/*! 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 66 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. What are the magnitude and direction of the force of the track on the car at the top of the hill if the car's speed is (a) \(11 \mathrm{~m} / \mathrm{s}\) and (b) \(14 \mathrm{~m} / \mathrm{s}\) ?

Short Answer

Expert verified
(a) 3698.67 N upward. (b) 1294.67 N downward.

Step by step solution

01

Identify the Given Values

Mass of the roller-coaster car, \(m = 1200 \, \mathrm{kg}\). Radius of the circular hill, \(r = 18 \, \mathrm{m}\). Car's speed \(v\) is given in two cases: (a) \(v = 11 \, \mathrm{m/s}\), (b) \(v = 14 \, \mathrm{m/s}\).
02

Determine the Centripetal Force

Centripetal force is needed to keep the car in circular motion. It is given by the formula \(F_c = \frac{mv^2}{r}\).
03

Calculate the Centripetal Force for (a) \(v = 11 \, \mathrm{m/s}\)

Substitute the given values into the formula: \[F_c = \frac{(1200 \, \mathrm{kg}) \times (11 \, \mathrm{m/s})^2}{18 \, \mathrm{m}} = \frac{1200 \times 121}{18} \, \mathrm{N} = 8073.33 \, \mathrm{N}\]
04

Calculate the Centripetal Force for (b) \(v = 14 \, \mathrm{m/s}\)

Substitute the given values into the formula: \[F_c = \frac{(1200 \, \mathrm{kg}) \times (14 \, \mathrm{m/s})^2}{18 \, \mathrm{m}} = \frac{1200 \times 196}{18} \, \mathrm{N} = 13066.67 \, \mathrm{N}\]
05

Determine the Gravitational Force

Gravitational force is given by the formula \(F_g = mg\), where \( g = 9.81 \mathrm{m/s^2} \). Substitute \(m = 1200 \mathrm{kg}\) to get: \[F_g = 1200 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} = 11772 \, \mathrm{N}\]
06

Calculate the Normal Force for (a) \(v = 11 \, \mathrm{m/s}\)

At the top of the hill, the normal force \(F_N\) and gravitational force \(F_g\) together provide the centripetal force. Thus, \(F_N + F_g = F_c\). Rearrange to find \(F_N\): \[F_N = F_c - F_g = 8073.33 \, \mathrm{N} - 11772 \, \mathrm{N} = -3698.67 \, \mathrm{N}\]. The negative sign indicates the force is upward.
07

Calculate the Normal Force for (b) \(v = 14 \, \mathrm{m/s}\)

Using \(F_N + F_g = F_c\), rearrange to find \(F_N\): \[F_N = F_c - F_g = 13066.67 \, \mathrm{N} - 11772 \, \mathrm{N} = 1294.67 \, \mathrm{N}\]. The positive value indicates the force is downward.

