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Consider there is a box of mass $m$ at rest on the floor. Most books give an example that we need to do a work of $mgh$ to lift the box $h$ upward.

If we analyze this work done, the external force acting on the box by us should be equal to the weight of the box. Therefore the net force is zero which in turn there is no acceleration. If there is no acceleration and the initial velocity of the box is also zero, how can the box move upward?

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4 Answers 4

up vote 25 down vote accepted

In introductory problems about work you're normally taught that it's force times distance:

$$ W = F \times x $$

and you treat the force as constant. If you look at the problem this way then you're quite correct that if the force is $F = mg$ then the box can't accelerate so it can't move. However a more complete way to define the work is:

$$ W = \int^{x_f}_{x_i} F(x) dx $$

The force $F(x)$ can be a function of $x$, and to get the work we integrate this force from the starting point $x_i$ to the final point $x_f$. Because $F(x)$ can vary we can make $F > mg$ at the beginning to accelerate the box then make $F < mg$ towards the end so the box slows to a halt again.

DavePhD comments that work is not a state function, and in general this is true. However in this case the work done is equal to the change in potential energy so as long as the box starts at $x_i$ at rest and ends at $x_f$ at rest we'll get the same work done regardless of the exact form of $F(x)$.

If you're really determined to have $F$ constant then start with $F > mg$ at the beginning and $F < mg$ at the end, then gradually reduce the initial value of $F$ and increase the final value to make the force more constant. This will cause the time taken to move the box from $x_i$ to $x_f$ to increase. The limit of this process is a completely constant value for $F$, in which case it takes an infinite time to move the box.

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yes, this is more correct than my answer. Giving the object kinectic energy would involve additional work on the object, but taking away the kinetic energy would then be negative work on the object. –  DavePhD Mar 11 '14 at 16:46
I'd say one notable exception where you can simply apply the first equation would be working against friction when the body is already in constant movement. Great answer though, in my degree we're often told the simplified version so this kind of questions pop up all the time. –  Francisco Presencia Mar 11 '14 at 19:11
"$\times$" may mislead to think of vector product, while work is a scalar product. Maybe change it to "$\cdot$"? –  Ruslan Mar 12 '14 at 9:13
@Ruslan: I used the $\times$ symbol because it's what non-nerds think of as the multiplication symbol. Hopefully we nerds will spot I used $F$ and $x$ instead of $\vec{F}$ and $\vec{x}$. –  John Rennie Mar 12 '14 at 9:22
Maybe consider using $y$ instead of $x$ because we are dealing exclusively with vertical motion. A newcomer may be confused when you say that work in this instance depends only on the values of $x_i$ and $x_f$, it would be much clearer IMO if the variable $y$ was used. –  Bryson S. Oct 11 '14 at 13:54

Newton's first law states that:

"An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

It is often called the law of inertia. So if You want to move an object with zero velocity, at first moment You have to apply a force slightly bigger than the weight of the box. When the box is in motion, the force You need to apply to make it move upwards is lower and equal to the box's weight (assuming that there is no air resistance).

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Please justify that answer. If lifting a box, assuming it's not stuck to the floor by a piece of discarded gum, why do you have to apply a force greater than W? Are you confused by static vs. dynamic friction when sliding a box along a surface? –  Phil Perry Mar 12 '14 at 16:26
If you apply a force equal to the weight, then the net force on the box is zero, and it's not going anywhere. But a force of $W + \epsilon$ will create a net force of $\epsilon$, so it will rise slowly. –  Henry Swanson Mar 12 '14 at 17:55

external force acting on the box by us should be a little more than the weight, otherwise indeed: no acceleration! So the mgh is really a lower limit. We need to accelerate. And, by the time we reach h, decelerate -- so we get this extra little bit of work back. But only in a physics sense.

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This is a question everyone asks at first because it intuitively seems like a contradiction. However, it is not.

Conceptual Examples

I think you are not far off but perhaps the third law is the one tripping you up, not the 1st... But anyway, here are some conceptual examples, which might help...

Example 1.

Consider the particle in the frame for a moment. Is it moving or is it still? Well, we know that:

  • A particle moving at velocity $v=0$ (in its own inertial frame) is at constant speed and constant acceleration because $$\frac{d v}{dt}=a$$

So if $v=0$ then it follows that $a=0$.

However, it is vital you make sure not to confuse this with a case when $a=0$, because in that case v could be $v=0$. Velocity might not be zero at all, the thing about constant acceleration is there is no change in speed because the individual forces acting on bodies in the system sum to zero.


Example 2.

  • A particle moving at velocity $v\approx c \approx 3\times 10^{8}\mathrm{ms^{-1}}$ (in its own inertial frame) is moving at constant acceleration, but it is definately moving and very, fast too! Though, it is likely to have negligible mass at that speed, don't worry about that for now. I am just trying to help you stop thinking of velocity and acceleration interchangeably (if that has been the source of the confusion)

Remember, we are talking about simple models involving conservation. So just because there is a reaction force in the system, that does not mean nothing in the system can move, but it does mean that net force, in the inertial frame of the system, $F_{net}=0$ which is not the same as velocity $v=0$ at all...


$$\frac{d p}{dt}=m\frac{d v}{dt}$$

Try to do some momentum conservation problems to help you get your head around the idea and recognise $a$, $v$, $x$ graphs of acceleration, velocity and displacement with respect to time, respectively.

What it means mathematically is that mass by the derivative of velocity is zero - or in other words: The change in momentum of the system is zero, which is different because the change in momentum is given by:

$$\frac{d p}{dt}=m\frac{d v}{dt}$$

Example 3

Imagine you are still for a moment and you find yourself in the path of a car driving towards you in a straight line at a constant speed of $20ms^{-1}$ you for some reason prefer to stay stationary (a pretty extreme hypothesis test!).

A collision happens between you and the car and you might expect to change your speed (from rest) pretty fast, and in the opposite direction on impact. You will do this at a ratio proportional to your initial speed and mass plus the speed and mass of the car equal to the final speed of you and the car (and once you get that. the next stage is getting familiar with varying mass problems - yay rocket science!)

$$m_1 u_1 + m_2 u_2=m_1 v_1 + m_2 v_2$$

Momentum is conserved: Alrthough you might be worse off than the car, this is so because the car has a larger mass

i.e. you go flying in one direction, because you are subject to the force of the car and the car is dented because of you but the net speed and mass of both of you combined is the same after the collision as it was beforehand

momentum $\frac{d p}{dt}=ma$:

$$\frac{d p}{dt}=m\frac{d v}{dt}=ma=F_{net}$$

The first law states that a body will move at constant speed and direction unless an external force causes the body to change speed and/or direction.

An external force is not (by definition) in the inertial frame of a body which is moving at constant speed (inside its own inertial frame as it goes...)

Incidentally, this idea of reference frames was first conceived by Galileo, when he came up with this notion of invariance.


These are simple models but generally, (I think) it is easier to appreciate and understand the mechanics when you get used to thinking of force as being a change in momentum rather than just thinking of it as $ma$:

Force is change in momentum and that the infinitesimal change in the velocity of a particle of mass, m is acceleration)

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protected by Qmechanic Mar 11 '14 at 23:36

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