Question

a block is attached to one end of a spring with the other end of the spring fixed to a wall. the block is vibrating horizontally on a frictionless surface and has a maximum displacement from the equilibrium point, a.
if the 3.4 - kg block is moving 15 cm/s half - way between the equilibrium position and a, what is the total energy of the system?
hint: at the maximum displacement, all the energy is elastic potential energy.
your answer:

Explanation:

Step1: Write the formula for total energy

The total energy \(E\) of a simple - harmonic oscillator is \(E=\frac{1}{2}kA^{2}\). At a position \(x = \frac{A}{2}\), the energy is \(E=\frac{1}{2}mv^{2}+\frac{1}{2}k(\frac{A}{2})^{2}\). Since \(E=\frac{1}{2}kA^{2}\), we can substitute \(kA^{2} = 2E\) into the energy equation at \(x=\frac{A}{2}\).

Step2: Substitute and solve for \(E\)

Substitute \(kA^{2}=2E\) into \(E=\frac{1}{2}mv^{2}+\frac{1}{2}k(\frac{A}{2})^{2}\). We get \(E=\frac{1}{2}mv^{2}+\frac{1}{8}kA^{2}\). Then substitute \(kA^{2} = 2E\) again: \(E=\frac{1}{2}mv^{2}+\frac{1}{8}(2E)\). Rearrange the equation: \(E-\frac{1}{4}E=\frac{1}{2}mv^{2}\), \(\frac{3}{4}E=\frac{1}{2}mv^{2}\).
Given \(m = 3.4\space kg\) and \(v=0.15\space m/s\) (since \(15\space cm/s=0.15\space m/s\)), we have \(\frac{3}{4}E=\frac{1}{2}\times3.4\times(0.15)^{2}\).
First, calculate \(\frac{1}{2}\times3.4\times(0.15)^{2}=\frac{1}{2}\times3.4\times0.0225 = 0.03825\).
Then solve for \(E\): \(E=\frac{0.03825\times4}{3}\).

Answer:

\(E = 0.051\space J\)