In the movie Superman, Lois Lane falls from a building and is caught by the diving superhero. Assuming that Lois, with a mass of \(50.0 \mathrm{~kg}\), is falling at a terminal velocity of \(60.0 \mathrm{~m} / \mathrm{s}\), how much force does Superman exert on her if it takes \(0.100 \mathrm{~s}\) to slow her to a stop? If Lois can withstand a maximum acceleration of \(7 g^{\prime}\) s, what minimum time should it take Superman to stop her after he begins to slow her down?

To determine the force exerted on Lois, we'll use Newton's second law of motion, which states that the force F is equal to mass m times acceleration a. In this situation, her mass is \(50.0kg\) and the acceleration is obtained by calculating the change in velocity (from her initial terminal velocity to a full stop) divided by the time it takes to slow her down. The initial terminal velocity is given as \(60.0m/s\), and the time is \(0.100s\). The formula for calculating force is: \(F = ma\) where \(F\) = force exerted on Lois \(m\) = mass of Lois \(a\) = acceleration We need to find acceleration first. We can use the formula: \(a = \frac{v_f - v_i}{t}\) where \(a\) = acceleration \(v_f\) = final velocity (0m/s, since she is stopped) \(v_i\) = initial velocity (60.0m/s, given as her terminal velocity) \(t\) = time (0.100s, given)

Now, we'll calculate the acceleration by substituting the given values: \(a = \frac{0 - 60.0}{0.100} = \frac{-60.0}{0.100}= -600.0 \mathrm{~m}/\mathrm{s}^2 \) Now that we have the acceleration, we can find the force exerted on Lois.

Using the force formula, we'll substitute the values for mass and acceleration: \(F = (50.0 \mathrm{~kg})(-600.0 \mathrm{~m}/\mathrm{s}^2) = -30000 \mathrm{~N}\) The force Superman exerts on Lois to bring her to a stop is \(30000 N\) in the upward direction.

Now, we will find the minimum time needed to stop Lois without exceeding her maximum acceleration. Lois' maximum acceleration can withstand is given as \(7g'\), where g is the gravitational acceleration, approximately \(9.81 \mathrm{~m}/\mathrm{s}^2\). Thus, her maximum acceleration is: \(max\_a = 7 \times 9.81 = 68.67 \mathrm{~m}/\mathrm{s}^2\) Now, we have to find the minimum time necessary. We can rearrange the acceleration formula to find the time: \(t = \frac{v_f - v_i}{a}\) where \(t\) = minimum time to stop Lois \(v_f\) = final velocity (0m/s, since she will be stopped) \(v_i\) = initial velocity (60.0m/s, given as her terminal velocity) \(a\) = maximum acceleration (68.67m/s^2, as calculated above)

Now we substitute the values into the formula to find the minimum time: \(t = \frac{0 - 60}{68.67} = \frac{-60.0}{68.67} ≈ -0.874 \mathrm{~s}\) Since time should be positive, we will take the absolute value: \(t = 0.874 \mathrm{~s}\) Therefore, Superman should take at least \(0.874\) seconds to stop Lois after he starts slowing her down to ensure her maximum acceleration isn't exceeded.