Problem statement

An object of mass m is hung on the end of a spring of relaxed length l and spring constant k. The object is pulled down until the stretch of the spring is so. What is the length of the spring when the object next comes to rest? (i.e. how high does the object rise before it comes to rest? What is the maximum speed that the object reaches, and where does it occur?

Diagrams.

Initial state ("before")                 Object at max height                   Object at max speed

Mark the reference level you choose for gravitational PE clearly on the diagrams.

My  reference level is the horizontal line drawn across all three diagrams. It is at the end of the relaxed spring.
(Other choices are possible, of course.)

Set-up of solution

The objects in the system are:

spring, earth, mass on spring

The forms of mechanical energy that this system possesses are :

kinetic, elastic potential and gravitational potential energy.

The kinetic energy is greatest when the total potential energy is least.

Fill in the first two columns of the table below. If you don't know the value of an entry in the top 3 rows, give it a name (algebraic symbol).
 Quantity Before(initial state) Object at max height Object at max speed stretch of spring s0 - s1 s  = mg/k height of object above reference level -s0 s1 - s = - mg/k speed of object 0 0 v ___________________ _____________________ _________________ _________________ Spring PE ks02/2 ks12/2 ks2/2 = (mg)2/2k Gravitational PE -mgs0 mgs1 -mgs = - (mg)2/k Kinetic energy 0 0 mv2/2 _____________________ _________________ _________________ Total mechanical energy ks02/2-mgs0 ks12/2 + mgs1 ks2/2 - mgs + mv2/2 = mv2/2 - (mg)2/2k

Solution to first problem.

Set the total energy in the before state equal to the total energy in the "max height" state. You will get one equation so you can solve for only one unknown. If you have more than one unknown in your equation, go back and see if you can fix something in your table. Your answer should contain the given quantities:

m, k , g, l , so

only.

ks02/2 - mgs0 = ks12/2 + mgs1
k(s02 - s12)/2 =  mg(s1 + s0)

So either s1 = - s0 or
k(s0 - s1)/2 =  mg
s1 = s0-2mg/k.
Since s1 = -s0   is the intial state, the answer we need is   s1 = s0-2mg/k.

Solution to second problem

The second part of the problem is harder. We need to fmd where the PE is least. So first write an expression for the PE in terms of the stretch of the spring s. Note this is not so or sf but just some arbitrary, so far unknown, value of s. Then find the derivative dU/ds and set it equal to zero to find the value of s that gives the minimum PE. Check that it really is a minimum (not a maximum, for example). Look at the answer -does it make sense? Once you have the answer, go back to your table, and fill in the final column. Set the totals in the bottom row equal, and solve for vmax.

The potential energy function is:

U(s) = ks2/2 - mgs
Differentiating, we get:
dU/ds = ks - mg
and setting this equal to zero, we fins that s=mg/k corresponds to minimum PE and maximum KE.  (A quick look at the second derivative d2U/ds2 = k > 0 confirms that this is a minimum of the PE. Note also that the net force on the object is zero. )  Then we find the speed from:
mv2/2 - (mg)2/2k = ks02/2-mgs0
v = sqrt(ks02/m-2gs0+mg2/k)