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Chapter 4: Problem 448

A spring with spring constant \(\mathrm{K}\) when stretched through $2 \mathrm{~cm}\( the potential energy is \)\mathrm{U}\(. If it is stretched by \)6 \mathrm{~cm}$. The potential energy will be...... (A) \(6 \mathrm{U}\) (B) \(3 \mathrm{U}\) (C) \(9 \mathrm{U}\) (D) \(18 \mathrm{U}\)

Short Answer

Expert verified
The potential energy of the spring when stretched by 6 cm is \( 9 \mathrm{U} \).

Step by step solution

01

Write the formula for potential energy of a spring

The potential energy U of a spring is given by the formula: \[ U = \frac{1}{2} kx^2 \] Here, U = potential energy k = spring constant x = extension of the spring
02

Find the relationship of the potential energy when the extension is doubled

When the spring is stretched by 2 cm, the potential energy is U. We can represent this situation as: \[ U = \frac{1}{2} k(2)^2 = 2k \] Now, the problem asks us what happens when the spring is stretched by 6 cm. We can represent this situation as: \[ U' = \frac{1}{2} k(6)^2 = 18k \]
03

Calculate the ratio of the new potential energy to the existing potential energy

To obtain the relationship between the new potential energy U' and the given potential energy U, we can divide the expression for U' by the expression for U: \[ \frac{U'}{U} = \frac{18k}{2k} = \frac{18}{2} = 9 \] Thus, the new potential energy U' is 9 times the initial potential energy U: \[ U' = 9U \] So, the correct answer is (C) \(9 \mathrm{U}\).

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Most popular questions from this chapter

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