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Chapter 7: Problem 16

How much energy can be stored in a spring with \(k=320 \mathrm{N} / \mathrm{m}\) if the maximum allowed stretch is \(18 \mathrm{cm} ?\)

Short Answer

Expert verified
Therefore, the amount of energy that can be stored in the spring is approximately \(5.184 J\).

Step by step solution

01

Convert centimeter to meter

The stretch given is in centimeters, but the spring constant is in N/m. We must ensure both quantities are in congruent units. So, convert 18 cm to meters. 1 cm = 0.01 meters, so \(18 cm = 18 * 0.01 m = 0.18 m\).
02

Substitute the values into the formula

Substitute 'k' and 'x' in the equation \(E_k = \frac{1}{2}kx^2\) with 320 N/m and 0.18 m respectively, then calculate.
03

Perform the calculations

On substituting, we get \(E_k = \frac{1}{2} * 320 N/m * (0.18 m)^{2}\). Solving this, we get \(E_k = 5.184 J\).

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