounts of energy on producing the force I push with. No useful work is done. None of the energy I spend goes into useful energy.

In your case, you can easily promise a delivery job done, but if you drive around and put the package back where you started, then noone would come and tell you that you did any useful work. You just wasted your fuel and time and effort. All that energy spent is lost as heat or other things; it is no spent as work done on the package.

But if I carry a heavy block up the stairs then I do some useful work - I move something somewhere else. My effort is not wasted. Sure, there is still a lot of heat generated, which is energy not spent on lifting the stone. That heat is wasted and did not help to do the work. The amount of useful work I did was the amount of energy needed to lift the block - any other energy spent is wasted and was not useful.

This is the idea of that sentence. You must supply exactly that amount of energy needed as work. If you apply any more, then... it is wasted. That's an issue with how the machine (or my body) produces the force needed. Another "machine" might be more efficient in producing the necessary force with less waste energy.

Steeven
  • 50,707
  • How can I calculate the useful spent energy? – Theoretical Feb 12 '18 at 08:04
  • @AsifIqubal Work is represented as the change in kinetic energy, as the other answer describes if no other forces do work on the object at the same time. Otherwise you can always find the work with the general force-times-displacement formula: $$W=Fx$$ – Steeven Feb 12 '18 at 09:08
  • 1
    Note that it is impossible to do physical work without "wasting" energy. – Peter - Reinstate Monica Feb 12 '18 at 10:49
  • 1
    @AsifIqubal That depends on the problem, i.e. on what you want to "get done": Heat a liter of water from t1 to t2 C, lift a weight n m, accelerate a Tesla roadster in vacuum to 12 km/s... all these tasks need a quantifiable amount of energy spent, a.k.a work. And always some more energy needs to be spent which is "wasted" by heating stuff up. Textbook examples and classroom experiments usually try to minimize losses by reducing friction, or ask explicitly to ignore it. – Peter - Reinstate Monica Feb 12 '18 at 10:53
  • @PeterA.Schneider If the task that you want to accomplish is heating stuff up (which it often is), then the energy that goes on heating stuff up is not wasted. – Mike Scott Feb 12 '18 at 11:35
  • Work is F times x over some path, (using dot product, not ordinary multiplication), but only when F is constant over the path, and the path is a straight line. If either F changes, or the path changes direction, you must sum the work done over each segment where F and the path are constant, or integrate Fdx over the entire path. – Charles Bretana Feb 12 '18 at 11:46
  • @MikeScott What you want to heat up is not stuff. Only stuff you don't care about is stuff. ;-) – Peter - Reinstate Monica Feb 12 '18 at 12:07
  • 2
    -1: this question misses the point focusing too much on English language and terminology rather than just applying the definitions and showing that if you correctly sum all the contributions, the calculations do match. – gented Feb 12 '18 at 16:02
1

Yes, work is done when you come back to the staring point, given your scenario. In this case work is done in overcoming friction, air resistance, generating heat, etc. When we say that no work is done when the displacement is zero, we think of the ideal scenario where there is no friction and air drag. Which is true. All things considered, if we stick to the ideal case, your work done is zero.

Arkadipta Sarkar
  • 171