Question

Two springs with spring constants \(k_{1}\) and \(k_{2}\) are connected in series. (a) What is the effective spring constant of the combination? (b) If a hanging object attached to the combination is displaced by \(4.0 \mathrm{cm}\) from the relaxed position, what is the potential energy stored in the spring for \(k_{1}=5.0 \mathrm{N} / \mathrm{cm}\) and $k_{2}=3.0 \mathrm{N} / \mathrm{cm} ?$ [See Problem \(83(\mathrm{a}) .]\)

Short answer

Answer: The potential energy stored in the springs is \(15 \, J\).

Step-by-step solution

Determine the effective spring constant of springs in series

Consider two springs with spring constants \(k_1\) and \(k_2\) connected in series. The effective spring constant \(K\) can be calculated using the formula: \[ \frac{1}{K} = \frac{1}{k_1} + \frac{1}{k_2} \] Rearrange the equation to solve for \(K\): \[ K = \frac{k_1k_2}{k_1 + k_2} \] We will use this equation to find the effective spring constant of the combination.

Calculate the effective spring constant for the given values

Now, we will substitute the given values \(k_1 = 5.0 \, N/cm\) and \(k_2 = 3.0 \, N/cm\) into the equation: \[ K = \frac{(5.0)(3.0)}{5.0 + 3.0} = \frac{15}{8} \, N/cm \] The effective spring constant of the springs in series is \(\frac{15}{8} \, N/cm\).

Calculate the potential energy stored in the spring for the given displacement

The potential energy \(U\) stored in a spring can be calculated using the formula: \[ U = \frac{1}{2} K x^{2} \] where \(K\) is the effective spring constant and \(x\) is the displacement from the relaxed position. We will use the effective spring constant found in step 2 and the given displacement of \(4.0 \, cm\) to calculate the potential energy stored in the spring.

Calculate the potential energy stored in the spring for the given displacement and effective spring constant

Now, substitute the effective spring constant \(K = \frac{15}{8} \, N/cm\) and the displacement \(x = 4.0 \, cm\) into the formula for the potential energy: \[ U = \frac{1}{2} \times \frac{15}{8} \times (4.0)^{2} = \frac{15}{8} \times 8 = 15 \, J \] The potential energy stored in the spring is \(15 \, J\).