Question

question 1 (6 points)
the potential energy stored in a spring, in joules, is given by the formula: ( p=\frac{1}{2}kl^{2} ), where ( k ) is the spring constant (a measure of the stiffness of the spring), and ( l ) is the amount by which the spring has been stretched, in metres.
a. determine ( k ) when ( p = 18 ) and ( l ) is 3.
b. determine ( l ) when ( p = 30 ) and ( k = 240 ).

Explanation:

Step1: Substitute values into formula for part a

Given \(P = 18\), \(L = 3\), substitute into \(P=\frac{1}{2}kL^{2}\). So \(18=\frac{1}{2}k\times3^{2}\).

Step2: Solve for \(k\) in part a

First, simplify \(3^{2}=9\). The equation becomes \(18=\frac{1}{2}k\times9\), or \(18 = \frac{9}{2}k\). Multiply both sides by \(\frac{2}{9}\): \(k=18\times\frac{2}{9}=4\).

Step3: Substitute values into formula for part b

Given \(P = 30\), \(k = 240\), substitute into \(P=\frac{1}{2}kL^{2}\). So \(30=\frac{1}{2}\times240\times L^{2}\).

Step4: Solve for \(L\) in part b

Simplify \(\frac{1}{2}\times240 = 120\). The equation is \(30=120L^{2}\). Divide both sides by \(120\): \(L^{2}=\frac{30}{120}=\frac{1}{4}\). Take square - root: \(L=\sqrt{\frac{1}{4}}=\frac{1}{2}\) (since \(L\gt0\) as it's a length).

Answer:

a. \(4\)
b. \(\frac{1}{2}\)