Questions about a spring and its principle

[Sangari Indiraj]

Summary: I am confused in when to use 1/2 e squared k

we studied in class that normally, the extension of the spring is directly proportional to the tension applied
and we did this question :
what is the work that has to be done for a spring to have an extension of length e
as the area of the force extension graph gives the work done, we get the formula as 1/2ke².
but in the next question :
A weightless spring is attached to a ceiling and a weight of 20 N is applied. Here the spring extend for 5cm. Calculate the spring constant.
solution1:
f= ek
k= 20 / 0.05
=400 N m^-1
solution 2
according to the energy conservation theorum,
mgh = 1/2 e² k
20* 0.05=1/2 * 0.05 * 0.05 *k
k=800 N m^-1

why am i getting two different answers? It will help if anyone explains. Also please don't use derivatives. I cannot understand them.

 

[topsquark]

Summary: I am confused in when to use 1/2 e squared k

we studied in class that normally, the extension of the spring is directly proportional to the tension applied
and we did this question :
what is the work that has to be done for a spring to have an extension of length e
as the area of the force extension graph gives the work done, we get the formula as 1/2ke².
but in the next question :
A weightless spring is attached to a ceiling and a weight of 20 N is applied. Here the spring extend for 5cm. Calculate the spring constant.
solution1:
f= ek
k= 20 / 0.05
=400 N m^-1
solution 2
according to the energy conservation theorum,
mgh = 1/2 e² k
20* 0.05=1/2 * 0.05 * 0.05 *k
k=800 N m^-1

why am i getting two different answers? It will help if anyone explains. Also please don't use derivatives. I cannot understand them.

Your energy conservation equation implies that we are going to drop the mass. So the spring length e is going to be longer than its equilibrium length. When the mass is just hanging then the forces cancel out and you have the equilibrium length of the spring. The two problems are different.

-Dan

 

[PeroK]

... if you drop the mass, then it will oscillate under simple harmonic motion. The mass will in fact have its maximum speed, hence maximum KE, at the equilibrium point. When it has zero velocity, it will have fallen twice as far.

If you lower the mass into the equilibrium position, then you have an additional external force doing negative work on the system- in order to stop the mass accelerating under gravity.

 

[sophiecentaur]

Also please don't use derivatives. I cannot understand them.

Perhaps it would be better to spend your time getting to grips with fairly elementary Calculus. You could then have the answer to your question (and many other basic questions), all on a plate.
This link shows an easy graphical method to show where the 'half' factor comes from. This diagram is explained fairly well in the link.

Visualise stretching the spring a tiny bit at a time. The force will not change noticeably so the work done will be kx times the small stretch Δx. The total work done will be the sum of all those small stretches, which is represented by the shaded part of the triangle. Area is half base times height.

 

[malawi_glenn]

same graph as above, but imagine there was a constant average force instead stretching the spring x. This average force would be kx/2 and the work done would be kx²/2.
You are probably familiar with this method in finding the distance traveled when you have constant acceleration.

 

[sophiecentaur]

same graph as above, but imagine there was a constant average force instead stretching the spring x. This average force would be kx/2 and the work done would be kx2/2.

I don't think you actually meant that. The force to keep the spring at x would be kx (not half). If that force were applied over a distance x then the work would be kx². That's twice as much as if the force is build up gradually.

 

[sophiecentaur]

You are probably familiar with this method in finding the distance traveled when you have constant acceleration.

I used to have terrible problems with that until I stopped thinking of a practical car engine. No one stressed the 'constant acceleration' enough to get through my thick skull.

 

[malawi_glenn]

I don't think you actually meant that. The force to keep the spring at x would be kx (not half). If that force were applied over a distance x then the work would be kx². That's twice as much as if the force is build up gradually.

 

[sophiecentaur]

View attachment 314949

View attachment 314950

A picture is worth a thousand words but you described stretching ‘the spring’ by x. Hence the confusion.

 

[Sangari Indiraj]

I don't think you actually meant that. The force to keep the spring at x would be kx (not half). If that force were applied over a distance x then the work would be kx². That's twice as much as if the force is build up gradually.

thankyou this is where am confused.