/*! 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 18 In In April 1974, John Massis of... [FREE SOLUTION] | 91Ó°ÊÓ

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In In April 1974, John Massis of Belgium managed to move two passenger railroad cars. He did so by clamping his teeth down on a bit that was attached to the cars with a rope and then leaning backward while pressing his feet against the railway ties. The cars together weighed \(700 \mathrm{kN}\) (about 80 tons). Assume that he pulled with a constant force that was \(2.5\) times his body weight, at an upward angle \(\theta\) of \(30^{\circ}\) from the horizontal. His mass was \(80 \mathrm{~kg}\), and he moved the cars by \(1.0 \mathrm{~m}\). Neglecting any retarding force from the wheel rotation, find the speed of the cars at the end of the pull.

Short Answer

Expert verified
The speed of the cars is approximately \(0.078 \text{ m/s}\).

Step by step solution

01

Determine John's Force in Newtons

Firstly, we need to calculate the force exerted by John in Newtons. John's mass is given as \(80 \text{ kg}\). The gravitational force acting on him due to his mass is \(mg\), where \(g \approx 9.8 \text{ m/s}^2\). So, this force is \(80 \times 9.8 = 784 \text{ N}\). According to the problem, he pulls with a force that is 2.5 times his body weight. Therefore, his pulling force in the direction of motion is \(2.5 \times 784 = 1960 \text{ N}\).
02

Resolve the Force into Horizontal Component

The force John exerts is directed at an angle \(30^{\circ}\) above the horizontal. We need the horizontal component of this force, as that's the portion contributing to moving the cars. The horizontal component \(F_{x}\) can be calculated using trigonometry: \(F_{x} = F \cos(\theta) = 1960 \cos(30^{\circ})\). The value of \(\cos(30^{\circ})\) is \(\frac{\sqrt{3}}{2}\), so \(F_{x} = 1960 \times \frac{\sqrt{3}}{2} \approx 1697 \text{ N}\).
03

Calculate the Work Done on the Cars

Work done by John on the railroad cars is given by the formula \(W = F_{x} \times d\), where \(d=1.0 \text{ m}\) is the distance moved. So, \(W = 1697 \times 1.0 = 1697 \text{ J}\).
04

Calculate the Final Kinetic Energy of the Cars

Since there is no other external force acting on the cars (neglecting the retarding force), the entire work done goes into the kinetic energy of the cars. Therefore, the final kinetic energy \(K\) of the cars is \(K = 1697 \text{ J}\).
05

Calculate the Final Speed of the Cars

The kinetic energy of an object is given by \(K = \frac{1}{2}mv^2\), where \(m = 700000 \text{ kg}\) (mass of the cars) and \(v\) is the final speed. Rearrange to solve for \(v\): \(v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 1697}{700000}}\). Calculate this to find \(v \approx 0.078 \text{ m/s}\).

