Question

(a) If forces of magnitude \(5.0 \mathrm{N}\) applied to each end of a spring cause the spring to stretch \(3.5 \mathrm{cm}\) from its relaxed length, how far do forces of magnitude \(7.0 \mathrm{N}\) cause the same spring to stretch? (b) What is the spring constant of this spring? (c) How much work is done by the applied forces in stretching the spring \(3.5 \mathrm{cm}\) from its relaxed length?

Short answer

a) The elongation caused by a 7.0 N force on the spring is approximately 4.9 cm. b) The spring constant of the spring is approximately 142.86 N/m. c) The work done by the applied forces in stretching the spring 3.5 cm from its relaxed length is approximately 0.088 J.

Step-by-step solution

Calculate the spring constant k with the given data

Using the given information, we can write Hooke's Law as: \(5.0 \mathrm{N} = k (3.5 \mathrm{cm})\) Convert the elongation from cm to m: \(3.5 \mathrm{cm} = 0.035 \mathrm{m}\) Then, we can solve for the spring constant k: \(k = \dfrac{5.0 \mathrm{N}}{0.035 \mathrm{m}} \approx 142.86 \mathrm{N/m}\)

Calculate the elongation caused by a 7.0 N force

Using the calculated value of k in Hooke's Law: \(7.0 \mathrm{N} = 142.86 \mathrm{N/m} \cdot x\) Now, we can solve for the elongation x: \(x = \dfrac{7.0 \mathrm{N}}{142.86 \mathrm{N/m}} \approx 0.049 \mathrm{m}\) Converting the elongation from m to cm: \(x \approx 4.9 \mathrm{cm}\) Therefore, a force of 7.0 N causes the spring to stretch 4.9 cm from its relaxed length.

Calculate the work done by the applied forces in stretching the spring 3.5 cm from its relaxed length

We can use the formula for work done on a spring: \(W = \dfrac{1}{2}kx^2\) Substituting the values of k and x: \(W = \dfrac{1}{2}(142.86 \mathrm{N/m})(0.035 \mathrm{m})^2\) Now, we can calculate the work done: \(W \approx 0.088 \mathrm{J}\) Therefore, the work done by the applied forces in stretching the spring 3.5 cm from its relaxed length is approximately 0.088 J.