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

Answer:

2.31J

Explanation:

the energy for a spring system is given by:

$E=\frac{1}{2} kx^2$

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:

$K_{1}=\frac{1}{2} (1530N/m)(0.0181m)^2=0.25J$

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

$x=5.79cm=0.0579m$

$K_{1}=\frac{1}{2} (1530N/m)(0.0579m)^2=2.56J$

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

$2.56J-0.25J=2.31J$