Why is work done defined the way it is? What's the intuition here? [duplicate]

I have a question I am struggling with for days. So, Work transfers energy from one place to another or one form to another. (at least as per Wikipedia)

Imagine 2 scenarios.

Scenario 1: A pen on which I am applying 1 N force over a course of 100m.

Scenario 2: An elephant on which I am applying 1 N force over a course of 100m.

Imagine frictionless surfaces!

In both scenarios, there would be 100 J of energy transfer as per "Work" formula. But what does it mean intuitively?

My doubt: In scenario 1 I would be applying 1 N force for say time t but in scenario 2 I would be applying same 1 N for time T and obviously,

T >>>>> t

So I am subjecting same force but for so different lengths of time. Why is "transfer of energy" defined as F.d, intuitively it seems more energy is expended in the Elephant's scenario since I am applying same force but for MORE time.

Can someone give me intuition which clarifies these scenarios?

• Possible duplicate: physics.stackexchange.com/q/277347 – Hritik Narayan Jun 26 '17 at 15:00
• @HritikNarayan My doubt is slightly different. I want to understand the time aspect in work done. I will anyways go through that question as well. – Sushant Gupta Jun 26 '17 at 15:02
• It certainly will take more time, because the elephant is heavier and hence has a much lesser acceleration. However, the end kinetic energy in both the scenarios will be the same. – Hritik Narayan Jun 26 '17 at 15:04
• It's defined that way because of the work-energy theorem which is a very useful principle in physics. I admit that definitions of work and energy are circular, but there's no precise way to define energy just like there's no real way to define mass. – xasthor Jun 26 '17 at 15:04
• @SushantGupta A way to see that time is irrelevant is to imagine a spring pressing against two walls. It can exert a 1N force forever against those walls, but surely it's not doing infinite work if it's not moving the walls since there's no change in system state. Sustaining a force requires no work at all unless you're doing it through some distance. – David Schwartz Jun 26 '17 at 19:00

Assuming a constant force, the time factor doesn't really matter when it comes to Energy.

The integral of $F$ over time gives you the momenta of the bodies, which need not be the same after whatever you've done.

The momenta would be equal if you applied the force over the same time interval, and the energies are same when the force is applied over the same space interval.

Also, remember that the total work done ($\int F.dl$) is the change in kinetic energy of the body the work is done on!

(The total biochemical energy expended is certainly much more in the case of the Elephant, as you're applying the same force over a much longer time. The energy gained by it will be the same as the energy gained by the pen, though.)

• Oh! You're talking about the Energy you expend, and not the energy gained by the elephant? Then yes. You'll definitely lose more biochemical energy pushing for a longer time. – Hritik Narayan Jun 26 '17 at 15:21
• So Work energy theorem is for Energy gained by the object on which work is done! Ah makes sense. Thank you SO much man. You might want to update you answer with this information. You literally clarified my misconception. Though you might want update answer so others can find it helpful, in case they too had the same doubt I had. – Sushant Gupta Jun 26 '17 at 15:23
• Makes sense. I can't be more thankful to you for clarifying my understanding on this concept. Can't tell how excited I am to learn more Physics. – Sushant Gupta Jun 26 '17 at 15:28
• Glad to help! Hope to see you here with better and better questions. – Hritik Narayan Jun 26 '17 at 15:29
• Yeah "work" has to be one of the most misunderstandable terms in physics, because it really is not used the way it is used in English. "Change in potential energy", how about? I will never forget a grade school class we had to take called "Physical Education Lab" trying to teach us about the science of bodily exercise - the teachers and textbook tried to use physics' definition of work, and persisted that descending a flight of stairs took the same work as climbing - I almost convinced him to give us a ride to the top of a hill for us to walk down for a workout. – Dronz Jun 26 '17 at 20:57

Intuitively, it means you imparted 100 J of energy to the object to bring it to that position.

Your confusion arises from two sources (as far as I can tell):

1. You are conflating Work with Power
2. The definition of Work you are using.

Work is always independent of time. It by definition, describes the difference in energy of a system between two states. Take your elephant and pen for example. We'll take the mass of the elephant, $m_e$, to be $1000 kg$ and the mass of the pen, $m_p$, to be $1 kg$. Now, as you stated, you exerted $1 N$ of force of each object for $100 m$. Finally, we'll assume their initial velocity was $0 m/s$. Using kinematics, we may conclude the following:

1. You applied a constant acceleration, $a = \frac{F}{m}$ of to each object. The Elephant accelerated to $\frac{1N}{1000kg} = 10^{-3} m/s^2$ and the pen accelerated to $\frac{1N}{1kg} = 1 m/s^2$.
2. At $100m$, each object will have accelerated to $v_f = \sqrt{v_i ^2 + 2 ad} = \sqrt{2ad}$. The Elephant accelerated to $\sqrt{2 \cdot 10^{-3} m/s^2 \cdot 10^2 m} = 0.447 m/s$. The pen accelerated to $\sqrt{2 \cdot 1 m/s^2 \cdot 10^2 m} = 14.1 m/s$.
3. It took $t = \frac{v_f}{a}$ seconds to bring each object to 100m. For the Elephant it took $\frac{0.447 m/s}{10^{-3} m/s^2} = 447s$. For the Pen it took $\frac{14.1 m/s}{1 m/s^2} = 14.1 s$.

Notice: Because the elephant was 100x more massive than the pen, it took significantly more time to push it 100m, and with a constant force it was traveling at one-tenth the velocity as the pen.

Now if we calculate the calculate the difference in kinetic energy to move the pen and elephant 100m: $$T_e = \frac{1}{2} m_e v_{ef}^2 - \frac{1}{2} m_e v_{ei}^2 = 0.5 \cdot 1000kg \cdot (0.447 m/s^2)^2 - 0.5 \cdot 1000kg \cdot (0 m/s^2)^2 = 100 J$$ $$T_p = \frac{1}{2} m_e v_{ef}^2 - \frac{1}{2} m_e v_{ei}^2 = 0.5 \cdot 1 kg \cdot (14.1 m/s^2)^2 - 0.5 \cdot 1 kg \cdot (0 m/s^2)^2 = 100 J$$

You'll derive that the kinetic energy is identical. According to the Work-Energy theorem, the difference in kinetic energy of the system is equal to the total work done on the system.

Still confused? Take a look at the power used to move the objects:

$$P = \frac{W}{t} = \frac{Fd}{t} = Fv$$

You'll find that the power used to move the elephant was 0.224 Watts and the power used to move the pen was 7 Watts. You still expended the same amount of energy for either scenario; but, you expended energy at a faster rate pushing the pen, than pushing the elephant. This may seam counter-intuitive; but, remember, you exerted 1 N of force on a friction-less surface. On surfaces with friction, we must exert enough force to overcome static friction before the object will move.

EDIT: corrected erroneous calculations.

• There's a factor of 10 problem (elephant is 1000x as heavy, not 100) – Jim Garrison Jun 27 '17 at 6:14
• Thanks; I've corrected it. Turns out my power calculation was off as well. – KareemElashmawy Jun 27 '17 at 15:10