Property of differentials that allow you to ignore this directional change?

[Whitishcube]

Hi, my question is sort of a general work problem.

using that work is equal to the integral from x initial to x final of F dot dl, I'm having trouble trying to visualize why this works for a spring.

assuming, for example, a spring is stretched from equilibrium, the force of the spring is going to be in the negative direction (which i understand, since the force by a spring is -kx).

this is my problem. why isn't dl also a negative value? shouldn't it also be negative, like the force vector? it is, after all, an infinitesimally small change in the negative direction of the displacement. This agrees with my intuitive sense of work, because since the force vector and the displacement vector are both in the same direction, the work is positive. However, when defining dl as a positive value, its like saying theyre in the opposite direction, and therefore the work would be negative. Is there something I'm missing? some property of differentials that allow you to ignore this directional change? thanks for the input.

 

[PhanthomJay]

The work done by the spring is negative in your example. The spring, as it is being stretched, is exerting a force acting in the negative direction, but as it is being stretched, it is moving in the positive direction. Hence, work is negative.

Suppose the spring was being compressed. The spring exerts a positive force, but is moving in the negative direction, so again, the work is negative in this example.

The plus and minus signs are always tricky to handle.

 

[Whitishcube]

Okay I don't think I was clear enough. In my example, the the spring is starting stretched and returning to equilibrium. So the work should be positive..

 

[PhanthomJay]

Okay I don't think I was clear enough. In my example, the the spring is starting stretched and returning to equilibrium. So the work should be positive..

OK, then yes, correct, work is positive. In the equation W = F dot dl, that implies that Work is the magnitude , which is always a positive number, of F times the magnitude of dl times the cos of the angle beteween the force and displacement. In your example, the angle is 0 (force and displacement in same direction), so the work is positive.

 

[Whitishcube]

Yeah I get the idea. I know that it is positive. But my question is how does the work integral play into this? F dot dl has a negative sign in it. If F is negative, then dl has to be positive. But my intuition is that dl should be negative because its the differential of the displacement (which is negative!). So why isn't dl negative?

 

[Whitishcube]

so this is what I'm trying to say.
I know that

$$W = \int_{x_i}^{x_f} \vec{F} \cdot d\vec{\ell}$$

and

$$\vec{F} = -kx\hat{\imath}$$

but here's my problem... the displacement is negative, so shouldn't it be

$$d\vec{\ell} = -dx \hat{\imath}$$
?

my reasoning is that the displacement is a negative vector, and therefore its differential should be negative. If the negative disappears when taking the magnitude of it, then the negative should disappear from the force vector too. What I don't get is where the negative sign comes from in

$$W = - \int_{x_i}^{x_f} kx dx$$

 

[PhanthomJay]

Yeah I get the idea. I know that it is positive. But my question is how does the work integral play into this? F dot dl has a negative sign in it. If F is negative, then dl has to be positive. But my intuition is that dl should be negative because its the differential of the displacement (which is negative!). So why isn't dl negative?

There is no negative in F.dl in this example. Actually, it's F.dx to avoid confusion in lettering.
F.dx = F(dx)(cos theta), where F and dx are always positive numbers, and since theta is zero degrees when the force and displacement are in the same direction, then F.dx = + F(dx). Integrating from x=0 to x = l , then work = + (1/2)(F)l^2.

If , instead, the spring is stretched starting from it's natural unstretched length and displaced a distance l, then, F.dx = F(dx)cos theta, and since the force and displacement are in opposite directions, theta is 180 degrees, F.dx = F(dx)(cos 180) = (F)(dx)(-1) = -F(dx). Integrating from x = 0 to x = l, W = - (1/2)Fl^2 (work done by spring is negative).

Another way of looking at this is using the equation that Work done by a conservative force (like a spring force) is the negative of its change in potential energy, that is W_s = - {PE_final - PE_initial}. In your original example, since the final PE is 0, then work is positive;in the 2nd example I gave, initial PE is 0, so work is negative.

Those plus and minus signs are tough, I'll tell you.

 

[Whitishcube]

okay, i think I am getting the idea now. there shouldn't be any negative signs on the differentials.

there is a negative sign, but it flips the limits on the integral. meaning that the negative sign in the force vector changes the limits around. in my example, x final is less than x initial, so the integral would be negative. however, the negative value in the force vector comes and flips those two limits, hence making the integral positive...

aha! i get it now!
thanks for the help!