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Chapter 7: Problem 25

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?

Short Answer

Expert verified
Answer: Superman exerts a force of 30,000 N on Lois Lane to bring her to a stop, and the minimum time required to stop her without exceeding her maximum acceleration is 0.874 seconds.

Step by step solution

01

Calculate the initial force exerted on Lois

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)
02

Calculate the acceleration

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.
03

Calculate 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.
04

Calculate the minimum time needed

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)
05

Calculate the minimum time

Now we substitute the values into the formula to find the minimum time: \(t = \frac{0 - 60}{68.67} = \frac{-60.0}{68.67} 鈮� -0.874 \mathrm{~s}\) 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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration
When discussing motion, acceleration plays a critical role. It tells us how quickly an object changes its velocity. In the scenario with Lois Lane, acceleration occurs as Superman decelerates her from a high-speed fall to a complete stop. This change is vital to analyze. The basic formula used for calculating acceleration is: \[ a = \frac{v_f - v_i}{t} \] Where:
  • \(a\) = acceleration
  • \(v_f\) = final velocity (for Lois, it is 0 m/s, as she stops)
  • \(v_i\) = initial velocity (in this case, 60.0 m/s)
  • \(t\) = time period over which this change occurs (0.100 seconds in this problem)
By substituting in the known values, we see that Lois experiences a dramatic change in velocity over a very short time frame. This scenario highlights how acceleration isn't just about speeding up but also slowing down.
Force Calculation
Newton's Second Law of Motion provides a clear path to understand forces involved in motion. This law is captured using the equation: \[ F = ma \] In the Lois Lane example, it's necessary to calculate the force Superman must exert to safely stop her fall. Here's what's involved: - **Mass** - Lois's mass impacts how much force is needed. In this case, her mass is 50.0 kg.- **Acceleration** - Previously calculated as \(-600.0 \mathrm{~m/s}^2\) from her initial terminal velocity to a stop.By multiplying mass and acceleration, we derive the force. Substituting the values gives: \[ F = (50.0 \mathrm{~kg})(-600.0 \mathrm{~m/s}^2) = -30000 \mathrm{~N} \] The negative sign indicates the direction is opposite to Lois's fall, meaning the force is upwards. This immense force underscores the strength Superman needs.
Terminal Velocity
Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. For Lois Lane, as she falls, air resistance reaches a point where it balances the force of gravity, leading her to fall at a steady terminal velocity of 60.0 m/s. Understanding terminal velocity involves noting a few critical factors:
  • **Mass of the object**: Heavier objects have a higher terminal velocity given more force is required to balance gravity.
  • **Air resistance**: This force opposes motion and depends on shape, size, and surface, explaining why falling objects reach a steady speed.
In Lois's case, knowing her terminal velocity helps Superman calculate the right amount of force and the required time to decelerate her safely when catching her. This showcases how physics principles guide heroic acts even in movies.

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Most popular questions from this chapter

Two bumper cars moving on a frictionless surface collide elastically. The first bumper car is moving to the right with a speed of \(20.4 \mathrm{~m} / \mathrm{s}\) and rear-ends the second bumper car, which is also moving to the right but with a speed of \(9.00 \mathrm{~m} / \mathrm{s} .\) What is the speed of the first bumper car after the collision? The mass of the first bumper car is \(188 \mathrm{~kg}\), and the mass of the second bumper car is \(143 \mathrm{~kg}\). Assume that the collision takes place in one dimension.

Consider two carts, of masses \(m\) and \(2 m\), at rest on a frictionless air track. If you push the lower-mass cart for \(3 \mathrm{~s}\) and then the other cart for the same length of time and with the same force, which cart undergoes the larger change in momentum? a) The cart with mass \(m\) has the larger change b) The cart with mass \(2 m\) has the larger change. c) The change in momentum is the same for both carts. d) It is impossible to tell from the information given.

Two Sumo wrestlers are involved in an inelastic collision. The first wrestler, Hakurazan, has a mass of \(135 \mathrm{~kg}\) and moves forward along the positive \(x\) -direction at a speed of \(3.5 \mathrm{~m} / \mathrm{s}\). The second wrestler, Toyohibiki, has a mass of \(173 \mathrm{~kg}\) and moves straight toward Hakurazan at a speed of \(3.0 \mathrm{~m} / \mathrm{s} .\) Immediately after the collision, Hakurazan is deflected to his right by \(35^{\circ}\) (see the figure). In the collision, \(10 \%\) of the wrestlers' initial total kinetic energy is lost. What is the angle at which Toyohibiki is moving immediately after the collision?

NASA has taken an increased interest in near Earth asteroids. These objects, popularized in recent blockbuster movies, can pass very close to Earth on a cosmic scale, sometimes as close as 1 million miles. Most are small-less than \(500 \mathrm{~m}\) across \(-\) and while an impact with one of the smaller ones could be dangerous, experts believe that it may not be catastrophic to the human race. One possible defense system against near Earth asteroids involves hitting an incoming asteroid with a rocket to divert its course. Assume a relatively small asteroid with a mass of \(2.10 \cdot 10^{10} \mathrm{~kg}\) is traveling toward the Earth at a modest speed of \(12.0 \mathrm{~km} / \mathrm{s}\). a) How fast would a large rocket with a mass of \(8.00 \cdot 10^{4} \mathrm{~kg}\) have to be moving when it hit the asteroid head on in order to stop the asteroid? b) An alternative approach would be to divert the asteroid from its path by a small amount to cause it to miss Earth. How fast would the rocket of part (a) have to be traveling at impact to divert the asteroid's path by \(1.00^{\circ}\) ? In this case, assume that the rocket hits the asteroid while traveling along a line perpendicular to the asteroid's path.

Here is a popular lecture demonstration that you can perform at home. Place a golf ball on top of a basketball, and drop the pair from rest so they fall to the ground. (For reasons that should become clear once you solve this problem, do not attempt to do this experiment inside, but outdoors instead!) With a little practice, you can achieve the situation pictured here: The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is \(0.0459 \mathrm{~kg}\), and the mass of the basketball is \(0.619 \mathrm{~kg}\). you can achieve the situation pictured here: The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is \(0.0459 \mathrm{~kg}\), and the mass of the basketball is \(0.619 \mathrm{~kg}\). a) If the balls are released from a height where the bottom of the basketball is at \(0.701 \mathrm{~m}\) above the ground, what is the absolute value of the basketball's momentum just before it hits the ground? b) What is the absolute value of the momentum of the golf ball at this instant? c) Treat the collision of the basketball with the floor and the collision of the golf ball with the basketball as totally elastic collisions in one dimension. What is the absolute magnitude of the momentum of the golf ball after these collisions? d) Now comes the interesting question: How high, measured from the ground, will the golf ball bounce up after its collision with the basketball?
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