Question

A block of wood is compressed 2.0 nm when inward forces of magnitude $120 \mathrm{N}$ are applied to it on two opposite sides. (a) Assuming Hooke's law holds, what is the effective spring constant of the block? (b) Assuming Hooke's law still holds, how much is the same block compressed by inward forces of magnitude \(480 \mathrm{N} ?\) (c) How much work is done by the applied forces during the compression of part (b)?

Short answer

Question: A block of wood compressed by an inward force of 120 N causes a compression of 2.0 nm. Assuming Hooke's Law holds, find the effective spring constant of the block. If the inward forces are increased to 480 N, find the new compression. Calculate the work done by the applied forces during this compression. Answer: The effective spring constant of the block is \(k = 6.0 \times 10^{10}\,\text{N/m}\). The new compression is \(x' = 8.0\,\text{nm}\). The work done by the applied forces during the compression is \(1.92\,\text{J}\).

Step-by-step solution

Find the spring constant

Using the given information, we know that the applied inward force is \(F = 120\,\text{N}\) and the compression is \(x = 2.0\,\text{nm}\) or \(2.0 \times 10^{-9}\,\text{m}\). Hooke's law states that \(F = kx\). We will solve for \(k\): $$k = \frac{F}{x} = \frac{120\,\text{N}}{2.0 \times 10^{-9}\,\text{m}} = 6.0 \times 10^{10}\,\text{N/m}.$$ The effective spring constant of the block is \(k = 6.0 \times 10^{10}\,\text{N/m}\).

Calculate the new compression

We are given a new applied inward force, \(F' = 480\,\text{N}\). We need to find the new compression (\(x'\)) using the known spring constant \(k = 6.0 \times 10^{10}\,\text{N/m}\) and Hooke's Law \(F = kx\). Solving for \(x'\): $$x' = \frac{F'}{k} = \frac{480\,\text{N}}{6.0 \times 10^{10}\,\text{N/m}} = 8.0 \times 10^{-9}\,\text{m}.$$ The new compression is \(x' = 8.0\,\text{nm}\).

Calculate the work done

To find the work done during the new compression, we will use the formula for work done on a spring: $$W = \frac{1}{2}kx'^2.$$ Substitute the known values of \(k\) and \(x'\): $$W = \frac{1}{2}(6.0 \times 10^{10}\,\text{N/m})(8.0 \times 10^{-9}\,\text{m})^2 = 1.92\,\text{J}.$$ The work done by the applied forces during the compression of part (b) is \(1.92\,\text{J}.\)