physics help please! has to do with spring potential energy?

Question

A vertical spring stores 0.962 J in spring potential energy when a 13-kg mass is suspended from it. You can now write two equations: one for the balance of forces on the mass, and another for the energy stored by the spring. This gives you two equations and two unknowns.

(a) How far is the spring stretched by the mass? (m)
(b) What is the spring constant of the spring? (N/m)
(c) By what factor does the spring potential energy change if the mass attached to the spring is multiplied by 4?

I thought I had to start off with 0.962 J = .5kd^2
but I'm not sure where to go from there...

any direction or help would be greatly appreciated. thanks!

Answer 1

First, you have a force balance equation for a hookean spring:

F=k*X where F=force, k=spring constant, X=spring extension

You also know that the force comes from a weight so F=m*g in this case.

Now you have the force balace equation:

m*g=k*X

Second, you have the energy equation for a stetched spring:

E=0.5*k*X^2

Now you have two equations containing two unknowns (k and X) all expressed in quantities that you do know.

Solve them algebraically.

(a) X=(2*E)/(m*g)

Do your own algebra to convince yourself. Then check the dimensions to make sure it makes sense. In physicals, dimension checks are REALLY useful to make sure you have the right answer. Finally plug in the number and compute.

(b) k=(m*g)^2/(2*E)

Do your own algebra to convince yourself. Check the dimension. Then plug in numbers to compute.

(c) Assume that the spring remains the same and the spring constant is the same, you will see from the force equation that spring extension is proportional to force, which is proportional to mass. Since mass increased to 4x, extension is going to increase to 4x. Now look at the energy equation. Energy is proportional to the square of spring extension. So since the spring extension is now 4x, the energy will now be 16x. So if the spring is the same and you increase the mass of the spring by factor of 4, the energy in the spring is going to increase by 16, square of the mass.

Answer 2

Energy stored in a spring is elastic potential energy.. Pe

a)
Pe = 1/2 k x ^2,
where k is the spring constant in N/m
and x is the distance stretched or compressed in m

the weight of the mass is 127.4 N
it is in equilibrium, so the forces are balanced. gravity down, the spring up.

Since a J = N*m

0.962 N*m / 127.4 N = 0.0076 m

b)
Since x = 0.0076 m, and we know Pe we can solve for k

k = (2 Pe) / x^2 which is 33310 N/m

c) by 4 as well.