# Why work done is needed when other physical quantities are available?

I have some question about work done. I understand the mathematical ways and examples which are floating all over on internet & books. But all these information not clearing the concepts of work done as well energy. Please clear following questions:

1. If there are other physical quantities available why work done is needed. what is special about work done that other quantities can not give us.
2. "$$W = \vec{F} \cdot \vec{d} = F \cdot d \cdot\cos\theta$$ " In this formula what are details of both quantities.
(a) Is it applied force which causes displacement e.g: Force $$\vec{F}$$ is applied on a box on horizontal surface displace $$\vec{d}$$ with a rope parallel or at some angle?

(b) Gravitational Force Field $$\vec{F}$$ and we throw a ball upward $$\vec{d}$$ against the direction of force. In this case we displace the object under the influence of field, object is totally rely on field.

(c) Electrostatic Force Field $$\vec{F}$$ and we displace $$\vec{d}$$ object in this case we move the object by our own accelerating and moving direction ,the path of object is defined by us under influence of field.

Please clear all these point wise.

• Can you clarify your question more? What do you mean "other physical quantities"? Aug 12 '20 at 10:02
• Hello @Tachyon209. Physical quantities means Force, momentum , acceleration , velocity etc.. I know the clear difference and usefulness of these quantities conceptually. but work and energy is not clear conceptually in physics way. If we have all these quantities why we need work and energy what is advantage we can not determine by which quantities i mentioned.
– 123
Aug 12 '20 at 10:07
• I answered your unsolved question , it might be a little long but hopefully it will clarify the significance of work . Sep 2 '20 at 8:50
• Hello all, If anyone has Keppler & Kolenkiw (1 edition). Pls share in pdf by any convenient way.
– 123
Sep 5 '20 at 10:35

I don't agree with your statement that the $$F$$ in the formula $$W=Fd\cos\theta$$ gives complete information about the motion and displacement of a body, rather, partial or sometimes doesn't.

The concept of work in physics is much more narrowly defined than the common use of the word. Work is done on an object when an applied force moves it through a distance. In our everyday language, work is related to expenditure of muscular effort, but this is not the case in the language of physics. A person that holds a heavy object does no physical work because the force is not moving the object through a distance. Work, according to the physics definition, is being accomplished while the heavy object is being lifted but not while the object is stationary.

Say, for example, a man is pushing a train (purposely I mentioned train because practically no man can move a train by pushing alone) and he is applying all the strength i.e. he is applying force but the train will not move. This means that force acting on a body doesn't imply that the body is in motion.

Concluding, work gives us the idea that to what extent the motion of the body is changed or to what extent the force applied is useful to alter the motion of the body.

• Hello @SarGe your conclusion statement gives me answer to one condition when the object is in rest and i can extend this idea to moving object if we alter force field to achieve the particular path $\vec{d}$. I understand if displacement is zero work is zero. In my 2$^{nd}$ point i have given 3 condition pls clear these. Thanks for useful answer.
– 123
Aug 12 '20 at 9:42
• @123, it can be explained by the same logic that gravitational or electrostatic force acting on a body doesn't imply that the body is in motion. You need the concept of "Work" there, too. Aug 12 '20 at 10:05
• [Reply for 2nd comment]@123, if we are applying a force on body which is still not moving implies that there is another force acting on the body in the opposite direction of the applied force. Aug 12 '20 at 10:09
• No, the formula for work $\displaystyle\int \vec F\cdot d\vec x$ is valid for any type of force. It is simplified as $F\cdot(d\cos\theta)$ when force is constant. Aug 12 '20 at 10:18
• We are taking dot product of two vectors which will be a scalar, since we are taking projection of path along the direction of force, so we get there $\cos\theta$. Aug 12 '20 at 10:27

I'm only going to answer this question, but if you needed more or extra explanation notify me:

"I still found no satisfactory answer to this question about why work done is needed. Why we create this physical quantity. What is the additional benefit to create this quantity which cannot calculate by e.g.: Force etc."

Well in theory we don't need anything else more than the Newton's laws in order to study the motion of any moving body (in classical mechanics era) although Newton's laws apply on a point particle but we can solve the problem of the motion of any real-life object by thinking about it as a "gathering" of a lot (maybe infinite) of point particles.

