Question

a spring has a spring of 120 newtons per meter. calculate the elastic potential energy stored in the spring when it is stretched 2.0 centimeters.

Answer 1

We know that \(F = kx\) for a linear spring. Now we need to come up with an expression for how much work we exerted on the spring over that 2.0 cm. Since energy is the potential to do work, and the spring is assumed to be perfectly elastic (it returns to the same equilibrium point after being compressed). Work is defined as: \[W = \int\limits\limits \bf{F} \it dx\]Substituting in our expression for the force of the spring, we get:\[W = \int\limits {\rm kx} dx\]Integrating:\[W = {1 \over 2} k (\Delta x)^2\]Substituting in our unknowns:\[W = {1 \over 2} 120 \left [{\rm N \over m} \right ](0.02 [\text{m}])\]

Answer 2

is that the answer

Answer 3

I'm afraid you'll have to multiply it out.

Answer 4

why dont we use w=kx.x in this problem?why do we integrate it??
i agree that potential energy of spring is 1/2kxsquare but why are we integrating it?

Answer 5

The reason we integrate come from the very definition of work, which is \[W = \int\limits{\bf F} dx\]If we failed to integrate, and simply used \[W = {\bf F}~ x^2\]we would be off by a factor of \(1 \over 2\).

Answer 6

but often while calculating work done by gravity we say it is mgh without integration?

Answer 7

Let's look at how \(W = mgh\) is derived. Starting with the basic definition of work, \[W = \int\limits {\bf F} dx\]and the definition of force, in terms of gravitational acceleration, \[{\bf F_g} = m g\]Now let's combine the two \[W = \int\limits mg dx\]Note that mass and gravity do not depend on x, therefore, integrating yields. \[W = mgx\]where x = h in the case of gravity.

Now let's notice that the definition of force for a spring does depend on x. Integration is necessary because of this fact. Integration is only not necessary when the force is constant and does not change with the direction we are considering for work. (In the above case, this direction is x.)