/*! 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 6 A friend claims that, as long as... [FREE SOLUTION] | 91Ó°ÊÓ

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A friend claims that, as long as he has his seat belt on, he can hold on to a 12.0 -kg child in a \(60.0 \mathrm{mi} / \mathrm{h}\) head-on collision with a brick wall in which the car passenger compartment comes to a stop in 0.050 0 s. Show that the violent force during the collision will tear the child from his arms. A child should always be in a toddler seat secured with a seat belt in the back seat of a car.

Short Answer

Expert verified
The force required to hold the child during the collision comes out to be -6436.8 N. The negative sign indicates the direction of the force, which is opposite to car's motion. If the friend cannot exert this force, the child will indeed be torn from his arms. Hence, it’s vital for a child to always be in a toddler seat secured with a seat belt in the back seat of a car.

Step by step solution

01

Convert the speed to m/s

The speed of the car is given in miles per hour. In order to make the calculation, it needs to be converted to the standard SI unit, which is meters per second (m/s). Use the conversion factor \(1 \: \mathrm{mi}/\mathrm{h} = 0.447 \: \mathrm{m/s}\). Therefore, \(60.0 \: \mathrm{mi}/\mathrm{h} = 60.0 \times 0.447 \: \mathrm{m/s} = 26.82 \: \mathrm{m/s}\) .
02

Calculate the acceleration

Acceleration (or in this case deceleration since the car is stopping) is simply the change in velocity divided by the time it takes for that change to occur. The initial speed of the car is 26.82 m/s and the final speed is 0 m/s. The formula to calculate acceleration is \(a = \frac {\Delta v} {\Delta t} = \frac {v_f - v_i} {t}\). The final velocity (\(v_f\)) is 0 m/s, the initial velocity (\(v_i\)) is 26.82 m/s and the time (\(t\)) is 0.0500 s. Substituting these values into the formula gives \(a = \frac {0 - 26.82} {0.0500} = -536.4 \: \mathrm{m/s^2}\). The acceleration is negative indicating deceleration.
03

Calculate the Force

Finally, apply Newton's second law (\(F=ma\)) to determine the force the person must exert to hold the child. Substituting the mass (12.0 kg) and the acceleration calculated (-536.4 m/s²) into the formula gives \(F = (12.0 \: \mathrm{kg}) \times (-536.4 \: \mathrm{m/s^2}) = -6436.8 \: \mathrm{N}\). The negative sign indicates the force is in the opposite direction of the motion of the car.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a cornerstone of classical mechanics. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. In mathematical terms, the law is represented as \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.

This simple yet powerful equation allows us to calculate how much force is required to change the motion of an object. For example, when a car abruptly stops in a collision, its passengers experience a rapid deceleration. Here, Newton's Second Law helps us find the force that will impact them.
  • The Law indicates that larger mass or greater acceleration results in higher force.
  • Force, mass, and acceleration are directly related; changing one affects the others.
Understanding this law is essential for analyzing any situation involving motion, like a car crash, where forces involved are critical for safety.
Conversions Between Units
When working with physics problems, converting units is often necessary to ensure consistency, especially when calculating forces or accelerations. Different measurement systems, such as SI (International System of Units) and Imperial systems, often necessitate conversions.

In our exercise, the car's speed was initially given in miles per hour. To solve the problem, this speed needed conversion to meters per second, the SI unit of speed. Using the conversion factor \(1 \: \text{mi/h} = 0.447 \: \text{m/s}\), converting becomes straightforward.
  • Miles per hour to meters per second: Multiply by 0.447.
  • Always make sure your final units match the required calculations.
  • SI units simplify calculations across many physics problems.
Maintaining the correct units throughout the problem helps prevent errors and ensures accurate results, ultimately helping you grasp and apply physics principles effectively.
Acceleration and Force Calculations
Calculating acceleration and force is crucial in understanding how objects move and interact with each other. Acceleration refers to the rate at which velocity changes, and it can be calculated with the formula \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time over which this change occurs.

Using this formula, we calculated the deceleration of the car as \(-536.4 \: \text{m/s}^2\). The negative sign indicates the car is slowing down. With acceleration known, force calculation using Newton's second law becomes simple: \( F = ma \). Here, applying the mass of the child (12 kg) gives the deep comprehension of the intense force in a collision.
  • Positive acceleration indicates speeding up; negative shows slowing down.
  • Force calculation requires accurate values for both mass and acceleration.
  • Understand both concepts to properly handle physics scenarios involving motion.
Mastering these calculations lets you predict and evaluate the effects of forces on bodies, pivotal in designing safe environments like vehicle restraints.

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Most popular questions from this chapter

A 75.0 -kg firefighter slides down a pole while a constant friction force of \(300 \mathrm{N}\) retards her motion. A horizontal 20.0 -kg platform is supported by a spring at the bottom of the pole to cushion the fall. The firefighter starts from rest \(4.00 \mathrm{m}\) above the platform, and the spring constant is \(4000 \mathrm{N} / \mathrm{m} .\) Find \((\mathrm{a})\) the firefighter's speed just before she collides with the platform and (b) the maximum distance the spring is compressed. (Assume the friction force acts during the entire motion.)

Two particles with masses \(m\) and \(3 m\) are moving toward each other along the \(x\) axis with the same initial speeds \(v_{i}\) Particle \(m\) is traveling to the left, while particle \(3 m\) is traveling to the right. They undergo an elastic glancing collision such that particle \(m\) is moving downward after the collision at right angles from its initial direction. (a) Find the final speeds of the two particles. (b) What is the angle \(\theta\) at which the particle \(3 m\) is scattered?

A 10.0 -g bullet is fired into a stationary block of wood \((m=\) \(5.00 \mathrm{kg}) .\) The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is \(0.600 \mathrm{m} / \mathrm{s} .\) What was the original speed of the bullet?

A glider of mass \(m\) is free to slide along a horizontal air track. It is pushed against a launcher at one end of the track. Model the launcher as a light spring of force constant \(k\) compressed by a distance \(x\). The glider is released from rest. (a) Show that the glider attains a speed of \(v=x(k / m)^{1 / 2} .\) (b) Does a glider of large or of small mass attain a greater speed? (c) Show that the impulse imparted to the glider is given by the expression \(x(k m)^{1 / 2} .\) (d) Is a greater impulse injected into a large or a small mass? (e) Is more work done on a large or a small mass?

A ball of mass \(0.150 \mathrm{kg}\) is dropped from rest from a height of \(1.25 \mathrm{m} .\) It rebounds from the floor to reach a height of \(0.960 \mathrm{m} .\) What impulse was given to the ball by the floor?

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