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Chapter 3: Problem 8

A car traveling at \(53 \mathrm{~km} / \mathrm{h}\) hits a bridge abutment. \(\mathrm{A}\) passenger in the car moves forward a distance of \(65 \mathrm{~cm}\) (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of \(41 \mathrm{~kg}\) ?

Short Answer

Expert verified
6825.68 N

Step by step solution

01

Convert velocity to meters per second

First, convert the car's speed from km/h to m/s. Given: \(53 \text{ km/h} = 53 \times \frac{1000}{3600} \text{ m/s} = 14.72 \text{ m/s} \)
02

Calculate deceleration

Use the formula for deceleration: \( v_f^2 = v_i^2 + 2ad \) where: \( v_f = 0 \text{ m/s} \) \( v_i = 14.72 \text{ m/s} \) \( d = 0.65 \text{ m} \) Rearrange for \(a\): \( 0 = (14.72)^2 + 2a(0.65) \) \( a = -\frac{(14.72)^2}{2(0.65)} \approx -166.48 \text{ m/s}^2 \)
03

Calculate the force

Use Newton's second law, \( F = ma \), to find the force: Given: \( m = 41 \text{ kg} \) \( a = -166.48 \text{ m/s}^2 \) \( F = 41 \times (-166.48) \approx -6825.68 \text{ N} \)
04

Find the magnitude of the force

The magnitude of the force is the absolute value of the result: \( | -6825.68 | = 6825.68 \text{ N} \)

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Key Concepts

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

Deceleration Calculation
When a moving object is brought to a stop, it experiences deceleration (negative acceleration). In the given problem, the car's initial speed is 53 km/h. To find the deceleration, we first need to convert this speed to meters per second. We use the formula:
\ 53 \text{ km/h} = 53 \times \frac{1000}{3600} \text{ m/s} = 14.72 \text{ m/s} \
Then we apply the kinematic equation:
\[ v_f^2 = v_i^2 + 2ad \] In our scenario:
  • Final velocity \( v_f \)= 0 m/s
  • Initial velocity \( v_i \) = 14.72 m/s
  • Distance \( d \)= 0.65 m
Rearranging the formula for acceleration \( a \):
\[ 0 = (14.72)^2 + 2a(0.65) \] \[ a = -\frac{(14.72)^2}{2(0.65)} \] \[ a \approx -166.48 \text{ m/s}^2 \] The negative sign indicates deceleration.
Newton's Second Law
Newton's second law of motion states that the force acting on an object is equal to the mass of that object multiplied by its acceleration: \[ F = ma \] For the calculated deceleration, we need to consider the mass of the passenger's upper torso, which is given as 41 kg. Using the deceleration value found earlier:
  • Mass \( m \) = 41 kg
  • Acceleration \( a \) = -166.48 m/s^2
We plug these values into the formula:
\[ F = 41 \times (-166.48) \] \[ F \approx -6825.68 \text{ N} \] The force exerted is negative, showing opposition to the passenger's motion.
Unit Conversion
In physics, unit conversion is essential for consistency. The initial speed was given in km/h, but standard SI units require meters per second (m/s) for calculations. Here's how we convert km/h to m/s:
  • 1 km = 1000 meters
  • 1 hour = 3600 seconds
So, to convert speed:
\[ 53 \times \frac{1000}{3600} = 14.72 \text{ m/s} \] Similarly, length may need conversion. In this problem, distance from cm to meters:
\[ 65 \text{ cm} = 0.65 \text{ m} \] Conversions ensure accuracy and compatibility in calculations.
Force Magnitude
Force magnitude refers to the absolute value of the force, disregarding direction. After computing the force using Newton's second law, we find:
\[ F \approx -6825.68 \text{ N} \] To report the magnitude of the force, take the absolute value:
\[ |F| = 6825.68 \text{ N} \] The magnitude gives us a sense of the force's strength, regardless of direction. In this scenario, the airbag exerts a significant force to decelerate the passenger safely.

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Most popular questions from this chapter

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