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Key Concepts

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

Classical Mechanics
Classical mechanics is the branch of physics that deals with the motion of objects and the forces acting upon them. This branch often involves analyzing situations where gravity, friction, and applied forces come into play. In the context of the problem with John Massis, we're exploring a situation where an individual uses force to overcome the weight of two enormous railroad cars.
Understanding classical mechanics helps us break down the problem into understandable parts; it lets us calculate forces, work, and energy.
Key components often examined in classical mechanics include:
  • Mass, which is the amount of matter in an object.
  • Forces, which can cause objects to accelerate as per Newton's laws.
  • Energy, which is needed to perform work.
  • Kinematics, which is the study of motion without considering its causes.
The principles of classical mechanics apply to a broad range of everyday situations, from the simple motion of a car to the more complex dynamics of Earth orbiting the sun.
Work and Energy
Work and energy are closely related concepts in physics. Work is defined as the amount of energy transferred by a force acting over a distance. When John Massis applied force over the distance by moving the railroad cars, he was doing work.
Several concepts interlink when we discuss work and energy in physics:
  • Work (W): This is calculated with the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force applied, \( d \) is the distance moved, and \( \theta \) is the angle between the force and direction of motion.
  • Kinetic Energy (K): This is the energy a body possesses due to its motion, expressed as \( K = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
Work done on an object manifests as a change in kinetic energy, according to the work-energy principle. Hence, when Massis pulled the cars, the work he did resulted in kinetic energy that increased their speed.
Newton's Laws
Newton's Laws of motion underpin most classical mechanics problems and give us a fundamental framework to understand how forces affect motion. In John Massis's scenario, these laws play a crucial role:
  • First Law (Law of Inertia): An object will remain stationary or in uniform motion unless acted upon by a force. This explains why the railroad cars initially didn't move until John applied the pulling force.
  • Second Law (F=ma): The force applied to an object is equal to the mass of the object times its acceleration. This principle helps us calculate the force John needed to accelerate the cars over a certain distance.
  • Third Law (Action and Reaction): For every action, there is an equal and opposite reaction. When John applied force on the cars, the cars exerted an equal and opposite force on him, rooted firmly by his stance.
These laws help provide a quantitative measure for predicting how objects will interact. They are profoundly pivotal in tasks involving pulling, pushing, lifting, or dropping objects.
Kinematics
Kinematics involves the description of motion, taking into account variables such as velocity, distance, and time, without considering the forces that cause this motion.
In John Massis's case, to fully understand the effect of his force and to quantify the motion of the railroad cars, we delve into kinematics:
  • Velocity ( v ): It's defined as the rate of change of position with time. From the problem, once we determined that all the work translated into kinetic energy, we could calculate the final velocity of the cars.
  • Distance ( d ): It's the scalar measure of 'how much ground' is covered, which, in the given scenario, turned out to be 1.0 ext{ m} as mentioned.
Kinematics serves as a fundamental tool in physics that allows us to analytically determine position, speed, and paths of moving objects without delving into the intricate details of the underlying forces.

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Most popular questions from this chapter

An interstellar ship has a mass of \(1.20 \times 10^{6} \mathrm{~kg}\) and is initially at rest relative to a star system. (a) What constant acceleration is needed to bring the ship up to a speed of \(0.10 c\) (where \(c\) is the speed of light, \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) relative to the star system in \(3.0\) days? (b) What is that acceleration in \(g\) units? (c) What force is required for the acceleration? (d) If the engines are shut down when \(0.10 c\) is reached (the speed then remains constant), how long does the ship take (start to finish) to journey \(5.0\) light-months, the distance that light travels in \(5.0\) months?

A \(40 \mathrm{~kg}\) girl and an \(8.4 \mathrm{~kg}\) sled are on the frictionless ice of a frozen lake, \(15 \mathrm{~m}\) apart but connected by a rope of negligible mass. The girl exerts a horizontal \(5.2 \mathrm{~N}\) force on the rope. What are the acceleration magnitudes of (a) the sled and (b) the girl? (c) How far from the girl's initial position do they meet?

If the \(1 \mathrm{~kg}\) standard body has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) at \(20.0^{\circ}\) to the positive direction of an \(x\) axis, what are (a) the \(x\) component and (b) the \(y\) component of the net force acting on the body, and (c) what is the net force in unit-vector notation?

(a) Figure \(5-47\) shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time \(t=0\), container 1 has mass \(1.30 \mathrm{~kg}\) and container 2 has mass \(2.80 \mathrm{~kg}\), but container 1 is losing mass (through a leak) at the constant rate of \(0.200 \mathrm{~kg} / \mathrm{s}\) At what rate is the acceleration magnitude of the containers changing at (a) \(t=0\) and (b) \(t=3.00 \mathrm{~s}\) ? (c) When does the acceleration reach its maximum value?

A \(0.150 \mathrm{~kg}\) particle moves along an \(x\) axis according to \(x(t)=-13.00+2.00 t+4.00 t^{2}-3.00 t^{3}\), with \(x\) in meters and \(t\) in seconds. In unit-vector notation, what is the net force acting on the particle at \(t=3.40 \mathrm{~s}\) ?

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