/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 You are the design engineer in c... [FREE SOLUTION] | 91Ó°ÊÓ

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You are the design engineer in charge of the crashworthiness of new automobile models. Cars are tested by smashing them into fixed, massive barriers at \(45 \mathrm{~km} / \mathrm{h}\). A new model of mass 1500 kg takes 0.15 s from the time of impact until it is brought to rest. \((a)\) Calculate the average force exerted on the car by the barrier. (b) Calculate the average deceleration of the car.

Short Answer

Expert verified
(a) The average force is -124,995 N. (b) The average deceleration is 83.33 m/s².

Step by step solution

01

Convert Speed to Meters per Second

The given speed of the car is 45 km/h. We need to convert this speed into meters per second (m/s) for ease of calculation. Use the conversion factor: \[1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}.\]Therefore, \[45 \text{ km/h} = \frac{45}{3.6} \approx 12.5 \text{ m/s}.\]
02

Determine the Change in Velocity

The initial velocity of the car is 12.5 m/s, and the final velocity is 0 m/s since the car comes to a stop. The change in velocity (\(\Delta v\)) is given by:\[\Delta v = 0 - 12.5 = -12.5 \text{ m/s}.\]
03

Calculate the Average Acceleration

The average acceleration (\(a\)) can be calculated using the formula:\[a = \frac{\Delta v}{\Delta t},\]where \(\Delta t = 0.15\) s.\[a = \frac{-12.5}{0.15} = -83.33 \text{ m/s}^2.\]
04

Calculate the Average Force

Using Newton's second law, the average force (\(F\)) exerted on the car by the barrier can be calculated as:\[F = ma,\]where \(m = 1500\) kg and \(a = -83.33\) m/s\(^2\).\[F = 1500 \times (-83.33) = -124995 \text{ N}.\]The negative sign indicates the force is in the opposite direction of motion.

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Key Concepts

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

Deceleration Calculation
Deceleration is a term used to describe how quickly an object reduces its speed. In the context of our car crash scenario, it refers to the rate at which the car slows down when it hits the barrier. Deceleration is essentially negative acceleration. To calculate deceleration, we use the formula:\[ a = \frac{\Delta v}{\Delta t}\]where:
  • \(\Delta v\) is the change in velocity
  • \(\Delta t\) is the time taken for the change
In our case, the car starts with a velocity of 12.5 m/s and comes to a stop, so \(\Delta v\) is \(0 - 12.5 = -12.5\) m/s. The time taken was 0.15 seconds. Plugging in these values, we get:\[ a = \frac{-12.5}{0.15} \approx -83.33 \text{ m/s}^2\]
Here, the negative sign indicates the velocity is decreasing, which means the car is decelerating. The magnitude of this value shows how rapidly the car stops once it hits the barrier.
Impact Force
When a car crashes into a barrier, it experiences an impact force, which is a crucial factor in determining the crashworthiness of a vehicle. Impact force can be enormous because it acts over a very short time interval during the crash. We calculate this force using the mass of the vehicle and its deceleration. According to Newton's second law of motion, the impact force is:\[F = ma\]where:
  • \(m\) is the mass of the car, which is 1500 kg
  • \(a\) is the acceleration (or deceleration) previously calculated as \(-83.33\) m/s²
Substituting these values, the impact force \(F\) is:\[F = 1500 \times (-83.33) = -124995 \text{ N}\]
The result is negative, showing that the force direction opposes the car's initial direction of travel. This large force indicates why crash design engineers focus on safely dissipating energy to protect passengers.
Newton's Second Law
Newton's second law of motion is a foundational principle in physics that helps explain how forces affect the motion of objects. It states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this is given as:\[F = ma\]For the car crash scenario, we applied this law to calculate the average force exerted on the car by the barrier. The car's mass \(m\) is 1500 kg, and the deceleration \(a\) we calculated earlier is \(-83.33\) m/s².Newton's second law is not just about motion but also about change. In this case, it helps to translate the drastic change from an initial speed of 12.5 m/s to a complete stop, resulting in the force exerted by the barrier. This powerful application of the law enables engineers to design safer vehicles and helps us understand the dynamics of real-world events like car crashes.

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Most popular questions from this chapter

(II) With what impulse does a 0.50 -kg newspaper have to be thrown to give it a velocity of \(3.0 \mathrm{~m} / \mathrm{s} ?\)

(III) Air in a \(120-\mathrm{km} / \mathrm{h}\) wind strikes head-on the face of a building \(45 \mathrm{~m}\) wide by \(65 \mathrm{~m}\) high and is brought to rest. If air has a mass of \(1.3 \mathrm{~kg}\) per cubic meter, determine the average force of the wind on the building.

A meteor whose mass was about \(2.0 \times 10^{8} \mathrm{~kg}\) struck the Earth \(\left(m_{\mathrm{E}}=6.0 \times 10^{24} \mathrm{~kg}\right)\) with a speed of about \(25 \mathrm{~km} / \mathrm{s}\) and came to rest in the Earth. (a) What was the Earth's recoil speed (relative to Earth at rest before the collision)? (b) What fraction of the meteor's kinetic energy was transformed to kinetic energy of the Earth? (c) By how much did the Earth's kinetic energy change as a result of this collision?

(II) \((a)\) A molecule of mass \(m\) and speed \(v\) strikes a wall at right angles and rebounds back with the same speed. If the collision time is \(\Delta t,\) what is the average force on the wall during the collision? \((b)\) If molecules, all of this type, strike the wall at intervals a time \(t\) apart (on the average) what is the average force on the wall averaged over a long time?

A hockey puck of mass \(4 m\) has been rigged to explode, as part of a practical joke. Initially the puck is at rest on a frictionless ice rink. Then it bursts into three pieces. One chunk, of mass \(m,\) slides across the ice at velocity \(v \hat{\mathbf{i}}\). Another chunk, of mass \(2 m\), slides across the ice at velocity \(2 v \hat{\mathbf{j}} .\) Determine the velocity of the third chunk.

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