Note : I'm only talking about theoretical foundations needed for solving such a system , surely we cannot do such heavy calculations in real life and that's the reason behind the development of Mechanics of Rigid bodies and Thermodynamics and so on , but as long as we are talking about the theoretical possibility of solving such systems , Newton's laws and the equation $$\vec{F} =$$ $$m\vec{a}$$ are all we need for solving a mechanical problem.

So the first motivations of defining something like Work was not conceptual, actually it was more a computational tool needed for solving harder problems (although later it lead to the concept of Energy and then got generalized beyond classical mechanics), I'll show you the math first and then explain its significance:

We know that for a particle with mass $$m$$ we have:

$$\vec{F}_{tot} = m\vec{a}=m\frac{\mathrm{d} \vec{v}}{\mathrm{d} t}$$ in which $$\vec{v}$$ is the velocity of the body which is $$\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}$$ and $$\vec{r}$$ is the position vector of $$m$$ .

Now dotting both sides by the differential of $$\vec{r}$$ , $$d\vec{r}$$ , and writing $$d\vec{r}$$ as $$\vec{v} dt$$: $$\vec{F}_{tot}\cdot {d\vec{r}} = m\frac{\mathrm{d} \vec{v}}{\mathrm{d} t}\cdot {\vec{v}dt}$$ $$\Rightarrow\vec{F}_{tot}\cdot {d\vec{r}} = m \vec{v}\cdot{d\vec{v}}$$ The right side of this equation is just $$m\frac{1}{2}d(v^{2})$$ therefore $$\vec{F}_{tot}\cdot {d\vec{r}} = \frac{1}{2}d(mv^{2})$$ Note that $$d\vec{r}$$ is the infinitesimal displacement of our point particle,so when you are considering bigger systems it will be the point where such a forces acts on it . now integrating over a curve that the point particle would take form the point $$\vec{r}_{i}$$ to $$\vec{r}_{f}$$ we have : $$\int_{i}^{f}\vec{F}_{tot}\cdot {d\vec{r}} = \frac{1}{2}mv^{2}_{f} - \frac{1}{2}mv^{2}_{i}$$

If we call the term $$\vec{F}_{tot}\cdot{d\vec{r}}$$ , $$dW$$ and writing $$\int {dW}$$ (note that this integral is in general path-dependent) as $$\Delta W$$ we would have: $$\Delta W = \frac{1}{2}mv^{2}_{f} - \frac{1}{2}mv^{2}_{i}$$ so we have found a scalar quantity which directly "connects" us to the (magnitude of) velocities rather than the other approach in which after solving the equation of motion we integrate $$\vec{a}$$ to find $$\vec{v}$$. but we had to integrate this other quantity called work so what's the point of this "pre-integrating" , well the good news is that for many important forces this integral is path-independent i.e. we don't have to evaluate it at all ! , for example in the case of a uniform gravitational field we just write $$mgh_{f}-mgh_{i}$$ now this really simplifies more complex problems that have such forces (even when they are not the only forces acting on our system) .

I recommend you to read the fourth chapter of An Introduction to Mechanics by Kleppner & Kolenkow(1st ed) which has very nice discussions on this subject and also comparisons of these two methods in solving the same problems , otherwise this answer would get even more lengthier than it is now !

– 123
Sep 3 '20 at 7:53
• I have downloaded Kleppner & Kolenkow (2nd ed). In this edition Chapter-5 is Energy. Thank you very much for your suggestion of this book. I am reading this a marvelous approach of book. explaining stories in short sentences.
– 123
Sep 3 '20 at 9:11
• @123 Sure i'd be happy to help you . I don't know much about the second edition but I've seen some omissions in it so I suggested the first one which is really good , I highly recommend the first 4 chapters of it even if you already know the material ! Sep 3 '20 at 12:14
• I can not find Kleppner 1-edition. If you have in pdf pls share with google drive or any other convenient way. Thanks this book is giving me almost answers which i was looking for.
– 123
Sep 3 '20 at 16:03
• After long long usage (old enough concept to understand) of work and energy idea. Is there any physical intuitive meaning of work and energy. Thanks you all.. Where you all guys are when i first asked this question. no one gave me the correct answer. Now this new six answers are real answers to my question. Specially you and J. Murray, I became hopeless to find my answer.
– 123
Sep 3 '20 at 16:14

Just to give a concrete example of a problem where work and energy are useful concepts, consider a ball placed on the following frictionless hill ($$x$$ is measured in meters): If the particle starts at rest at the top of the hill and is given a gentle nudge to the right, what speed will it approach as it escapes off to the right?

