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Chapter 6: Problem 9

A spring has a spring constant of \(80 \mathrm{~N} / \mathrm{m}\). How much potential energy does it store when stretched by \(1.0 \mathrm{~cm} ?\) a) \(4.0 \cdot 10^{-3}\) J b) \(0.40 \mathrm{~J}\) c) 80 d) \(800 \mathrm{~J}\) e) \(0.8 \mathrm{~J}\)

Short Answer

Expert verified
Answer: (a) \(4.0 \cdot 10^{-3}\) J

Step by step solution

01

Identify the given information

We are given: - spring constant (k) = \(80 \mathrm{~N/m}\) - stretch length (x) = \(1.0 \mathrm{~cm}\)
02

Convert units if necessary

Since the spring constant is given in \(\mathrm{N/m}\), we need to convert the stretch length to meters: \(x = 1.0 \mathrm{~cm} \times \dfrac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.01 \mathrm{~m}\)
03

Apply Hooke's Law formula

Using the formula for potential energy (PE) stored in a stretched spring: \(PE = \dfrac{1}{2}kx^2\) Plug in the values of k and x: \(PE = \dfrac{1}{2}(80 \mathrm{~N/m})(0.01 \mathrm{~m})^2\)
04

Solve for the potential energy

Calculate the potential energy stored: \(PE = \dfrac{1}{2}(80 \mathrm{~N/m})(0.0001 \mathrm{m}^2) = 0.004 \mathrm{J}\)
05

Compare the result with the given options

The calculated potential energy is \(0.004 \mathrm{J}\) which is the same as \(4.0 \cdot 10^{-3} \mathrm{J}\). Therefore, the correct answer is (a) \(4.0 \cdot 10^{-3}\) J.

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