91Ó°ÊÓ

To find the acceleration at maximum extension, we can use the Hooke's Law formula, which states that the force exerted by a spring is proportional to its displacement from equilibrium: \(F = -kx\) Where \(F\) is the force exerted by the spring, k is the spring constant, and x is the displacement from equilibrium. Now, using Newton's second law, we can find the acceleration (\(a\)) of the mass (\(m\)) at maximum extension: \(F = ma\) Combining Hooke's Law and Newton's second law, we have: \(-kx = ma\) Now, when the extension is maximum, the displacement is also maximum (\(x_{max}\)). So, we can rewrite and rearrange the equation to find the acceleration at maximum extension \(a_{max}\): \(a_{max} = -\frac{kx_{max}}{m}\) In this case, \(k = 25\,\text{N/m}\) and \(m = 1.0\,\text{kg}\). Since we haven't found the maximum extension yet, we will leave \(x_{max}\) in the equation.

To find the maximum extension, we can use the conservation of mechanical energy with no friction or air resistance as follows: \(E_{total} = E_{elastic} + E_{gravitational} = \frac{1}{2} kx_{max}^2 + \frac{1}{2}m \dot{x}_{max}^2 + mgx_{max}\) In this case, the total mechanical energy is equal to the potential energy stored in the spring when the body is at its maximum extension (\(x_{max}\)): \(\frac{1}{2} kx_{max}^2 = mgx_{max}\) Divide both sides by \(x_{max}\): \(\frac{1}{2} kx_{max} = mg\) Now, we can solve for the maximum extension: \(x_{max} = \frac{2mg}{k}\) Now, plug in the given values: \(x_{max} = \frac{2(1.0\,\text{kg})(9.8\,\text{m/s}^2)}{25\,\text{N/m}}\) Calculate the maximum extension: \(x_{max} = 0.784\,\text{m}\) Now that we have the maximum extension, we can plug this back into the equation for acceleration at maximum extension: \(a_{max} = -\frac{(25\,\text{N/m})(0.784\,\text{m})}{1.0\,\text{kg}}\) Calculate the acceleration at maximum extension: \(a_{max} = -19.6\,\text{m/s}^2\) So, the magnitude of the acceleration at maximum extension is 19.6 m/s², and the maximum extension of the spring is 0.784 m.