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Suppose you use a rope to hoist a box of tools vertically at constant speed v. The rope exerts a constant upward force of magnitude F_up on the box, and gravity exerts a constant downward force (the weight of the box). No other forces act on the box. To raise the box twice as fast, the force of the rope on the box would have to have:- A) the same magnitude $F_{up}$ B) a magnitude of $2F_{up}$

The answer seems to be (A), and I don't understand why! It says, if it's constant velocity, both upward and downward force = 0. Since the downward force is always the same, the upward force must be the same, too, regardless of the box's speed.

My take on that: But how are you gonna change the speed of the box without exerting force? If the force is the same, while there is one in the other direction, there would be no change of speed.

So, can you explain to me the correct answer?

EDIT: "Consider ... (1) starting the motion, (2) maintaining the motion, and (3) slowing the toolbox to a stop ..." (1) Starting. The object needs force greater than weight to go upward. (2) Maintaining. The object needs force equal to the weight to maintain motion upward. (3) Slowing. The object needs force less than weight for it to stop.

Ummm.. I don't know what to conclude with that? I understand one important thing: It can't be $F_{up} = W$, or in other words $F_{up} = F_{down}$, because there would be no acceleration in any direction. If there is no acceleration at any direction, I can't get the velocity to be twice as fast, or any amount faster.

EDIT: Thank you all. @Manishearth I could imagine this, but the book doesn't mention the "jerk", nothing about it in the question or in the explanation for the answer. (This is an example in the book). That's why I got confused. Anyway, again, thank you all for the clarification.

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Consider that the whole toolbox hoisting scenario includes three phases (1) starting the motion, (2) maintaining the motion, and (3) slowing the toolbox to a stop when it reaches the height of the presumed workman, and note that you are only examining phase 2. How might phases 1 and 3 differ in your two cases? Does that resolve your confusion? –  dmckee Feb 26 '12 at 23:55
    
BTW---By starting with what looked like a homework question you triggered my "close with extreme prejudice" reflex, but I think that this is a good question focused on conceptual matters. –  dmckee Feb 26 '12 at 23:57
    
I suspect that you are finding the answers you wrote in the edit to be really simple (almost trivial), but here's the thing ... that's it. The point of the problem is that sometimes physics is really simple. Give yourself a round of applause and move on. –  dmckee Feb 27 '12 at 1:03
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2 Answers 2

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Well, for the main duration of pulling it up, the force will be the same. You will have to apply "jerks" twice as strong (by 'jerk' I mean the common usage, not the abusive usage or the physics term for $\dot{a}$). Let's break it into three portions:

Initial Jerk

Initially, when you start to pull it, the box will receive a 'jerk', or 'impulse'. You will be applying an extremely large time varying (net) force $F$ in an extremely short time $\delta t$, such that the change in momentum of the body $\Delta p=m\Delta v=\int\limits_0^{\delta t} Fdt$. From this, one can see that if we apply a force twice as strong varying in the same manner, the change in velocity will also be twice as much. So in this leg, the force is twice as strong. Actually not exactly twice as strong, we need the net force to be twice as strong, but gravity is negligible here.

So in this leg, we have changed the speed of the block.

Pulling it up

Now, the body has already reached a velocity $v$. By Galileo's principle the body will continue to move with this velocity if the net force is zero. So here, we apply a force equal to the weight of the body.

Note that if we are considering viscous drag and friction, then the force will change, but won't be double.

Slowing it to a stop

This situation is pretty much the same as the first leg, so again the force here is twice as strong.

In conclusion, the force needs to be double only when you are making it reach the velocity $2v$, and slowing it back down. At all other points it is the same.

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As I understand it, I will have to first apply a force $F_{up}$ for some time $t$ such that the box accelerates upwards until it achieves a velocity $v$. Once I achieve $v$, I only need to apply enough force to match gravitational pull $= -Fg$. Assuming friction-less perfect worlds, there is no net force on the box and it continues at velocity $v$ forever. To move the velocity up to $2v$, I will now have to apply an additional $F_{up}$ for time $t$ to achieve $2v$ after which I again only need to apply $-Fg$.

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