This is a straightforward question, but it would be a nightmare to solve using Newton's 2nd law.

You'd have to calculate the tangent vector to each point of the hill and find the component of the gravitational force along this vector just to set up your complicated differential equation, which probably doesn't have an analytic solution anyway. Once you had this equation and its solution, you'd have to find the velocity by taking a derivative, and then you'd need to take the limit as $$t\rightarrow\infty$$. This process would require a significant amount of mathematical skill and knowledge, and would probably take even a very motivated undergraduate quite a while to finish.

Alternatively, you could note that (i) the gravitational force is the only force doing work on your particle, and (ii) the gravitational force is conservative, so

$$\frac{1}{2}mv_f^2 = mg(h_i-h_f) \implies v_f = \sqrt{2(9.8\text{ m/s})(5\text{ m})}\approx 9.89\text{ m/s}$$

• Thanks @J. Murray to give me another beautiful idea. But using this way we can only only calculate magnitude of velocity (speed),displacement (distance), magnitude of acceleration etc.. we are unable to track the direction of particle when using energy. In Work done if Force vector and position vector it also gives us scalar. (1) If we want to measure Force Vector using work formula $\frac{W}{\vec{r}} = \vec{F}$ we cannot divide vector. What is the solution. (2) What is the meaning of scalar value of work.
– 123
Sep 3 '20 at 8:20
• @123 The function $\mathbf r(t)$, which gives the position of the particle as a function of time, contains all the dynamical information about the particle and its motion for every moment in time. However, we often don't need this much information. If I throw a ball in the air at some speed $v$ and want to know how high it goes, I don't need to know exactly how long it will take to reach the top of its arc. Sep 3 '20 at 15:22
• After long long usage (old enough concept to understand) of work and energy idea. Is there any physical intuitive meaning of work and energy. Thanks .. Where you all guys are when i first asked this question. no one gave me the correct answer. Now this new six answers are real answers to my question. Specially you and Kia.J.
– 123
Sep 3 '20 at 16:09

1."If there are other physical quantities available why work done is needed. what is special about work done that other quantities can not give us."

Work done is mathematically defined as the scalar product of the force $$\vec F$$ and the displacement $$\vec s$$. So
$$W=\vec F.\vec s$$ for constant forces. In case of variable forces we say something similar(I have just broken down the scalar product):$$W=\int F.dx +\int F.dy +\int F.dz$$ Here I am basically describing that total work done on an object is the sum of the work done by the individual forces acting on it resolved in coordinate axes of your choice.

So, what is special about work done that cannot be described by other physical quantities?

Work done as you might have guessed provides a relation between a force and the 'displacement of the object'(not necessarily caused by the force itself). That's why, we take the $$\vec F$$ and 'scale' it(stretching or compressing) to the $$\vec s$$ (or vice versa, but this makes more sense). You can imagine that we do this to show: just exactly 'how much' does this force contribute to the change in position of the object. We don't care about how fast the object changes its position(that's power), we just want to know what this force is doing in a system.

You can check, that no other physical quantities give us this relation, and the reason why it's needed in the first place is that: All forces contribute to net acceleration(that's what it means to be a force), but having information about its relation with displacement can tell us if individual forces are 'taking away' from the system, or 'putting something into' the system; this can be explained by the principle of energy, which tells us, not just about the current state of an object, but also about how this object will behave and interact with other objects in the future--bonus points because it is conserved all over the universe with NO EXCEPTIONS, it might as well be a base quantity!

