Question

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1. 1.04 / 2.08 points

suppose that 6 j of work is needed to stretch a spring from its natural length of 28 cm to a length of 45 cm.
(a) how much work (in j) is needed to stretch the spring from 36 cm to 40 cm? (round your answer to two decimal places.)
(b) how far beyond its natural length (in cm) will a force of 15 n keep the spring stretched? (round your answer one decimal place.)

Explanation:

Step1: Find the spring - constant \(k\)

The work done in stretching a spring from its natural length \(x = 0\) to a displacement \(x\) is given by \(W=\int_{0}^{x}F(s)ds=\int_{0}^{x}ksds=\frac{1}{2}kx^{2}\). The spring is stretched from its natural length of \(28\) cm to \(45\) cm, so the displacement \(x=(45 - 28)\text{ cm}=17\text{ cm}=0.17\text{ m}\), and \(W = 6\text{ J}\). Using \(W=\frac{1}{2}kx^{2}\), we have \(6=\frac{1}{2}k(0.17)^{2}\). Solving for \(k\):
\[k=\frac{2\times6}{(0.17)^{2}}=\frac{12}{0.0289}\approx415.22\text{ N/m}\]

Step2: Calculate the work done in stretching from \(36\) cm to \(40\) cm

The natural - length is \(28\) cm. The displacements are \(x_1=(36 - 28)\text{ cm}=8\text{ cm}=0.08\text{ m}\) and \(x_2=(40 - 28)\text{ cm}=12\text{ cm}=0.12\text{ m}\).
The work done \(W=\int_{x_1}^{x_2}ksds=\frac{1}{2}k(x_2^{2}-x_1^{2})\).
Substitute \(k = 415.22\text{ N/m}\), \(x_1 = 0.08\text{ m}\), and \(x_2 = 0.12\text{ m}\) into the formula:
\[W=\frac{1}{2}\times415.22\times((0.12)^{2}-(0.08)^{2})\]
\[W = 207.61\times(0.0144 - 0.0064)\]
\[W=207.61\times0.008\]
\[W = 1.66\text{ J}\]

Answer:

\(1.66\)