How to evaluate the work done by spring without applying calculus

[Nousher Ahmed]

In my textbook, the formula used to evaluate the value of work done by horizontally placed spring has been evaluated by applying the rules of integration. I want to evaluate it with another simple way except calculas. I want to learn the simplest way to evaluate it in such a way that even a boy of 12 years old can understand it without facing any difficulty.

 

[cnh1995]

Well, restoring force of a spring is proportional to it's displacement x from the origin. Energy stored in spring∝restoring force*displacement.
So, without calculus, you can only say that the spring energy is proportional to the square of displacement i.e. E=ax². But to calculate a, you should use calculus which yields a=k/2 and exact formula becomes E=½kx². I see no other way.

 

[nrqed]

In my textbook, the formula used to evaluate the value of work done by horizontally placed spring has been evaluated by applying the rules of integration. I want to evaluate it with another simple way except calculas. I want to learn the simplest way to evaluate it in such a way that even a boy of 12 years old can understand it without facing any difficulty.

The work done is given by the area under the curve in the graph of the force versus position. The force is kx (or -kx, it depends on whether you want the work done on the spring or by the spring). So the graph of force versus position is a straight line with slope k. The work done between, say x=0 and $x= x_f$ is the area of a right angle triangle with sides $x_f$ and $k x_f$. The area of that triangle is therefore $1/2 k x_f^2$.

 

[sophiecentaur]

I want to evaluate it with another simple way except calculas.

I wonder why. Calculus should be embraced as a way of answering questions and not rejected in favour of simpler approaches. As it happens, the energy in an ideal spring is represented by the area under a triangle but you can't go any further without calculus.
Not all intergrations are fiendishly difficult and there are books to help you when they start to get grisly.
A twelve year old boy would probably not have come across calculus at School but it would be a good idea to make him aware of it. It's easy to step around the calculus for linear relationships - spring stretching, constant acceleration etc.

 

[ZapperZ]

In my textbook, the formula used to evaluate the value of work done by horizontally placed spring has been evaluated by applying the rules of integration. I want to evaluate it with another simple way except calculas. I want to learn the simplest way to evaluate it in such a way that even a boy of 12 years old can understand it without facing any difficulty.

This is the 2nd post you've made on a similar topic, i.e. looking for a way to do something without using calculus. So do you have a natural aversion against calculus? And who is this 12-year old that you are trying to impress?

Zz.

 

[nrqed]

This is the 2nd post you've made on a similar topic, i.e. looking for a way to do something without using calculus. So do you have a natural aversion against calculus? And who is this 12-year old that you are trying to impress?

Zz.

Wow... I took the initial post as meaning that the OP wanted to teach some physics to a young student without using calculus (and I personally do not find it unreasonable at al to not have to introduce calculus at that age).
I guess I should be more mean spirited and assume that people are trying to impress students with their knowledge when they ask questions. But hold on...would it not have been more impressive to the student to go full calculus to do this simple problem? So why ask to do something without calculus if the goal is to impress? I am confused :-)

There is almost always a more mathematically sophisticated way to do simple physics problem but it does not mean that it is necessary or better. One can do quantum mechanics using symplectic geometry and fiber bundles but why on Earth would one insist on doing it this way when teaching for the first time quantum mechanics?

 

[Chestermiller]

If x is the amount that the spring is compressed, the force to compress it is kx. To start with, x = 0 and the force is zero. At the end of the compression, the force is kx. So the average force during the compression is kx/2. The displacement is x, so the amount of work is (kx/2)x=kx²/2.