When a \(99.5-\mathrm{g}\) slice of bread is inserted into a toaster, the toaster's ejection spring is compressed by \(7.50 \mathrm{~cm}\). When the toaster ejects the toasted slice, the slice reaches a height \(3.0 \mathrm{~cm}\) above its starting position. What is the average force that the ejection spring exerts on the toast? What is the time over which the ejection spring pushes on the toast?

To find the work done by the ejection spring on the toast, we can use Hooke's law, which states that the force exerted by a spring is proportional to its displacement from its equilibrium position. The formula for Hooke's law is: \(F = -kx\) where \(F\) represents the force, \(k\) represents the spring constant, and \(x\) represents the displacement from the equilibrium position. We don't have information about the spring constant, but we can use the work done by the spring (using the work-energy theorem) to find the spring force exerted on the toast. The work-energy theorem states that the work done on an object is equal to its change in kinetic energy, which can be written as: \(W = \Delta KE\) In this case, the toast starts at rest and ends at rest (at the highest point it reaches), so its initial and final kinetic energies are both zero. Thus, we have: \(W = \frac{1}{2} kx^2\) Now, we need to relate the work done by the spring to the potential energy gained by the toast. Since the toast reaches a height of \(3.0 \mathrm{~cm}\) above its initial position, it gains gravitational potential energy, which can be written as: \(\Delta PE = mgh\) where \(\Delta PE\) represents the change in potential energy, \(m\) is the mass of the toast, \(g\) is the acceleration due to gravity, and \(h\) is the height gained. Now, we can set the work done by the spring equal to the change in potential energy: \(\frac{1}{2} kx^2 = mgh\)

To find the average force exerted by the ejection spring, we can solve the equation we derived in Step 1 for the spring constant \(k\). First, we can plug in the given values for the mass (\(m = 99.5 \mathrm{~g}\)), the height (\(h = 3.0 \mathrm{~cm}\)), and the spring compression (\(x = 7.5 \mathrm{~cm}\)), and noting that \(g = 9.81 \mathrm{~m/s^2}\). \(\frac{1}{2} k(0.075 \mathrm{~m})^2 = (0.0995 \mathrm{~kg})(9.81 \mathrm{~m/s^2})(0.03 \mathrm{~m})\) Now, we can solve for \(k\): \(k = \frac{2 \times (0.0995 \mathrm{~kg})(9.81 \mathrm{~m/s^2})(0.03 \mathrm{~m})}{(0.075 \mathrm{~m})^2} \approx 187.40 \mathrm{N/m}\) With the spring constant \(k\) found, we can now calculate the average force exerted by the ejection spring using Hooke's law: \(F = kx = (187.40 \mathrm{N/m})(0.075 \mathrm{~m}) \approx 14.055 \mathrm{N}\) So, the average force exerted by the ejection spring on the toast is approximately \(14.055 \mathrm{N}\).

To calculate the time over which the ejection spring pushes on the toast, we will again use the work-energy theorem and the kinematic equations. Since the toast starts at rest and reaches a height of \(3.0 \mathrm{~cm}\), we can write a kinematic equation for its vertical motion: \(v^2 = v_0^2 + 2gh\) where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(h\) is the height. Since the initial velocity is 0, we can use the final velocity: \(v^2 = 2gh\) The toast's final velocity can also be found using the average force exerted by the spring and the toast's mass: \(v = \frac{Ft}{m}\) where \(t\) is the time over which the ejection spring pushes on the toast. Now, we can set these two expressions for the final velocity equal to each other: \(\frac{Ft}{m} = \sqrt{2gh}\) Solving for the time \(t\), we have: \(t = \frac{m \sqrt{2gh}}{F} = \frac{(0.0995 \mathrm{~kg})\sqrt{2(9.81 \mathrm{~m/s^2})(0.03 \mathrm{~m})}}{14.055 \mathrm{N}} \approx 0.02224 \mathrm{s}\) Thus, the time over which the ejection spring pushes on the toast is approximately \(0.02224 \mathrm{s}\).