# In calculating work done by a constant force over a constant distance, why doesn't the subject's initial velocity matter? [duplicate]

Assume a point-mass $m$ is travelling in a straight line, and a force $F$ will act on $m$ (in the same direction as $m$'s velocity) over a constant distance $d$; why doesn't $m$'s velocity matter to the calculation of work done on $m$ by $F$? Work is defined such that, in this example, the work done by $F$ on $m$ is equal to $Fd$, but it seems that if $m$ were moving slower, it would spend more time in the field, allowing $F$ more time to act on $m$, thereby doing more work. In fact, if $m$'s velocity were very great, it would hardly spend time in $F$'s field at all (so very little work done). Maybe I misunderstand work; can someone address this confusion of mine?

## marked as duplicate by knzhou, Jon Custer, Yashas, GiorgioP, ZeroTheHeroMar 22 at 2:57

• Side note - this is why if you pull a tablecloth fast enough, everything will stay in place on the table. – Señor O Apr 11 '14 at 14:51
• @SeñorO actually I think that has more to do with impulse, and also with the limited capacity of static friction. – David Z Apr 11 '14 at 16:24

Well, you simply need to accept that work is given by Force time Distance, and it doesn't matter how long it takes.

For example, the work done on a mass $m$ lifted a distance $h$ against gravity with an acceleration $g$ is given by:$$W=F\times h=mgh$$

If you are told that someone is going to drop a $1$ kilogram mass on your head from a height of $10$ metres, you may well have a lot of urgent questions, but how long the evil dropper took to get the weight up there is likely not one of them.

In the case of your example, suppose you have an object with mass $m$ travelling at velocity $v_o$, when a force $F$ is applied for a distance $D$, after which it is travelling at a velocity $v_f$, having experience an acceleration $a$.

The definition of the various constant acceleration equations give us:$$v_f^2=v_o^2+2aD$$ Multiply by $m$, divide by $2$, and we get:$$\frac12 mv_f^2=\frac12 mv_o^2+maD=\frac12 mv_o^2+FD$$The LHS is the final kinetic energy, and the RHS is the initial kinetic energy plus the work done.

• It is good that you emphasize definition. A mathematical model becomes a physics model by imposing laws/postulates definitions that connect the mathematics to measurements. – anna v Apr 11 '14 at 5:34
• The fact that kinetic energy is proportional to $v^2$ seems like the kind of explanation OP is looking for. Although you accelerate more if it was going slow, each unit of speed is worth less energy, than when speed is higher. – Cruncher Apr 11 '14 at 13:58
• @Cruncher, precisely so! :-) – BlueBomber Apr 11 '14 at 19:01

Well, the reason it doesn't matter is that work is defined as

$$W = \int\vec{F}\cdot\mathrm{d}\vec{s}$$

so if you keep the force the same and the distance the same, this remains the same, regardless of what you do with the initial velocity.

Of course, that definition probably isn't particularly satisfying. So consider this: when an object is subject to a force, the rate at which that force changes its energy is given by the power,

$$P = \vec{F}\cdot\vec{v}$$

And since power is the rate at which the force transfers energy, the total work done will be the integral of power over time,

$$W = \int P\,\mathrm{d}t$$

or $W = P\Delta t$ for constant power.

When an object moves fast, then yes, it spends less time in the region with the force ($t = d/v$), but also that force produces more power $P = Fv$. These two effects cancel out exactly:

$$W = P\Delta t = (Fv)\biggl(\frac{d}{v}\biggr) = Fd$$

So the work done is the same either way.

• Unfortunately, appealing to power (whose definition in my physics tests all depends on the definition of work!) isn't particularly satisfying to me, either. I think the other responses that attempt to clarify by drawing a connection to kinetic energy more closely address my confusion. Thanks for the answer, though, I do appreciate it. – BlueBomber Apr 11 '14 at 18:58

As you describe, the definition of work is just: $W=F d$.

What you are confusing maybe is the rate of work $P$ and the force $F$. When you move fast, $P=Fv$ is larger, however the travelling time is shorter. let's consider we are moving in a constant velocity. Then:

$$W=Pt=Fvt=Fd$$

Independent of velocity.

