Question

A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as measured from equilibrium) would result in 100 J being stored in this case?

Answer 1

Answer: 14.14 cm

Step-by-step explanation:

Here is an easier way to understand this.

The formula for spring potential energy is

Us = 1/2k (deltax)^2

k = spring constant

if you plug in 12 cm and 72 J

(and eliminate the k constant, because we're going to be looking at the same spring and it will cancel out (finding it will make it unnecessarily complicated, we want to find a ratio here; what compression would result in 100 J of potential energy?)

however,

you can see here that

72 = 1/2 (12.0)^2

and

72 = 72

so this is a 1:1 ratio directly related as according to HOOKES LAW

if you plug in

100 = 1/2(delta x)^2

you can easily solve for the one variable, and get 14.14 cm

Answer 2

Answer:14.14 cm

Step-by-step explanation:

Given

Spring Compression
`x=12 cm`

Potential energy Stored in spring
`=72 J`

Suppose k is the spring constant of spring

Potential Energy of spring is given by
`=(kx^2)/(2)`

`(k(0.12)^2)/(2)=72`

`k(0.12)^2=144`

`k=10,000 N/m`

`k=10 kN/m`

for 100 J energy

`(k(x_0)^2)/(2)=100`

`10* 10^3\cdot (x_0)^2=200`

`(x_0)^2=2* 10^(-2)`

`x_0=0.1414`

`x_0=14.14 cm`