/*! 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 15 A \(1200-\mathrm{kg}\) trailer i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(1200-\mathrm{kg}\) trailer is hitched to a \(1400-\mathrm{kg}\) car. The car and trailer are traveling at \(72 \mathrm{km} / \mathrm{h}\) when the driver applies the brakes on both the car and the trailer. Knowing that the braking forces exerted on the car and the trailer are \(5000 \mathrm{N}\) and \(4000 \mathrm{N}\), respectively, determine \((a)\) the distance traveled by the car and trailer before they come to a stop, \((b)\) the horizontal component of the force exerted by the trailer hitch on the car.

Short Answer

Expert verified
(a) 57.8 m; (b) 288 N.

Step by step solution

01

Convert velocity to meters per second

The initial velocity of the car and trailer is given as \(72 \text{ km/h}\). To convert this to meters per second, use the formula: \[ \text{Velocity in m/s} = \frac{\text{Velocity in km/h} \times 1000}{3600} \]. Calculating, \(72 \text{ km/h} = \frac{72 \times 1000}{3600} = 20 \text{ m/s} \).
02

Calculate deceleration of car and trailer

The force-exerted results in decelerations. Use Newton's second law, \( F = ma \), to find deceleration \(a\) for each:- Car: \[ a_1 = \frac{F_1}{m_1} = \frac{5000 \text{ N}}{1400 \text{ kg}} = 3.57 \text{ m/s}^2 \]- Trailer: \[ a_2 = \frac{F_2}{m_2} = \frac{4000 \text{ N}}{1200 \text{ kg}} = 3.33 \text{ m/s}^2 \]These are effective accelerations for their respective masses while braking.
03

Calculate effective combined deceleration

For the entire system (car and trailer), calculate the combined mass, which is \(2600 \text{ kg}\). Determine the net force:\[ F_{net} = -5000 \text{ N} - 4000 \text{ N} = -9000 \text{ N} \]Use \( F = ma \) to find the combined deceleration: \[ a = \frac{F_{net}}{m_{total}} = \frac{-9000 \text{ N}}{2600 \text{ kg}} = -3.46 \text{ m/s}^2 \]
04

Calculate stopping distance

Use the kinematic equation \( v^2 = u^2 + 2as \) where \( u = 20 \text{ m/s} \) (initial velocity), \( v = 0 \text{ m/s} \) (final velocity), and \( a = -3.46 \text{ m/s}^2 \). Solve for \( s \) (the stopping distance):\[ 0 = (20)^2 + 2(-3.46)s \]\[ 0 = 400 - 6.92s \]\[ 6.92s = 400 \]\[ s = \frac{400}{6.92} \approx 57.8 \] m.
05

