1. A 0.50 kg block oscillates on the end of a spring with a spring constant of 150N/m If the system has an energy of 8.01, then what is the amplitude of the oscillation?

## Step 1: <br />The total energy of the system is given by the formula \(E = \frac{1}{2} kA^2\), where \(E\) is the energy, \(k\) is the spring constant, and \(A\) is the amplitude.<br /><br />## Step 2:<br />We can rearrange the formula to solve for the amplitude: \(A = \sqrt{\frac{2E}{k}}\).<br /><br />## Step 3:<br />Substitute the given values into the formula: \(A = \sqrt{\frac{2*8.01}{150}}\).<br /><br />## Step 4:<br />Calculate the value inside the square root: \(\frac{2*8.01}{150} = 0.1068\).<br /><br />## Step 5:<br />Take the square root of the result: \(A = \sqrt{0.1068} = 0.33\) m.