• I would like to add that if you are moving slower in the presence of a single force $F$, the corresponding impulse is greater: Your change in momentum is greater. – BMS Apr 11 '14 at 5:43
• Yes, my answer is considering the work done by a particular force $F$. The velocity is unchanged because there must be some other forces balance this particular $F$. In presence of a single force $F$, the velocity will not be constant, this demonstration should be modified by using the difference between final and initial kinetic energy as other answers did. – an offer can't refuse Apr 11 '14 at 7:55
• Thanks for your answer, but I don't find appeals to power or its connection to work satisfactory, but others may. @BMS, that's very interesting... Maybe I am confusing the concepts of impulse and work. – BlueBomber Apr 11 '14 at 19:04

As you note, for a constant force acting on an object which moves in one direction, the work done is equal to $Fd$. One can see from the equation that work is not dependent on time, but only on force and displacement. In order to conceptualize this, you could think about the energy involved in the situation you describe.

When work is applied by an external force to the point mass, the kinetic energy of the mass will change such that $\Delta K=W$. For a slower moving point mass, a force applied for a certain distance will act for a longer time than a force applied to a faster moving particle, as you note. It would seem, as a result, that the effect the force has on the faster mass would be less than the affect had on the slower mass, and in a way this is true. The change in velocity of the faster moving point mass will be smaller than the change in velocity of the slower moving mass, because of the difference in time that the force is applied. However, the change in kinetic energy of both cases will be identical. Because of the work kinetic energy theorem: $$K_i + W=K_f$$we can say that if the change in kinetic energy for both cases is equivalent, then the work done in both cases must also be equivalent. No matter the amount of time the force is applied the work will be the same. In order to convince yourself of this, make up a problem for yourself where you apply a certain force to a particle over a certain distance, then calculate the resulting kinetic energy and velocity for both a particle moving slowly, and a particle moving quickly. You will find that though the change in velocity depends on the time the force is applied, the change in kinetic energy and the work are independent of this.

• Kinetic energy is the key. Why are the other answers getting votes? – C. Towne Springer Apr 11 '14 at 5:16
• It would seem the other answers are concerned more with the definition of work. I got the feeling that the OP was already familiar with the fact that time is not involved in the mathematical definition of work, so I went with a more conceptual approach. The other answers are perfectly fine answers, I just felt as if the answer 'Because it's defined that way,' wasn't really what the OP was looking for. – wgrenard Apr 11 '14 at 5:56
• OP here, and you're answer has come closest to expressing exactly my question/confusion. You're correct in assuming that I asked the question already knowing the definitions the other answers provided, and "because it's defined that way" isn't a satisfactory answer. Thanks for your explanation. I did try to compare kinetic energies, but got lost in equations and integrals (it's not that simple, surprisingly, and my physics text has no problem similar to this). I may give it another shot. – BlueBomber Apr 11 '14 at 18:54
• If you keep everything in one dimension using a constant force you shouldn't need integrals. – wgrenard Apr 11 '14 at 19:24

Without any math and considering only Newtonian model here, I would say that

if you move the inertial system at the same speed and direction as your mass point is moving, than you have no initial movement of the mass point

and the total force used for acceleration will be the same as if you calculated or measured it in the original inertial system.

Work done is also defined as change in kinetic energy of the body.

Since F is constant force so F/m=a is a constant acceleration of m.

So, $$v^2-u^2=2ad$$or$$mv^2/2-mu^2/2=2mad/2$$which is the work done by the force. The body has travelled d distance with accleration a in the force field assuming u was a constant velocity when it entered the field and v is final velocity when it leaves the field. Even if it had some acceleration say, a', then also$$v^2-u^2=2(a+a')d$$So the right hand side always remains constant because a,a',d and m are constants. So work done is same for all velocities. These things comes from the definition only which is $$Fd$$ So whether body is moving or at rest, if a constant force acts and body moves a distance d in the direction of F work done is$$Fd$$

Lots of good answers here, but most of them are pretty mathematical and not very intuitive. Lets consider a realistic example.

You're on the moon with a six shooter and some extra bullets. You are in a uniform field, and you make two point masses travel through the same distance by dropping a bullet with one hand and firing at the lunar surface with the other.

Both bullets speed up as they approach the lunar surface, under the influence of lunar gravity. Since the force is the same and the distance is the same, the work done by the moon's gravity on the two bullets is the same, despite the fact that the fired bullet went through that meter of gravitational field orders of magnitude faster. But despite that, the kinetic energy gained by both bullets is the same and the potential energy lost by both is the same.

Does that make it intuitively any more clear?