Question

Suppose a force of \(15 \mathrm{N}\) is required to stretch and hold a spring \(0.25 \mathrm{m}\) from its equilibrium position. a. Assuming the spring obeys Hooke's law, find the spring constant \(k\) b. How much work is required to compress the spring \(0.2 \mathrm{m}\) from its equilibrium position? c. How much additional work is required to stretch the spring \(0.3 \mathrm{m}\) if it has already been stretched \(0.25 \mathrm{m}\) from its equilibrium position?

Short answer

Answer: The spring constant, \(k\), is \(60 \mathrm{N/m}\). The work required to compress the spring \(0.2 \mathrm{m}\) from its equilibrium position is \(1.2 \mathrm{J}\). The additional work required to stretch the spring \(0.3 \mathrm{m}\) if it has already been stretched \(0.25 \mathrm{m}\) from its equilibrium position is \(3.075 \mathrm{J}\).

Step-by-step solution

a. Finding the spring constant (\(k\)) #

We are given the force and displacement required to stretch the spring. According to Hooke's Law, we can find the spring constant \(k\) as follows: \(F = kx\) where \(F\) is the applied force, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. We are given \(F = 15 \mathrm{N}\) and \(x = 0.25 \mathrm{m}\), so we can solve for \(k\): \(k = \frac{F}{x} = \frac{15 \mathrm{N}}{0.25 \mathrm{m}} = 60 \mathrm{N/m}\) The spring constant, \(k\), is \(60 \mathrm{N/m}\).

b. Calculating the work required to compress the spring#

The work \(W\) needed to compress a spring can be expressed as: \(W = \frac{1}{2}kx^2\) Now, we want to know the work required to compress the spring \(0.2 \mathrm{m}\) from its equilibrium position. We already found the spring constant \(k = 60 \mathrm{N/m}\) in the previous step. We are given the new displacement \(x'= 0.2 \mathrm{m}\). Let's plug the values into the formula: \(W = \frac{1}{2}(60 \mathrm{N/m})(0.2 \mathrm{m})^2 = \frac{1}{2}(60 \mathrm{N/m})(0.04 \mathrm{m^2}) = 1.2 \mathrm{J}\) The work required to compress the spring \(0.2 \mathrm{m}\) from its equilibrium position is \(1.2 \mathrm{J}\).

c. Calculating the additional work required to stretch the spring further #

We want to find the additional work required to stretch the spring \(0.3 \mathrm{m}\) if it has already been stretched \(0.25 \mathrm{m}\). Let's first find the total work required for stretching the spring \(0.55 \mathrm{m}\) from its equilibrium position. We will use the work formula again: \(W_{total} = \frac{1}{2}k(x+x')^2\) Now, let's calculate the initial work, which was done to stretch the spring \(0.25 \mathrm{m}\) from its equilibrium position: \(W_{initial} = \frac{1}{2}kx^2\) Finally, we will find the difference between the total work and the initial work, which will give us the additional work: \(W_{additional} = W_{total} - W_{initial} = \frac{1}{2}k(x+x')^2 - \frac{1}{2}kx^2\) Plugging in the values for \(k\), \(x\), and \(x'\): \(W_{additional} = \frac{1}{2}(60 \mathrm{N/m})((0.25 + 0.3) \mathrm{m})^2 - \frac{1}{2}(60 \mathrm{N/m})(0.25 \mathrm{m})^2 = 4.95 \mathrm{J} - 1.875 \mathrm{J} = 3.075 \mathrm{J}\) The additional work required to stretch the spring \(0.3 \mathrm{m}\) if it has already been stretched \(0.25 \mathrm{m}\) from its equilibrium position is \(3.075 \mathrm{J}\).