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What exactly is work? My book confuses me:

a force can lift an object to a height h, or it can accelerate an object through gravity. In all these cases, a force displaces an object and change the object's total energy.

The examples it gives confuse me. On one hand, there is lifting and on, accelerating through gravity.

I can imagine how lifting will change total energy (kinetic + potential). It essentially gives it more potential energy, and does not take from it any kinetic energy. But just dropping an object and let gravity do the work does not seem to me to increase total energy. Because the total amounts of kinetic and potential energy should be equal. If just let drop, the object will gain kinetic and lose some potential. Not really anything added to the total energy.

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  • $\begingroup$ For a good picture on work see here. Feynman explains it better then any of us ever will, although it might be a little to advanced. $\endgroup$ – john Jul 22 '15 at 21:40
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Work is transfer of energy from one system to another OR transformation of energy from one form to another. Either way, work does not create energy.

When I lift an object, I am transferring energy from my body/muscles to the object-earth system. The energy goes into potential energy of the object-earth system because the separation between the object and the earth increases.

When I drop an object, the energy stays in the object-earth system, but is transformed from potential energy to kinetic energy. The gravitational force does the work, i.e. produces the transformation of energy from one form to the other.

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  • $\begingroup$ I tend to agree more with transfer rather than 'spending' of energy. It preserves the only fact we know about energy - that it is conserved. Other than that we know absolutely nothing else as to what energy is $\endgroup$ – docscience Jul 23 '15 at 4:15
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    $\begingroup$ Well put, I think this is a good explanation, without the use of any mathematics. $\endgroup$ – john Jul 23 '15 at 13:58
  • $\begingroup$ Sorry but you are wrong. "Work is transfer of energy from one system to another OR transformation of energy from one form to another." This is incorrect. Work is the transfer of energy from somewhere to the kinetic energy. The work of a force $f$ obeys: $W_{f} = \Delta T_{f}$, where $\Delta T_{f}$ is the increment of kinematic energy due the force $f$. What happens is the if we have multiple forces, on force can delivary energy into kinematic energy of the body and other force can absorb energy from the same kinematic energy. $\endgroup$ – Nogueira Mar 7 '16 at 9:43
  • $\begingroup$ @Nogueira If I lift an object against gravity at constant speed, there is no change in anything's kinetic energy, yet work is being done. Also, if work were defined as changing kinetic energy, the work-kinetic energy theorem would have no content. I'm not sure what "kinematic energy" is. $\endgroup$ – pwf Mar 10 '16 at 19:55
  • $\begingroup$ You are right when you say that the Kinect energy duo not change. But the kinetic energy until play the hole of "channel" for the energy exchange between the hand and the potential energy. If you can't have the opportunity to change the kinetic energy then work can't be used to exchange energy. $\endgroup$ – Nogueira Mar 10 '16 at 23:37
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Work is defined as $\displaystyle\int F(d) \cdot d$ where $F(d)$ is the net force acting on the object. That's it, that's the definition.

You might ask; why do scientists talk about energy and work so much? The answer is that experimental science shows that the energy in a system is constant, and thus energy is a useful concept which can be talked about in a meaningful sense.

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  • $\begingroup$ Isn't that definition pretty specific? What about the work it takes to compress a gas? The formula you wrote for this application of work is meaningless. There must be a more general way to define it. $\endgroup$ – docscience Jul 23 '15 at 4:06
  • $\begingroup$ Your notation is a bit unclear, but I think you have an extra factor of d. @docscience: It takes a force to compress a gas (e.g. the force pushing on a piston in a cylinder if that's how the gas is being confined), and integrating that force along the path it's acting over is actually meaningful. Whether that's the definition or just a mathematical way of accounting for the work is a philosophical question, but it can certainly be computed. $\endgroup$ – pwf Jul 23 '15 at 5:26
  • $\begingroup$ @docscience How else would you define work to be able to compute it? $dW = \vec{F}\cdot d\vec{r}$ seems pretty general and fundamental to me. I'm interested in a more fundamental computational definition if it's possible. $\endgroup$ – Bill N Jul 23 '15 at 14:25
  • $\begingroup$ @BillN A counterexample - the thermodynamic work of a Carnot cycle where work is determined by the contour integration of a pressure-volume loop: $$W=\oint P dV$$ $\endgroup$ – docscience Jul 23 '15 at 15:13
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    $\begingroup$ Doesn't $dW = PdV$ arise from $dW = F\ dr = PA\ dr \rightarrow P dV$? $\endgroup$ – Bill N Jul 23 '15 at 16:21
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You're completely correct, letting an object fall under the influence of gravity will not change its kinetic+potential energy, it just transforms potential energy into kinetic energy, while leaving the total constant.

