Work by a Spring Force - Understanding Uinitial and Ufinal

[hquang001]

Here it writes Wby spring= Uinitial - Ufinal. So is it equal to opposite of the change in potential energy ?

 

[Richard R Richard]

You can ask yourself, The work of the spring, on who is done? Well the increase or decrease in energy of that system depending on whether the potential of the decrease or increases, why? because when you include the spring as part of the system and you assume that the energy only transforms and does not escape from the system.

$E_f = W_r + E_i$

That is, $\Delta E = W_r$

But the one that delivers energy from the spring does so as a function of decreasing its potential and vice versa

$W = -\Delta U_r$

Then the conservation of mechanical energy holds

$\Delta E + \Delta U_r = 0$

 

[PeroK]

Here it writes Wby spring= Uinitial - Ufinal. So is it equal to opposite of the change in potential energy ?

Think of potential energy as "money in the bank"; and, work done as money you spend.

Money spent = initial balance - final balance

 

[vanhees71]

You can also simply use the equation of motion,
$$m \ddot{x}=-k x=-\frac{k}{2} \frac{\mathrm{d}}{\mathrm{d} x} (x^2).$$
Now multiply this equation with $\dot{x}$ and integrate the result from $t_1$ to $t_2$. What do you get?

 

[hquang001]

You can also simply use the equation of motion,
$$m \ddot{x}=-k x=-\frac{k}{2} \frac{\mathrm{d}}{\mathrm{d} x} (x^2).$$
Now multiply this equation with $\dot{x}$ and integrate the result from $t_1$ to $t_2$. What do you get?

Sorry but this is quite complicated for me, I've never seen this

 

[hquang001]

Think of potential energy as "money in the bank"; and, work done as money you spend.

Money spent = initial balance - final balance

So work done equal negative change of potential energy
and in case of kinetic energy, work done is equal to the change of kinetic energy
am i correct ?

 

[PeroK]

So work done equal negative change of potential energy
and in case of kinetic energy, work done is equal to the change of kinetic energy
am i correct ?

I never liked these general soundbites about work and energy. I prefer to think in terms of concrete examples:

A block sliding down a frictionless ramp; a block sliding down a ramp with friction.

I see the rules as reflecting the phenomena, not the other way round.

 

[Richard R Richard]

So work done equal negative change of potential energy
and in case of kinetic energy, work done is equal to the change of kinetic energy
am i correct ?

It would only be true if there are no other possible potentials of other forces or interactions that could be altered in the system, that is why I understand that @PeroK does not like the idea of defining it that way very much and I share his concern.
So if no other potential is altered and you also maintain idealized statement conditions such as no friction, then $\Delta K + \Delta W_r = 0$

 

[hquang001]

It would only be true if there are no other possible potentials of other forces or interactions that could be altered in the system, that is why I understand that @PeroK does not like the idea of defining it that way very much and I share his concern.
So if no other potential is altered and you also maintain idealized statement conditions such as no friction, then $\Delta K + \Delta W_r = 0$

Oh ok
Thank you

 

[sophiecentaur]

So is it equal to opposite of the change in potential energy ?

It's probably a good idea to think in terms of Energy being conserved (ignore any friction within the spring, which can change things anyway). Energy that's transferred from the extended spring (change in potential Energy) must all turn up in the device it's driving. If you do the process slowly enough (quasi-statically) then all the Energy from ('work done by ') the spring has to be 'done on' the driven device and you can ignore any Kinetic Energy.

Sorry but this is quite complicated for me, I've never seen this

The equations that have been written here will only apply to a simple, ideal spring (Hooke's Law) and the appropriate equations would actually be a lot more complicated, But there is no real need ( at least for starters) to understand the Maths in order to get the basic idea. People can get very bothered when Integral Signs turn up and can lose the message. Having said that, it's not possible to get much further without getting into the Maths so I am not offering you a free pass to avoid Maths for long.