7.4 Hooke's Law

MIT 8.01 Classical Mechanics, Fall 2016
View the complete course: ocw.mit.edu/8-0...
Instructor: Dr. Peter Dourmashkin
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu

[@k1lg0re23]

I really appreciate that these videos are offered to the public for free, I hear so many people saying that they think 'education should be free' and I tell them that it already is, and yet it is saddening to me that there aren't more people taking advantage of the opportunity in front of them. There are only 1,153 views on this video as of the time that I am commenting..

[@vivekvisupandey5393]

Yes ...I am totally agree with you friend. These lectures are from world's finest institute MIT. Those who dream to get into MIT can benefited by these lectures. MIT is offering for free it is a noble work it is doing.

[@mohfa1806]

thank you for those graet great lessons... i salute you dr

[@veronicanoordzee6440]

So whether you extend or compress the spring it's potential energy (PE) goes up. But why does the calculation give another answer? Let's extend the spring. So we compare x>0 with x=0. We know that W = dPE = integral of F(x).dx = integral of - k.x.dx = - 1/2. x^2. The PE has gone down according to this calculation!?!? Can someone explain this?

[@adarshkulkarni9991]

Let us say the spring is extended by 1 cm. Now calculate the P.E stored in spring .You get 0.5 times the stiffness . So P.E is always increasing because it was zero (we assume)at rest position (or when there is no external force).
So now we compare it with compression x

[@JohnSmith-lt1rz]

In both cases the potential energy is positive (it increases as we expect). The PE is the work that YOU do in order to setup the system. Don’t forget that work is a dot product of your applied force and the displacement vector. In both cases your applied force is in the SAME direction as the displacement of the mass. Said another way, in both cases the angle between your applied force and the displacement vector is =0. So, the dot product of applied force and displacement is positive in both cases. Integrating these positive work elements gives 1/2kx^2 for both cases.
@veronica, what you’ve calculated above is actually the work done by the spring, which is indeed negative. When you are doing positive work on the mass, the spring is doing negative work. The PE given to the mass comes from the work that YOU do on the mass. So integral from x=0 to x, of F(x).dx = integral of kx dx=1/2kx^2