Why, when extending a spring, is it important not to accelerate it, for instance when hanging a weight from a spring, so that \( \vec{F} = -k \vec{x} \) ? Acceleration showed up in another thread recently, whose title eludes me for the moment. Anyway, this is true if dv/dt = 0 and v is constant, when obviously there will be a small amount of acceleration initially. Is it possible to keep this initial change from v=0 to v=const, small enough it can be ignored so it, you know, "vanishes"? Why is this a requirement? Sorry, this is pretty much a question for undergrads.
Please re-write the OP in a way that a human being may understand what is being assumed and what is being asked.
I think the easy answer is that the function \( F(\vec{x}) \) whose integral is the work done has to be linear over the domain, i.e. have a constant gradient. Otherwise the system will do work which isn't integrable? But I also believe this leads to a deeper reason, which is to do with covariance, since \( m \vec{g} \) is also constant over the domain and \( m \vec{g} = k \vec{x} \) at equilibrium.
I'm not sure why Example 8.3 concentrates on the work done by an external force, rather than on the work done by the spring. Depending on the acceleration, the external force must vary, but that doesn't change the work done by the spring in extending a particular distance.
If the acceleration is zero, the spring will have work done "on" it. If the spring does work by oscillating or "waggling" during the extension it won't be part of the integral for W, unless you extend the domain to (x,y,z)?
The work done by the spring is \(W = \int kx dx = \frac{1}{2}k\Delta x^2\) where \(\Delta x\) is the net amount by which the spring's length changes. This is independent of any external force applied or how the spring "waggles".
What's the work done by a weight hanging from a spring when x=const? The spring does work linearly if dv/dt = 0, which extends the spring along a single axis. If the spring moves along the y or z axes it will do work which won't extend the spring in the x direction. Likewise if it oscillates in the x direction only, the condition \( m \vec{g} = k \vec{x} \) won't be an equilibrium, the spring will do work (or gravity will) until x=const.
I think what it really means is constant displacement and adiabatic transport equals constant gradient of F(x), we "follow" \( \vec{x} \) with W. As long as we choose the x coordinate, kx varies with acceleration and we want to extend the spring so the work done "varying" the system is a minimum--it has to gain only the same energy, if possible, as the equivalent work done by (the integral of) mg over dx.
SInce W=∫F⋅dl, the answer is obviously none. No motion ⇒ no work. If dv/dt=0 the net force is identically zero, so once again there is zero work. Yet another way to look at this is via the work energy theorem, which states that work is equal to the change in kinetic energy. If dv/dt=0 there is no change in kinetic energy, hence no work.
Well, that's the whole point of the exercise. Lower the weight at constant velocity so the work done is equal to the change in energy, which is equal to the change in tension in the spring. But the weight and the spring have momentum during the extension. There is a change in potential in the spring, because work is done. When an object reaches terminal velocity in the atmosphere during free-fall, no work is being done on it?
Zero. No motion means no work, as D H said. The work done by the spring, as I said before, depends only on the starting and ending configurations of the spring, because the spring force is conservative. It doesn't matter one bit how the spring gets from the initial position to the final position. The work done is independent of the path.
No motion means no work is being done on the spring. What's the work done extending the spring by the weight which is now hanging suspended, if you assume the spring was not extended initially. Perhaps instead assume the question you and D H have answered was a sidetrack, and start again: And the easy answer is, it isn't important, unless you want the system to only gain the same potential as the displacement horizontally. That is, so the work done "by" the system equals the work done "on" the system by the field.
You said that it doesn't matter what path the system takes, but specified that the integral is in the x direction. So does it matter if the spring is extended by the weight, and no longer moves along the "x=vertical"; it looks like that's what I should have said in #13 instead of horizontal--if it moves horizontally the potential doesn't change. If it "orbits" in a way that stays on the same horizontally aligned plane, it won't do any work extending the spring, will it?
Which question is kind of the guttes, as it were, to the whole question of work and energy. In free-fall an object in a viscous medium will have its motion damped, so it eventually reaches a constant velocity after a time t. In other words the damping is a function of time and viscosity. Work is a function of space, or a space integral, so since a body in free-fall motion at v=const is moving through space it is doing work, or is having work done on it. Likewise if a spring is extended over time, at v=const, work is done by the spring and on the spring--these in fact are equivalent to a choice of coordinates, whereas momentum integrals, or kinetic energy, are time-coordinates. If the spring acts like a rigid connection once it is extended, so kx = mg, then kinetic motion in a horizontal plane does no work in the field conventionally, because there is no further extension/contraction of the spring in this case, and it stays at a constant tension. In other words, the system gains energy without increasing the tension in the spring, and you need to change the coordinates to an orbital frame of reference. However, if the weight is swinging instead it will follow a curved path and accelerate, which will do work on the spring, so it will stretch and relax as well making the weight's trajectory curve upward. So orbital motion makes this curvature vanish, so to speak.
Parallel transport is what rotates a vector when the transport is over a curved surface. So if a vector is rotated it has an equivalent amount of transport 'information', which is kept in the Riemann tensor. The hanging weight model is indirectly connected to what happens to a material object--an observer--as it follows a path through spacetime, and undergoes arbitrary velocity boosts and rotations. As the following diagram attempts to show, the boosts are like intrinsic curvature and rotations are like extrinsic curvature in this 'worldline' that is following the local gradient. Please Register or Log in to view the hidden image! http://en.wikipedia.org/wiki/Newtonian_motivations_for_general_relativity It says acceleration is equivalent to the local rotation of the velocity vector, which rotation is due to local 'parallel' transport of position/momentum.
