/*! 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 48 A roller-coaster car has a mass ... [FREE SOLUTION] | 91Ó°ÊÓ

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A roller-coaster car 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 \(18 \mathrm{~m}\), its speed is not changing. At the top of the hill, what are the (a) magnitude \(F_{N}\) and (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 (c) \(F_{N}\) and (d) the direction if \(v=\) \(14 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) \(F_{N} = -3686.67 \text{ N}\) (upward); (c) \(F_{N} = 1306.67 \text{ N}\) (downward).

Step by step solution

01

Understanding the forces

When the roller coaster car is at the top of a hill, two main forces act on it: the gravitational force (\(F_{ ext{gravity}} = m imes g\)) pulling it down and the normal force (\(F_{N}\)) from the track acting on the car, which also acts downward when the car is over the top of a hill.
02

Applying Newton's Second Law at the hill's top

Since the car is in circular motion at the top of the hill, the net force towards the center (downward) must equal the centripetal force required to keep it moving in a circle: \(F_{ ext{net}} = m imes a_{ ext{centripetal}}\) where \(a_{ ext{centripetal}} = \frac{v^2}{r}\). Therefore, \(mg + F_{N} = m \times \frac{v^2}{r}\).
03

Calculate \(F_{N}\) for \(v = 11 \text{ m/s}\)

Substitute the given values into the equation: \(1200 imes 9.8 + F_{N} = 1200 \times \frac{11^2}{18}\). Simplifying, we have \(11760 + F_{N} = 8073.33\). Solving for \(F_{N}\), we find that \(F_{N} = 8073.33 - 11760 = -3686.67 \text{ N}\). The negative sign indicates that the direction is upward.
04

Calculate \(F_{N}\) for \(v = 14 \text{ m/s}\)

Similarly, substitute the new speed into the equation: \(1200 imes 9.8 + F_{N} = 1200 \times \frac{14^2}{18}\). Simplifying, we have \(11760 + F_{N} = 13066.67\). Solving for \(F_{N}\), we find that \(F_{N} = 13066.67 - 11760 = 1306.67 \text{ N}\). The positive sign indicates that the direction is downward.

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Key Concepts

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

Normal Force
Normal force is a crucial concept when discussing objects in contact with surfaces, like the roller-coaster car on the track. It is the force exerted by a surface to support the weight of an object placed on it. This force acts perpendicular to the surface of contact.
  • When the car is at the top of the hill, the normal force not only supports the car's weight but also contributes to the centripetal force required to keep the car moving in its circular path.
  • In this problem, the normal force is acting in the same direction as gravity—both forces point downwards at the hill's top.
When calculating the normal force, you must account for all other forces acting in the system. If the net force (sum of gravitational force and normal force) exceeds the centripetal force required at the hill's top, the normal force must compensate by acting upwards to reduce the net force closer to the necessary centripetal force. It's essential to understand this balancing act to correctly determine both the magnitude and direction of the normal force.
Circular Motion
Circular motion involves objects travelling in a circular path at a constant or varying speed. In the case of the roller coaster at the hill's top, the car is moving in circular motion due to the combined effect of gravitational force and the normal force.
  • Circular motion is characterized by the need for a centripetal force that points toward the center of the circle.
  • This force keeps the object moving in its circular path and is given by the equation: \[ F_{\text{centripetal}} = m \times \frac{v^2}{r} \]
Here, \( m \) represents the mass of the object, \( v \) its velocity, and \( r \) the radius of the circle. At the top of the hill, the net force required for circular motion is the sum of the gravitational force \( m \times g \) and the normal force \( F_N \). Understanding these interactions is key to mastering circular motion.
Newton's Second Law
Newton's Second Law of Motion lays the foundation for understanding how forces affect motion. It states that the net force acting on an object is equal to the product of its mass and its acceleration: \[ F_{\text{net}} = m \times a \]
When applying this to circular motion, the acceleration involved is the centripetal acceleration: \[ a_{\text{centripetal}} = \frac{v^2}{r} \]In our roller coaster scenario, Newton's Second Law helps us determine the net force required to keep the car moving in a circle at the hill's top.
  • The equation ties the gravitational force and normal force into the necessity of centripetal force for circular motion.
  • Thus, the sum of the gravitational force and the normal force needs to equal the centripetal force derived from the car's velocity and the hill's radius: \[ mg + F_N = m \times \frac{v^2}{r} \]
By using Newton's Second Law, we can calculate whether these forces are adequate for maintaining circular motion or if adjustments in the normal force's magnitude or direction are required.

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Most popular questions from this chapter

Calculate the ratio of the drag force on a jet flying at 1000 \(\mathrm{km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a prop- driven transport flying at half that speed and altitude. The density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\) at \(10 \mathrm{~km}\) and \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\) at \(5.0 \mathrm{~km}\). Assume that the air- planes have the same effective cross-sectional area and drag coefficient \(C\).

A worker pushes horizontally on a \(35 \mathrm{~kg}\) crate with a force of magnitude \(110 \mathrm{~N}\). The coefficient of static friction between the crate and the floor is 0.37. (a) What is the value of \(f_{s, \max }\) under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the Fig. \(6-2\) first worker's \(110 \mathrm{~N}\) push to move the crate? (e) If, Problem instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

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A student of weight 667 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force \(\vec{F}_{N}\) on the student from the seat is \(556 \mathrm{~N}\). (a) Does the student feel "light" or "heavy" there? (b) What is the magnitude of \(\vec{F}_{N}\) at the lowest point? If the wheel's speed is doubled, what is the magnitude \(F_{N}\) at the (c) highest and (d) lowest point?

ssM 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) \(334 \mathrm{~N},(\mathrm{c}) 445 \mathrm{~N}\), and \((\mathrm{d}) 556 \mathrm{~N}\). 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?

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