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What I understand is that work is not the same as a car using gas or a crane lifting a car high up into the air. Let's use the crane as an example. And let me write out a few lines from the book.

$$W_{\Sigma F} ~=~ \Delta E_k ~=~ \frac{1}{2}mv^2 - \frac{1}{2} mv_0^2. $$

So basically this means that there really isn't any work done by the crane. The start speed for the car being lifted is 0, and so is the end speed. It makes me cringe, because I want to write out the energy the crane used to lift the car. Same as the potential energy for that given car at that height. And the same goes for a car driving up a mountain at a constant speed.

$$E ~=~ E_0 + W_A.$$
This means that the mechanical energy ($E_p + E_P$) is the same after a change if you add whatever work is done by other forces. If I apply this to the crane lifting $E_0$ would be zero because there is no potential at height=0. Which implies the mechanical energy at the top is equal to the work done by the crane. But this thinking is wrong. I can't see the difference between a guy pushing a box something with a specific force a specific distance at a constant speed and the crane lifting this car under the same parameters. I think work is only applied when something is being accelerated, but then again the book shows example of how much work is done when picking up a backpack.

Ironically I can get around my lack of understanding and do the tasks needed doing, but I would very much like to understand this properly before my practical exam next week.

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Where did you get the first equation? It certainly doesn't hold in general, and I suspect that it was derived under a particular set of condition. Probably accelerating a mass without friction on a horizontal surface. Go back to the definition of work $\int \vec{F} \cdot \mathrm{d}\vec{x}$. –  dmckee May 17 '12 at 0:04
    
This question is not a question: the formula you write is wrong. Work is just the change in energy, it shouldn't be called something else, but it is. –  Ron Maimon May 17 '12 at 0:26
    
This chapter assumes movement only along a road or up and down. So we never need the cosine part of the work formula. Even so the book clearly states that there is only under two circumstances that physical labor doesn't count as work; when the movement is 0, and when the force is 90 degrees in respect to direction of movement. The origin of my confusion was an online chapter test at the site of the book, that's where it said that the work done by the crane was 0. I will look long and hard at this today. –  Algific May 17 '12 at 7:15
    
Specifically the question states that the work done on the car is 0 during the movement of the car and then it states that this is due to the kinetic energy being 0 at the start and 0 at the end. Is this wrong? In that case it's just that stupid question at the site that threw me off. –  Algific May 17 '12 at 7:22
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For a car moving horizontally (that is, at 90 degrees to the force of gravity) it's correct, but it's not correct for a crane lifting the car because the movement is not perpendicular to the force (gravity) acting on it. If it lifts the car $h$ metres, the work done is $gh$. –  Nathaniel May 17 '12 at 7:56
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2 Answers

Think of the following problem: An ice-skater starts at rest in the middle of a frozen lake. They speed up to a maximum velocity $v_f$ and then they drift over the ice without friction. After some frictionless sliding the skater goes over a region of ice with sand on it. The ice-with-sand surface has certain friction with the skate blades. The skater slides for a few meters and then stops due to this friction.

What is the minimum work done by the skater of mass $m$ to reach the velocity $v_f$?

What is the work done by the friction force in the region of ice covered with sand?

If you understand that, you will see that a crane is doing the work of the skater, to accelerate the body it is carrying until a given velocity (and it has to compensate for friction on top of that) and then it has to do extra work to stop it. If the crane has no internal friction and there is no friction with the environment, then everything is symmetric and the answer is quite easy, let say $W$. In the case of friction, nothing can be said in general but that the work $W_{fr} \geq W$.

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btw, if you can get this article, it may help –  JuanPi Jun 8 '12 at 16:30
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I think you're confused simply by not interpreting the Work-Energy Theorem correctly.

The work-energy theorem states:

$K_f - K_i = W_e + W_n + W_c$

...where $K_f - K_i$ is the change in kinetic energy of the body, $W_e$ is the work done by external forces, $W_n$ is the work done by non-conservative forces (lossy forces, e.g. friction, air resistance) and $W_c$ is the work done by conservative forces (e.g. gravity, force due to extension of a spring, etc.) Conservative forces are called so because they only depend on the position of an object.

Let's take your car and crane example. The work done by gravity is $W_c$. The crane is doing externally providing the force to lift the car. So the work done by it is $W_e$. As gravity is a conservative force, $W_c = -mgh$. (this negative sign is important). How do I know the work done is negative? That's because $mg$ or the force of gravity pulling the car down is opposite to the direction of displacement of the car, which is upwards. Gravity is trying to pull the car in the other way we want to pull it.

And, the change in $K$ is $K_f - K_i$, which as you said, is $0$.

Therefore, applying the WE Theorem:

$K_f - K_i = W_e + W_n + W_c$

As $K_f - K_i = 0$ and no non-conservative forces such as friction are assumed to be acting on the body, $W_n = 0$.

$0 = W_e + W_c$

$W_e = -W_c$

Hence,

$W_e = -(-mgh)$

or

$W_e = mgh$

So, we can see that the crane is doing work, equal to $mgh$ in magnitude!

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