However, OTHER forces could cause the kinetic energy to increase without changing the potential energy. Imagine a flat ice rink, and a puck sitting in the middle of it. Initially, the puck has zero kinetic energy, and some potential energy depending on how high up your ice rink is. Then you whack it with a stick, and it goes sliding off! The puck is still at the same height, but now it has kinetic energy: the force (hitting it with a hockey stick) caused it to gain energy.

Force could also cause the puck to LOSE kinetic energy. For example, if the puck is sliding towards a goal and a goalie stops it, the force on the puck has SLOWED THE PUCK DOWN, thus decreasing its kinetic energy.

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  • $\begingroup$ Am I correct in thinking that the convention is, when a defenseman slaps the stationary puck we say that the defenseman has done work on (i.e., transferred energy to) the puck, and when the goalie stops the shot, we say that the puck has done work on the goalie? $\endgroup$ – Solomon Slow Jul 22 '15 at 21:55
  • $\begingroup$ That is correct $\endgroup$ – Jahan Claes Jul 23 '15 at 13:41
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Work is transfere of any type of energy into kinetical energy. $$ T=\frac{mv^2}{2} $$ where $v$ is the velocity of the particle in some referential frame and $m$ is the mass of the body.

When the body is lifted in a gravitational field a force $F$ is required to balance the gravitational force $F_g$. During the lift, the force $F_1$ are doing a positive work $W_{F_1}=|W_{F_1}|$ and gravity are doing a negative work $W_{F_g}=-|W_{F_g}|$. At any instant of time, the kinematic energy of the body is equal to (assuming that the body are taken from rest):

$$ T = \Delta T _{1}= W_{F_1} + W_{F_g} = |W_{F_1}| -|W_{F_g}| $$

Then, during the lifting, the work done by the force $F$ need to be greater than graviational work. When you finish the lifting, the boddy is stoped, and this is an acceleration so we need a force to do that. Now a force $F_2$ is need to stop the body. This force make a work as well, need to enrase the kinetic energy:

$$ W_{F_2} = \Delta T _{2} = - \Delta T_{1} = -|W_{F_1}| + |W_{F_g}| $$

then $$ W_{F_1} + W_{F_2} = - W_{F_g} $$

This means, the work done by $F_1$ and $F_2$ is equal to the the negative of work done by gravity. By conservation of energy, the work done by gravity is equal the energy taked from the gravitational field.

$$ \Delta E_{g} = W_{F_1} + W_{F_2} $$

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Work is a definition of the expenditure of energy over time.

That's it.

It's not contained in the definition of any other physical quantity. One could make lots of examples that generate work with chemical reactions (internal combustion engines, explosives), mechanical dynamics (piston firing), electromechanical transformation (electric motor), and so on. Even potential energy in a gravity field like the height of a mass over the centre of the earth.

You are confusing closed system energy. Your closed system would NOT increase or decrease energy. It would transform from one to another form. What that poorly written book passage is saying I believe is that the total POTENTIAL energy changes by converting to other forms like kinetic (motion) and others like mechanical energy( sound waves emitted) and so on. Anything else is breaking the first law of thermodynamics. Energy is conserved and cannot be created nor destroyed.

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    $\begingroup$ No, your definition is closer to the definition of power. The amount of work done has no direct relationship to time. (Also, what is "expenditure" here?) $\endgroup$ – pwf Jul 23 '15 at 0:09
  • $\begingroup$ Yes - what does it mean to 'spend' energy. Since you cannot create or destroy energy, spending energy must just be a way to say you've moved it from some region of space to another. And in the process gained the advantage of moving, heating or compressing something. Right? $\endgroup$ – docscience Jul 23 '15 at 4:10

protected by Qmechanic Jul 22 '15 at 23:10

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