/*! 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 74 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
The net force on the passenger is approximately 542.36 N and the speed of the car is approximately 43.99 m/s.

Step by step solution

01

Find the Magnitude of the Net Force

To find the magnitude of the net force on the passenger, use the Pythagorean theorem. The magnitudes of the horizontal and vertical components of the force are given. \text{Net Force} = \sqrt{F_h^2 + F_v^2}where:\(F_h = 210\text{ N}\) (horizontal force)\(F_v = 500\text{ N}\) (vertical force)Then:\(\text{Net Force} = \sqrt{(210)^2 + (500)^2}\ = \sqrt{44100 + 250000}\ = \sqrt{294100}\ \approx 542.36\text{ N}\)
02

Identify the Centripetal Force

Because the horizontal component of the force is responsible for providing the centripetal force, use:\( F_c = F_h = 210 \text{ N} \)
03

Apply Newton's Second Law for Circular Motion

Recall the formula for centripetal force: \( F_c = m \frac{v^2}{r} \)where:\(m = 51.0 \text{ kg}\) (mass of the passenger)\(r = 470 \text{ m}\) (radius of the circle) Then,\( 210 = 51.0 \frac{v^2}{470} \)
04

Solve for the Speed

Rearrange the equation to solve for \(v\): \( v^2 = \frac{210 \times 470}{51.0} \)\( v^2 = 1935.29\)\( v = \sqrt{1935.29}\ \approx 43.99 \text{ m/s}\)

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Key Concepts

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

Net Force Calculation
In physics, the net force is the sum of all the forces acting on an object. In our exercise, the net force includes both horizontal and vertical components. To find the net force acting on the passenger, we use the Pythagorean theorem. Given two perpendicular forces, the formula for the net force (\text{Net Force}) is: \ \text{Net Force} = \ \ \text{Net Force} = \ \sqrt(F_h^2 + F_v^2) \ where:\ \(F_h = 210\ N\) (horizontal force) \ \(F_v = 500\ N\) (vertical force). \ Substituting the values, we get: \ \ \text{Net Force} = \ \sqrt((210)^2 + (500)^2) = \ \sqrt(44100 + 250000) = \ \sqrt(294100) \ \ Net Force \approx 542.36\ N. \ By calculating this, we find that the net force acting on the passenger is approximately 542.36 N. This value helps in understanding the total influence on the passenger due to various forces in motion.
Centripetal Force
Centripetal force is a crucial concept when dealing with circular motion. It is the force that keeps an object moving in a circular path, directed towards the center of rotation. In our exercise, the horizontal component of the force (\text{F_h = 210 N}) provides the necessary centripetal force (\text{F_c}). \ To recall, the formula for centripetal force is: \ \ F_c = m \ \frac(v^2)(r) \ where: \ \(m\) = 51.0 kg (mass of the passenger), and \ \(r\) = 470 m (radius of the circle). \ Substituting the values into the formula, we get: \ \ \ 210 = \frac(51.0 \ \times v^2)(470). \ Solving this equation provides further insight on circular motion parameters, mainly the speed of the car which is derived thereafter. Understanding centripetal force helps us connect the force with the motion and velocity of the car.
Newton's Second Law
Newton's second law states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. It can be expressed as: \ \ F = ma \ where: \ \(F\) = net force (N) \ \(m\) = mass (kg) \ \(a\) = acceleration (m/s^2). \ For circular motion, the centripetal acceleration is given by the equation: \ \ a = \ \frac(v^2)(r). \ Applying this in our context, we can further isolate the speed (\text{v}) of the railway car. Replacing the needed variables into the centripetal force equation, \ F_c = m \ \frac(v^2)(r) \ allows us to solve for speed (\text{v}). Rearranging terms, we get: \ \ v^2 = \ \frac(210 \ \times 470)(51.0) \ \ Solving, we find: \ v^2 = 1935.29 \ v \ \approx 43.99\ m/s. \ Thus, Newton's second law provides a comprehensive framework to understand forces, acceleration, and their impact on motion. This law is essential for analyzing and solving many physics problems related to motion.

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Most popular questions from this chapter

A \(40 \mathrm{~kg}\) slab rests on a frictionless floor. A \(10 \mathrm{~kg}\) block rests on top of the slab (Fig. 6-58). The coefficient of static friction \(\mu^{\text {stat }}\) between the block and the slab is \(0.60\), whereas their kinetic friction coefficient \(\mu^{\text {kin }}\) is \(0.40\). The \(10 \mathrm{~kg}\) block is pulled by a horizontal force with a magnitude of \(100 \mathrm{~N}\). What are the resulting accelerations of (a) the block and (b) the slab?

A \(1000 \mathrm{~kg}\) boat is traveling at \(90 \mathrm{~km} / \mathrm{h}\) when its engine is shut off. The magnitude of the frictional force \(\vec{f}^{\text {kin }}\) between boat and water is proportional to the speed \(v\) of the boat; \(\vec{f}^{\mathrm{kin}}=(70 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}) \vec{v}\). Find the time required for the boat to slow to \(45 \mathrm{~km} / \mathrm{h}\).

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