/*! 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 64 A high-speed railway car goes ar... [FREE SOLUTION] | 91Ó°ÊÓ

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A high-speed railway car goes around a flat, horizontal circle of radius \(470 \mathrm{~m}\) at a constant speed. The magnitudes of the horizontal and vertical components of the force of the car on a \(51.0 \mathrm{~kg}\) passenger are \(210 \mathrm{~N}\) and \(500 \mathrm{~N},\) respectively. (a) What is the magnitude of the net force (of all the forces) on the passenger? (b) What is the speed of the car?

Short Answer

Expert verified
(a) Net force is 542.27 N. (b) Speed is 70.64 m/s.

Step by step solution

01

Understand the Components of Forces

The problem provides two force components: a horizontal component of \(210 \text{ N}\), and a vertical component of \(500 \text{ N}\). These forces act on a passenger of mass \(51 \text{ kg}\) who is undergoing circular motion.
02

Calculate the Net Force

The net force is the vector sum of the horizontal and vertical forces. This can be calculated using the Pythagorean theorem: \ \[ F_{\text{net}} = \sqrt{F_{\text{horizontal}}^2 + F_{\text{vertical}}^2} = \sqrt{210^2 + 500^2} = \sqrt{44100 + 250000} = \sqrt{294100} \approx 542.27 \text{ N} \]
03

Apply Newton's Second Law

In circular motion, the net force is the centripetal force required to keep the passenger moving in a circle. According to Newton's second law, \(F_{\text{net}} = m \cdot a\). Here, \(a = \frac{v^2}{r}\), where \(v\) is the speed of the car and \(r\) is the radius of the circle. Therefore, \ \[ 542.27 = 51 \cdot \frac{v^2}{470} \]
04

Solve for Speed

Rearrange the equation from the previous step to solve for \(v\): \ \[ 542.27 = 51 \cdot \frac{v^2}{470} \] \ \[ v^2 = \frac{542.27 \cdot 470}{51} \approx 4991.54 \] \ \[ v = \sqrt{4991.54} \approx 70.64 \text{ m/s} \]
05

Final Answer

(a) The magnitude of the net force on the passenger is approximately \(542.27 \text{ N}\). (b) The speed of the car is approximately \(70.64 \text{ m/s}\).

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Key Concepts

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

Centripetal Force
Whenever an object moves in a circle, there is a need for a force that keeps it in its path. This force is called the centripetal force. In our exercise, the passenger in the railway car is moving in a circular path with the help of centripetal force.
This force always points towards the center of the circle. It ensures that the object does not fly off into a straight line.

Here are some key points about centripetal force:
  • It is always perpendicular to the direction of motion.
  • The magnitude depends on the mass of the object, the speed of the object, and the radius of the circle.
  • In this case, the centripetal force is the net force that arises from the horizontal component of the force of the car on the passenger.
The centripetal force is essential to maintaining circular motion, without it, an object would move in a straight line due to inertia.
Newton's Second Law
Newton's Second Law of Motion is fundamental to understanding how forces work. It states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it's expressed as \( F = m \cdot a \).
In our scenario, this law helps us relate the net force to the centripetal force required to keep the passenger in circular motion. The acceleration here is centripetal acceleration \( a = \frac{v^2}{r} \), where \( v \) is the speed of the railway car and \( r \) is the radius of the circle.

Key insights from this law in this context:
  • The net force acting on the passenger provides the centripetal force needed for circular motion.
  • The magnitude of this force can tell us how fast the car is moving, by solving for \( v \).
  • This relationship allows us to solve problems that involve calculating speeds and forces in circular paths.
Using Newton's Second Law, we can determine that the net force is not something separate but directly connected to the motion of the passenger.
Vector Components
Every force can be broken down into vector components that align with specific axes, like horizontal and vertical axes. This is a crucial step in analyzing forces in physics.
In the problem, the force exerted on the passenger by the car has both a horizontal and a vertical component. These components can be thought of as the legs of a right triangle where the hypotenuse is the net force. Hence, using the Pythagorean theorem helps in finding the resultant force.
  • The horizontal component in this example is part of maintaining the circular path.
  • The vertical component typically counteracts weight, ensuring the passenger's vertical motion is controlled.
  • Using vector components facilitates calculations by allowing us to deal with simpler, one-dimensional forces.
Understanding vector components allows one to better grasp how forces cooperate and oppose each other to maintain balance and motion. It provides a geometrical representation that is simple to visualize and calculate.

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Most popular questions from this chapter

A filing cabinet weighing \(556 \mathrm{~N}\) rests on the floor. The coefficient of static friction between it and the floor is \(0.68,\) and the coefficient of kinetic friction is \(0.56 .\) In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) \(222 \mathrm{~N}\), (b) \(334 \mathrm{~N},\) (c) \(445 \mathrm{~N},\) and (d) \(556 \mathrm{~N}\). For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

A roller-coaster car at an amusement park has a mass of \(1200 \mathrm{~kg}\) when fully loaded with passengers. As the car passes over the top of a circular hill of radius \(18 \mathrm{~m},\) assume that its speed is not changing. At the top of the hill, what are the (a) magnitude \(F_{N}\) and \((\mathrm{b})\) direction (up or down) of the normal force on the car from the track if the car's speed is \(v=11 \mathrm{~m} / \mathrm{s} ?\) What are \((\mathrm{c}) F_{N}\) and (d) the direction if \(v=14 \mathrm{~m} / \mathrm{s} ?\)

A slide-loving pig slides down a certain \(35^{\circ}\) slide in twice the time it would take to slide down a frictionless \(35^{\circ}\) slide. What is the coefficient of kinetic friction between the pig and the slide?

A child weighing \(140 \mathrm{~N}\) sits at rest at the top of a playground slide that makes an angle of \(25^{\circ}\) with the horizontal. The child keeps from sliding by holding onto the sides of the slide. After letting go of the sides, the child has a constant acceleration of \(0.86 \mathrm{~m} / \mathrm{s}^{2}\) (down the slide, of course). (a) What is the coefficient of kinetic friction between the child and the slide? (b) What maximum and minimum values for the coefficient of static friction between the child and the slide are consistent with the information given here?

A 3.5 kg block is pushed along a horizontal floor by a force \(\vec{F}\) of magnitude \(15 \mathrm{~N}\) at an angle \(\theta=40^{\circ}\) with the horizontal (Fig. 6-19). The coefficient of kinetic friction between the block and the floor is \(0.25 .\) Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

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