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Key Concepts

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

roller-coaster physics
Roller-coaster physics is all about understanding the forces at play as the ride zips through its track. A roller-coaster car is subjected to various forces that govern its motion and ensure it adheres to the track. The mass of the car, the shape of the track, gravitational forces, and speed are all crucial.
As the car moves over a hill or through a loop, different forces come into play including gravitational force and centripetal force. These forces need to be balanced to keep the car safely on the track and to ensure the thrilling experience for the riders.
For instance, at the top of a hill, the weight of the car tries to pull it down due to gravity, while the track exerts an upward force to support it. Balancing these forces correctly is a delicate act that engineers need to master.
circular motion
Circular motion refers to the movement of an object along a circular path. When a roller-coaster car moves over the top of a circular hill, it is in circular motion. A key characteristic of circular motion is that the direction of the velocity of the moving object constantly changes, even if the speed remains constant.
This changing direction is caused by a continuous acceleration towards the center of the circle. This is referred to as centripetal (center-seeking) acceleration. It is important to understand that this acceleration does not change the speed of the object, but continuously changes its direction. Thus, even if a roller-coaster maintains a consistent speed as it moves around a curve or a hill, it is still accelerating due to the continuous change in direction.
centripetal force
Centripetal force is the force that is required to keep an object moving in a circular path. Without this force, an object would simply move in a straight line due to inertia. For a roller-coaster car moving over a hill, the centripetal force is provided by the combined effect of the gravitational force and the normal force exerted by the track on the car.
The formula for centripetal force is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is the velocity of the car, and \( r \) is the radius of the hill. This formula shows that centripetal force increases with both the mass of the car and the square of its speed. Therefore, the faster the car moves and the heavier it is, the greater the centripetal force required to keep it in circular motion.
gravitational force
Gravitational force is the force of attraction between any two bodies that have mass. On Earth, this force pulls objects towards the center of the planet. The strength of this force is given by the formula \( F_g = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, approximately \( 9.81 \mathrm{m/s^2} \).
For a roller-coaster car at the top of a hill, the gravitational force acts downward. This force is crucial when calculating the net force acting on the car, especially in assessing whether the track needs to exert an additional force upwards (normal force) to keep the car on its circular path. When moving over a hill, the car's gravitational force opposes its centripetal force requirement, and understanding this balance is key to designing safe and exciting roller-coaster rides.

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Most popular questions from this chapter

An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed \(1100 \mathrm{~N}\). The coefficient of static friction between the box and the floor is \(0.35 .\) (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?

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 frictional force on the crate from the floor? (b) What is the maximum magnitude \(f_{\max }^{\text {stat }}\) of the static frictional force under the circumstances? (c) Does the crate move? (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 first worker's \(110 \mathrm{~N}\) push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

Two blocks of weights \(3.6 \mathrm{~N}\) and \(7.2 \mathrm{~N}\) are connected by a massless string and slide down a \(30^{\circ}\) inclined plane. The coefficient of kinetic friction between the lighter block and the plane is \(0.10 ;\) that between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string. (c) Describe the motion if, instead, the heavier block leads.

Two horizontal forces act on a \(2.0 \mathrm{~kg}\) chopping block that can slide over a frictionless kitchen counter, which lies in an \(x y\) plane. One force is \(\vec{F}_{A}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}+(4.0 \mathrm{~N}) \hat{\mathrm{j}}\). Find the acceleration of the chopping block in unit-vector notation when the other force is (a) \(\vec{F}_{B}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-4.0 \mathrm{~N}) \hat{\mathrm{j}}\), (b) \(\vec{F}_{B}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}+\) \((4.0 \mathrm{~N}) \hat{\mathrm{j}}\), and \((\mathrm{c}) \vec{F}_{B}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-4.0 \mathrm{~N}) \hat{\mathrm{j}}\).

Luggage is transported from one location to another in an airport by a conveyor belt. At a certain location, the belt moves down an incline that makes an angle of \(2.5^{\circ}\) with the horizontal. Assume that with such a slight angle there is no slipping of the luggage. Determine the magnitude and direction of the frictional force by the belt on a box weighing \(69 \mathrm{~N}\) when the box is on the inclined portion of the belt for the following situations: (a) The belt is stationary. (b) The belt has a speed of \(0.65 \mathrm{~m} / \mathrm{s}\) that is constant. (c) The belt has a speed of \(0.65 \mathrm{~m} / \mathrm{s}\) that is increasing at a rate of \(0.20 \mathrm{~m} / \mathrm{s}^{2}\). (d) The belt has a speed of \(0.65 \mathrm{~m} / \mathrm{s}\) that is decreasing at a rate of \(0.20 \mathrm{~m} / \mathrm{s}^{2}\). (e) The belt has a speed of \(0.65 \mathrm{~m} / \mathrm{s}\) that is increasing at a rate of \(0.57 \mathrm{~m} / \mathrm{s}^{2}\).

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