91Ó°ÊÓ

Hooke's Law states that the force exerted by a spring is proportional to its displacement from the equilibrium position. Mathematically, this can be represented as: \(F = k \Delta x\) Where: - \(F\) is the force exerted by the spring - \(k\) is the spring constant - \(\Delta x\) is the displacement from the equilibrium position In this problem, the baby's weight acts as the force exerted on the elastic cord. The baby's weight is given by: \(F = mg\) Where: - \(m\) is the mass of the baby, \(6.8\,\mathrm{kg}\) - \(g\) is the acceleration due to gravity, \(9.81\,\mathrm{m/s^2}\) We can calculate the force (or weight) of the baby: \(F = (6.8\,\mathrm{kg})(9.81\,\mathrm{m/s^2}) = 66.708\,\mathrm{N}\) Now we can find the spring constant by rearranging Hooke's Law and substituting the known values: \(k = \frac{F}{\Delta x} = \frac{66.708\,\mathrm{N}}{0.2\,\mathrm{m}} = 333.54\,\mathrm{N/m}\)

When the mother pulls the seat down by \(8.0\,\mathrm{cm} = 0.08\,\mathrm{m}\), the potential energy stored in the spring is: \(PE_{spring} = \frac{1}{2} k \Delta x^{2} = \frac{1}{2} (333.54\,\mathrm{N/m})(0.08\,\mathrm{m})^2 = 0.534\,\mathrm{J}\) As the baby is released from rest, the total initial energy is the potential energy in the spring: \(E_{total} = 0.534\,\mathrm{J}\) When the baby reaches its maximum speed, all the potential energy in the spring will have been converted to kinetic energy: \(KE = \frac{1}{2} mv^2\) Solving for the maximum speed (\(v\)), we get: \(v = \sqrt{ \frac{2KE}{m}} = \sqrt{ \frac{2(0.534\,\mathrm{J})}{6.8\,\mathrm{kg}}} = 0.453\,\mathrm{m/s}\)

The period (\(T\)) of oscillation in a spring-mass system is given by: \(T = 2 \pi \sqrt{\frac{m}{k}}\) Substituting the mass and spring constant in the formula, we get: \(T = 2 \pi \sqrt{\frac{6.8\,\mathrm{kg}}{333.54\,\mathrm{N/m}}} = 0.860\,\mathrm{s}\) Now, we can answer both parts of the exercise: (a) The period of the motion is \(0.860\,\mathrm{s}\). (b) The maximum speed of the baby is \(0.453\,\mathrm{m/s}\).