Calculate force on trailer hitch

The trailer hitch exerts a force determined by the difference in decelerations. Calculate the force on the hitch using \( F_h = m_2 (a_1 - a_2) \), where:- \(a_1\) is the car's deceleration \((3.57 \text{ m/s}^2)\),- \(a_2\) is the trailer's deceleration \((3.33 \text{ m/s}^2)\), - \(m_2 = 1200 \text{ kg}\).Substituting:\[ F_h = 1200 (3.57 - 3.33) = 1200 \times 0.24 = 288 \text{ N}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 fundamental principle in physics, connecting the concept of force with the motion of an object. It is often expressed using the formula:\[ F = ma \]where:
  • \( F \) is the force applied to the object.
  • \( m \) is the mass of the object.
  • \( a \) is the acceleration produced on the object.
This law reflects how much an object will accelerate when a certain force is applied. In our exercise, it is used to find the deceleration of both the car and trailer when braking. The force exerted by the brakes leads to negative acceleration (deceleration), slowing them down to stop. Understanding this relationship helps us determine how external forces like braking affect motion, crucial for calculations involving stopping distances and safety in automotive contexts.
Kinematic Equations
Kinematic equations are essential tools in mechanics to describe and analyze motion without explicitly considering the forces that cause it. They relate parameters like initial velocity, final velocity, acceleration, time, and displacement. A common equation used is:\[ v^2 = u^2 + 2as \]where:
  • \( v \) is the final velocity.
  • \( u \) is the initial velocity.
  • \( a \) is the acceleration (or deceleration).
  • \( s \) is the displacement.
In the scenario of the car and trailer applying brakes, we use this equation to determine the stopping distance. By knowing initial speed and deceleration, we can calculate how far the vehicle travels before it comes to a complete stop. This is crucial for safety evaluations like determining safe following distances on roads.
Braking Force
Braking force is the force applied to slow down or stop a moving vehicle. It directly influences the rate at which a vehicle decelerates. The braking force must overcome the vehicle's motion to bring it to rest. In the exercise, the car and trailer have distinct braking forces applied:
  • The car has a braking force of \( 5000 \text{ N} \).
  • The trailer has a braking force of \( 4000 \text{ N} \).
These forces must be analyzed individually to determine their respective effects on motion. The differences in force are crucial to understanding how each component of a system, like a car and trailer, interact and how they individually and collectively respond to brakes. The knowledge of braking force helps in engineering safer and more efficient braking systems by selecting appropriate materials and mechanisms.
Deceleration Calculation
Deceleration calculation is a specific application of Newton's second law to determine how quickly an object slows down when an opposing force is applied. The calculation utilizes the formula for acceleration \( a = \frac{F}{m} \), where force \( F \) is in the opposite direction of motion, hence the negative sign.
  • For the car, the deceleration \( a_1 \) is calculated as \( \frac{5000 \text{ N}}{1400 \text{ kg}} = 3.57 \text{ m/s}^2 \).
  • For the trailer, the deceleration \( a_2 \) is \( \frac{4000 \text{ N}}{1200 \text{ kg}} = 3.33 \text{ m/s}^2 \).
Furthermore, the effective deceleration when considering both masses is vital. Here, it is determined by calculating the combined mass and total braking force, giving us a holistic view of how the entire vehicle system slows down. Understanding deceleration informs us on safe stopping distances and impacts on vehicle dynamics during braking events.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Collar \(A\) has a mass of \(3 \mathrm{kg}\) and is attached to a spring of constant \(1200 \mathrm{N} / \mathrm{m}\) and of undeformed length equal to \(0.5 \mathrm{m} .\) The system is set in motion with \(r=0.3 \mathrm{m}, v_{\theta}=2 \mathrm{m} / \mathrm{s},\) and \(v_{r}=0 .\) Neglecting the mass of the rod and the effect of friction, determine the radial and transverse components of the velocity of the collar when \(r=0.6 \mathrm{m} .\)

A 500-kg communications satellite is in a circular geosynchronous orbit and completes one revolution about the earth in 23 h and 56 min at an altitude of 35 800 km above the surface of the earth. Knowing that the radius of the earth is 6370 km, determine the kinetic energy of the satellite.

The Mars Pathfinder spacecraft used large airbags to cushion its impact with the planet's surface when landing. Assuming the space-craft had an impact velocity of \(18.5 \mathrm{m} / \mathrm{s}\) at an angle of \(45^{\circ}\) with respect to the horizontal, the coefficient of restitution is 0.85 and neglecting friction, determine \((a)\) the height of the first bounce, (b) the length of the first bounce. (Acceleration of gravity on Mars \(=3.73 \mathrm{m} / \mathrm{s}^{2}\) )

A spacecraft of mass \(m\) describes a circular orbit of radius \(r_{1}\) around the earth. \((a)\) Show that the additional energy \(\Delta E\) that must be imparted to the spacecraft to transfer it to a circular orbit of larger radius \(r_{2}\) is $$\Delta E=\frac{G M m\left(r_{2}-r_{1}\right)}{2 r_{1} r_{2}}$$ where \(M\) is the mass of the earth. ( \(b\) ) Further show that if the transfer from one circular orbit to the other is executed by placing the spacecraft on a transitional semielliptic path \(A B,\) the amounts of energy \(\Delta E_{A}\) and \(\Delta E_{B}\) which must be imparted at \(A\) and \(B\) are, respectively, proportional to \(r_{2}\) and \(r_{1}\) : $$\Delta E_{A}=\frac{r_{2}}{r_{1}+r_{2}} \Delta E \quad \Delta E_{B}=\frac{r_{1}}{r_{1}+r_{2}} \Delta E$$

A 2-kg collar is attached to a spring and slides without friction in a vertical plane along the curved rod ABC. The spring is undeformed when the collar is at C and its constant is 600 N/m. If the collar is released at A with no initial velocity, determine its velocity (a) as it passes through B, (b) as it reaches C.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.