Question

if 320 j of work is done on a spring with a spring constant of 730 n/m, how far will it stretch? 0.87 m 0.94 m 0.61 m 0.58 m

Explanation:

Step1: Recall the formula for work done on a spring

The work done on a spring is given by \(W=\frac{1}{2}kx^{2}\), where \(W\) is the work done, \(k\) is the spring constant, and \(x\) is the displacement (stretch in this case).

Step2: Solve the formula for \(x\)

Starting with \(W = \frac{1}{2}kx^{2}\), we can solve for \(x\). First, multiply both sides by \(2\) to get \(2W=kx^{2}\). Then divide both sides by \(k\): \(x^{2}=\frac{2W}{k}\). Take the square - root of both sides: \(x=\sqrt{\frac{2W}{k}}\).

Step3: Substitute the given values

We are given that \(W = 320\space J\) and \(k=730\space N/m\). Substituting these values into the formula \(x=\sqrt{\frac{2\times320}{730}}\).
First, calculate the numerator: \(2\times320 = 640\). Then, \(\frac{640}{730}\approx0.8767\). And \(\sqrt{0.8767}\approx0.94\space m\).

Answer:

0.94 m