Two springs with spring constants \(k_{1}\) and \(k_{2}\) are connected in parallel. (a) What is the effective spring constant of the combination? (b) If a hanging object attached to the combination is displaced by \(2.0 \mathrm{cm}\) 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 \(83(\mathrm{b}) .]\)

To calculate the effective spring constant of the combination of two springs connected in parallel, we'll use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement (or deformation) from its equilibrium (or relaxed) position: \(F = kx\). When two springs are connected in parallel, the forces exerted by each spring add up: \(F_\text{total} = F_1 + F_2 = k_1x_1 + k_2x_2\). Since the springs are connected in parallel and have the same displacement, \(x_1 = x_2 = x\). Thus, \(F_\text{total} = k_\text{eff}x\), where \(k_\text{eff}\) is the effective spring constant. Combining these equations, we get \(k_\text{eff}x = k_1x + k_2x\). Divide both sides by \(x\) to get the effective spring constant, \(k_\text{eff}\): \(k_\text{eff} = k_1 + k_2\)

To find the potential energy stored in the spring when displaced by 2.0 cm, we'll use the elastic potential energy formula: \(U = \frac{1}{2} kx^2\). We already have the displacement \(x=2.0 \,\text{cm}\), and we need to find the effective spring constant. Using the values given for \(k_1\) and \(k_2\), we find the effective spring constant: \(k_\text{eff} = k_1 + k_2 = 5.0\, \mathrm{N/cm} + 3.0\, \mathrm{N/cm} = 8.0\, \mathrm{N/cm}\) Now, plug the effective spring constant and displacement into the potential energy formula: \(U = \frac{1}{2} k_\text{eff}x^2 = \frac{1}{2}(8.0\, \mathrm{N/cm})(2.0\, \mathrm{cm})^2\) Calculate the potential energy: \(U = \frac{1}{2}(8.0\, \mathrm{N/cm})(4.0\, \mathrm{cm}^2) = 16\,\text{J}\) Thus, the potential energy stored in the spring when it is displaced by 2.0 cm is 16 J.