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Chapter 6: Problem 23

How much work does it take to stretch a spring with \(k=200 \mathrm{N} / \mathrm{m}\) (a) \(10 \mathrm{cm}\) from equilibrium and (b) from \(10 \mathrm{cm}\) to \(20 \mathrm{cm}\) from equilibrium?

Short Answer

Expert verified
The work done to stretch the spring (a) \(10 \mathrm{cm}\) from equilibrium is \(1 \mathrm{J}\) and (b) from \(10 \mathrm{cm}\) to \(20 \mathrm{cm}\) from equilibrium is an additional \(1 \mathrm{J}\).

Step by step solution

01

Calculate Work done for Part (a)

First, we calculate the work done, \(W\) to stretch a spring a distance of \(x = 10 \mathrm{cm} = 0.1 \mathrm{m}\) from equilibrium. The spring constant \(k\) is given as \(200 \mathrm{N/m}\). We substitute these values into our work done formula: \[W = \frac{1}{2}(200 \mathrm{N/m})(0.1 \mathrm{m})^2\]
02

Evaluate expression from Step 1

Simplify the expression calculated in step 1: \[W = \frac{1}{2}(200 \mathrm{N/m})(0.1 \mathrm{m})^2 = 1 \mathrm{J}\]. So, it takes \(1 \mathrm{J}\) of work to stretch the spring \(10 \mathrm{cm}\) from equilibrium.
03

Calculate Work done for Part (b)

Next, we calculate the work done to stretch the spring from \(10 \mathrm{cm}\) to \(20 \mathrm{cm}\) from equilibrium. This is essentially the work done to move the spring an additional \(10 \mathrm{cm} = 0.1 \mathrm{m}\) from the position in part (a). We substitute these values into our work done formula: \[W = \frac{1}{2}(200 \mathrm{N/m})(0.1 \mathrm{m})^2\]
04

Evaluate expression from Step 3

Simplify the expression calculated in step 3: \[W = \frac{1}{2}(200 \mathrm{N/m})(0.1 \mathrm{m})^2 = 1 \mathrm{J}\]. So, it takes an additional \(1 \mathrm{J}\) of work to stretch the spring from \(10 \mathrm{cm}\) to \(20 \mathrm{cm}\) from equilibrium.

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