$$(A)$$Hopefully I have described the purpose of $$F$$ and $$d$$ in the formula. In example (a), there are multiple forces acting on the system: Tension, gravitation, possibly friction, normal force by the box. You are correct, applied force at an angle does cause displacement; but does all of it goes into displacing the object? Of course not! Clearly a part of it goes into countering gravity and part of it is parallel. Is the part countering gravity doing work? No. This is not just because $$cos\theta$$ is $$0$$ at $$\pi/2$$ radians, but because it just makes sense! This part of the force neither takes away from the system, nor puts in anything. Think this one through.

$$(B)$$In part (b), we use the aforementioned reasoning again! Gravitational force field is indeed the only force acting on the system during the projectile motion, but its initial upward displacement comes from the external force applied by us. Therefore as it is going upwards(and gravity is slowing it down), gravitational force field is taking away from the system; the $$\vec s$$ and $$\vec F_g$$ are opposite in direction; which implies, work done by gravity is negative for the first part of the motion. But as it comes down, gravity contributes to motion, and work done is positive!

$$(C)$$In part (c), the exact same reasoning can be used. I will let you work this one out.

HINT: Once again, an applied force and a field force is acting on the system. Therefore, individual work done and total work done will be different.

• Hi @Reet Jaiswal, Thanks for your answer. What is in your answer it is really good. But after long period of time of thinking i have also reached the same idea what you discussed. Once again thanks for your support. But this arises more questions that counter this idea. Like we calculate displacement when Net Force is not equal 0. We can easily calculate every single parameter displacement, velocity, acceleration etc..
– 123
Sep 3 '20 at 8:05
• There is no extra benefit of creating work done. If only gravity is acting and i want to move object along the surface at some angle. Object doesn't move until applied force $F_{y} = F_{g}$. Then $F_{x}$ causing displacement. Why we need to multiply $F_{x}$ with displacement. No need after this information.
– 123
Sep 3 '20 at 8:07
• Thank you I understand what you ask and I probably have an answer somewhere at the back of my head but I don't think I can put it in better words than some other answers here:) Anyway I hope you have found your answer. Sep 3 '20 at 18:32
– 123
Sep 3 '20 at 18:38
• One last point I would still add though would be that work done is an EXTREMELY efficient method of fusing together Kinematics and Dynamics of newtonian bodies: where we would need at least 5 to 6 equations comprising 'F=ma' and the 3 equations of motion to reach at the answer, like finding final velocity or something, work-energy theorem provides wonderfully simplified explanations, majorly because computing scalars is just easier, and also because it is great at isolating forces in a system, as the uniformity of force fields really sits well with the idea of not knowing all the vectors Sep 3 '20 at 18:40

(b) Gravitational Force Field F⃗ and we throw a ball upward d⃗ against the direction of force. In this case we displace the object under the influence of field, object is totally rely on field.

Actually you do not displace this object, i.e. you do not apply constant force in it's whole movement trajectory. You just generate initial thrust, giving to object initial kinetic energy. Then this kinetic energy is gradually reduced by Earth gravitational field, until object reaches maximum height possible. If body initial speed is not greater or equal to escape velocity, then this body will fallback to Earth surface. So upon reaching maximum height, gravity will do a full-stop negative work to body, so :

$$E_k - W = 0$$

Substituting kinetic energy definition and work done by gravity gives :

$$\frac {m{v_o}^2}{2} - F_{grav}\cdot h = 0$$

From there you can express maximum height $$h$$ until which object will go upwards.

1 more thing here to understand in work.. That if the angle 'theta' is perpendicular than work done is 0....that's I mean wrong... Let's say when we carry say 50 kg load and stand at one point than we haven't done any work.... Irony... But we will feel tired.... 😬... Yeah to understand it properly we need to be more practical....

• Hello user, in this formula angle between force and displacement matter. because component of applied force parallel to the displacement (let say x-component) is useful other component (y-component) is not contributing in motion it is wastage.
– 123
Aug 12 '20 at 9:48
• The 'doing work' that we use in everyday language is not the same as what 'work' means in Physics. Sep 2 '20 at 2:22
• @user272129 Physics can be as practical as YOU want it to be. It is the literal description of the universe, and if you can't explain some phenomenon, then it means you have failed in describing it, not that physics itself is impractical Sep 2 '20 at 2:37