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Chapter 5: Problem 14

In a head-on collision, a car stops in 0.10 s from a speed of \(14 \mathrm{m} / \mathrm{s} .\) The driver has a mass of \(70 \mathrm{kg},\) and is, fortunately, tightly strapped into his seat. What force is applied to the driver by his seat belt during that fraction of a second?

Short Answer

Expert verified
The force applied to the driver by his seat belt during the collision is -9800 N, acting opposite to his initial motion.

Step by step solution

01

Find the deceleration

Since the driver comes to a stop from a speed of 14 m/s in 0.10 s, we can find the rate of his deceleration by using the formula \(a = \frac{{\Delta v}}{{\Delta t}}\), where \(a\) is acceleration, \({\Delta v}\) is the change in velocity and \({\Delta t}\) is the change in time. Here the change in velocity, \( \Delta v = v_f - v_i = 0 - 14 = -14 m/s\), as the initial velocity \(v_i = 14 m/s\) and the final velocity \(v_f = 0 m/s\). The change in time, \( \Delta t = 0.10 s\). Substituting these values into the equation, we have \(a = \frac{{-14 m/s}}{{0.10 s}} = -140 m/s^2\). The acceleration is negative because it is a deceleration.
02

Compute the force

Next, we can compute the force applied to the driver. According to Newton's second law, \(F = ma\), where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration. By substituting \(m = 70 kg\) and \(a = -140 m/s^2\), we get \(F = 70 kg \times -140 m/s^2 = -9800 N\). The force is negative, indicating that it acts in the opposite direction to the initial motion of the driver.

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Key Concepts

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

Deceleration
When a car comes to a sudden stop, like in a head-on collision, it undergoes deceleration. Deceleration is simply acceleration in the opposite direction of motion, slowing something down.

To calculate deceleration, we use the formula \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity, and \( \Delta t \) is the time taken for that change. In our example, the car's velocity drops from 14 m/s to 0 m/s in 0.10 seconds.

Substituting these values, we get \( a = \frac{-14}{0.10} = -140 \text{ m/s}^2 \). The negative sign denotes that this is a deceleration, a force acting to stop the vehicle.
  • Initial velocity \( v_i = 14 \text{ m/s} \)
  • Final velocity \( v_f = 0 \text{ m/s} \)
  • Time \( \Delta t = 0.10 \text{ s} \)
  • Deceleration \( a = -140 \text{ m/s}^2 \)

Understanding deceleration is crucial for evaluating forces in sudden stops, such as in our example.
Force Calculation
Newton's Second Law of Motion helps us find the force exerted on an object through the formula \( F = ma \). Here, \( F \) stands for force, \( m \) is mass, and \( a \) is acceleration or deceleration.

In the case of the driver experiencing deceleration, we use this law to determine the force the seatbelt applies to him. Given a mass \( m = 70 \text{ kg} \) and a deceleration \( a = -140 \text{ m/s}^2 \), the force is calculated as:
  • Substituting the values: \( F = 70 \times -140 \)
  • Force \( F = -9800 \text{ N} \)
The negative sign indicates the force direction is opposite to the initial movement.

This force is crucial, as it determines how much pressure the seatbelt applies, protecting the driver during the crash. It's a clear illustration of how safety devices rely on physical laws to keep us safe.
Mass and Velocity
Understanding both mass and velocity is essential when analyzing motion. Mass is a measure of how much matter is in an object and affects how much force is needed to change its motion. In our scenario, the driver's mass is 70 kg.

Velocity is the speed of an object in a specified direction. In the example, the initial velocity (before the car stops) is 14 m/s.

Both mass and velocity are critical in calculating forces during motion changes.
  • Higher mass = more force needed to change motion.
  • Higher velocity = more change needed for stopping, increasing impact force.

These parameters are vital in ensuring the safety and efficiency of mechanisms like car seatbelts, which protect passengers by controlling the force applied during sudden stops.

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Most popular questions from this chapter

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about \(1 \mathrm{m}\) as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. By contrast, an unrestrained occupant keeps moving forward with no loss of speed (Newton's first law!) until hitting the dashboard or windshield, as we saw in Figure \(4.2 .\) These are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about \(5 \mathrm{mm}\) a. A \(60 \mathrm{kg}\) person is in a head-on collision. The car's speed at impact is \(15 \mathrm{m} / \mathrm{s}\). Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys. b. Estimate the net force that ultimately stops the person if he or she is not restrained by a seat belt or air bag. c. How do these two forces compare to the person's weight?

\(\mathrm{A} 50 \mathrm{kg}\) box hangs from a rope. What is the tension in the rope if a. The box is at rest? b. The box has \(v_{y}=5.0 \mathrm{m} / \mathrm{s}\) and is speeding up at \(5.0 \mathrm{m} / \mathrm{s}^{2} ?\)

\(\mathrm{A} 2200 \mathrm{kg}\) truck has put its front bumper against the rear bumper of a \(2400 \mathrm{kg}\) SUV to give it a push. With the engine at full power and good tires on good pavement, the maximum forward force on the truck is \(18,000 \mathrm{N}\). a. What is the maximum possible acceleration the truck can give the SUV? b. At this acceleration, what is the force of the SUV's bumper on the truck's bumper?

Your forehead can withstand a force of about \(6.0 \mathrm{kN}\) before fracturing, while your cheekbone can only withstand about \(1.3 \mathrm{kN}\). a. If a 140 g baseball strikes your head at \(30 \mathrm{m} / \mathrm{s}\) and stops in \(0.0015 \mathrm{s},\) what is the magnitude of the ball's acceleration? b. What is the magnitude of the force that stops the baseball? c. What force does the baseball apply to your head? Explain. d. Are you in danger of a fracture if the ball hits you in the forehead? In the cheek?

A simple model shows how drawing a bow across a violin string causes the string to vibrate. As the bow moves across the string, static friction between the bow and the string pulls the string along with the bow. At some point, the tension pulling the string back exceeds the maximum static friction force and the string snaps back. This process repeats cyclically, causing the string's vibration. Assume the tension in a 0.33 -m-long violin string is \(50 \mathrm{N}\), and the coefficient of static friction between the bow and the string is \(\mu_{\mathrm{s}}=0.80 .\) If the normal force of the bow on the string is \(0.75 \mathrm{N},\) how far can the string be pulled before it slips if the string is bowed at its center?
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