How was it?

[qshyrn]

Er what exactly is the question?

well there was a spring (vertical) when u put a 2 kg mass on it, it extends by 0.4m

and?

it said wats the spring k

[fdsfsgdfgdf]

Er what exactly is the question?

well there was a spring (vertical) when u put a 2 kg mass on it, it extends by 0.4m

and?

it said wats the spring k

  what is the magnitude of change in energy of figure a [unstreched spring] and figure b [spring moves downwards and is stretched by .4m]

[lachymm]

I dont understand how the formula f=kx and strain energy of 1/2kx^2 can give different answers unless there is a loss of energy

[schnappy]

An 'ideal' spring...

You needed to first find k, the spring constant, then explain the change in energy as the 2kg mass is lowered from the spring causing it to extend 0.4m.

The second question did say the change in the system or the change in the spring?

[Cthulhu]

Well if it were me I'd set it out


I made a negative for good measure to cancel out the - in the RHS
then just plug and play and you'd get



Whats so hard about that?

Then for the change in energy:





Edit: Sorry about that folks.

[Greggler]

it said find the change of total energy in the spring - mass system

[schnappy]

So presumerable from the mass not being extended, to then being extended 0.4m. So you should lose EGP, and gain Estrain... that's how I did it to get 4J change.

[fdsfsgdfgdf]

Well if it were me I'd set it out


I made a negative for good measure to cancel out the - in the RHS
then just plug and play and you'd get



Whats so hard about that?

Then for the change in energy:





Or am I missing something?

the problem is some people including my self used Us=GPE to find the constant, hence the variation in the change in energy

[schnappy]

I believe the alternate response was that the change is 0J since that EGP must become strain energy, and since it's an ideal spring there are no losses.

[Cthulhu]

Well if it were me I'd set it out


I made a negative for good measure to cancel out the - in the RHS
then just plug and play and you'd get



Whats so hard about that?

Then for the change in energy:





Or am I missing something?

the problem is some people including my self used Us=GPE to find the constant, hence the variation in the change in energy

Err.. I made a woopsie.... REVISED BELOW:



I made a negative for good measure to cancel out the - in the RHS
then just plug and play and you'd get



Whats so hard about that?

Then for the change in energy:






Sorry about that folks.

[rubiks]

I believe the alternate response was that the change is 0J since that EGP must become strain energy, and since it's an ideal spring there are no losses.

thats what I did. Maybe VCAA will be nice and say both answers are right!

[appianway]

The gravitational potential is NOT converted to strain energy completely. This is as if the mass had NOT been lowered gently, it would have been oscillating at that point as it would also have kinetic energy, however, the hand "forces" against this and hence it has no kinetic energy. So you have to include the new elastic strain energy and subtract the gravitational potential lost.

[Blakhitman]

But the only force acting on the block is still weight?

[fdsfsgdfgdf]

what i dont get is where does the 4J go? this is an ideal situation.

[appianway]

It's lowered gently so while it's being lowered, the weight force is NOT the only force acting.