Question

A spring with k = 15.3 N/cm is initially stretched 1.81 cm from its equilibrium length. a) How much more energy is needed to further stretch the spring to 5.79 cm beyond its equilibrium length?

Answer 1

2.31J

Step-by-step explanation:

the energy for a spring system is given by:

`E=(1)/(2) kx^2`

where
`k` is the spring constant:
`k=15.3N/cm=1530N/m` and
`x` is the distance stretched from the equilibrium position.

In the first case
`x=1.81cm=0.0181m`

so the energy to stretch the spring 1.81cm is:

`K_(1)=(1)/(2) (1530N/m)(0.0181m)^2=0.25J`

and for the second case, the energy to stretch the spring 5.79cm:

`x=5.79cm=0.0579m`

`K_(1)=(1)/(2) (1530N/m)(0.0579m)^2=2.56J`

so to answer a) we must find the difference between these energies:

`2.56J-0.25J=2.31J`