/*! 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 131 A roller coaster is towed up an ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 131

A roller coaster is towed up an incline at a steady speed of $0.50 \mathrm{m} / \mathrm{s}$ by a chain parallel to the surface of the incline. The slope is \(3.0 \%,\) which means that the elevation increases by \(3.0 \mathrm{m}\) for every \(100.0 \mathrm{m}\) of horizontal distance. The mass of the roller coaster is \(400.0 \mathrm{kg}\). Ignoring friction, what is the magnitude of the force exerted on the roller coaster by the chain?

Short Answer

Expert verified
Answer: The force exerted by the chain is 117.72 N.

Step by step solution

01

Calculate the angle of incline

Since the slope is given as \(3.0 \%\), we can find the tangent of the angle of incline, which is the ratio of elevation to horizontal distance. \(\tan(\theta) = \frac{3.0}{100.0}\). Now we can find the angle of incline using the arctangent function: \(\theta = \arctan\left(\frac{3}{100}\right)\).
02

Calculate the gravitational force acting on the roller coaster

The gravitational force acting on the roller coaster, \(F_g\), can be calculated using \(F_g = mg\), where \(m\) is the mass of the roller coaster, and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)). So, \(F_g = 400.0 \mathrm{kg} \times 9.81 \mathrm{m/s^2} = 3924 \mathrm{N}\).
03

Calculate the component of gravitational force parallel to the incline

Since the force exerted by the chain opposes the component of gravitational force parallel to the incline, which we can find as \(F_{g\parallel} = F_g \sin(\theta)\), using the angle we found in Step 1. We find: \(F_{g\parallel} = 3924 \mathrm{N} \times \sin(\arctan(0.03)) = 117.72 \mathrm{N}\).
04

Apply Newton's second law

The force exerted by the chain, \(F_c\), must equal the component of gravitational force parallel to the incline since the roller coaster is moving at a constant speed (no acceleration). Using Newton's second law \(\Sigma \boldsymbol{F} = m \boldsymbol{a}\), we can write: \(F_c - F_{g\parallel} = ma \implies F_c = F_{g\parallel}\), since the acceleration is 0. \(F_c = 117.72 \mathrm{N}\) Therefore, the magnitude of the force exerted on the roller coaster by the chain is \(117.72 \mathrm{N}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 6.0 -kg block, starting from rest, slides down a frictionless incline of length \(2.0 \mathrm{m} .\) When it arrives at the bottom of the incline, its speed is \(v_{\mathrm{f}} .\) At what distance from the top of the incline is the speed of the block \(0.50 v_{\mathrm{f}} ?\)
You want to push a \(65-\mathrm{kg}\) box up a \(25^{\circ}\) ramp. The coefficient of kinetic friction between the ramp and the box is \(0.30 .\) With what magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed?
Tamar wants to cut down a large, dead poplar tree with her chain saw, but she does not want it to fall onto the nearby gazebo. Yoojin comes to help with a long rope. Yoojin, a physicist, suggests they tie the rope taut from the poplar to the oak tree and then pull sideways on the rope as shown in the figure. If the rope is 40.0 m long and Yoojin pulls sideways at the midpoint of the rope with a force of \(360.0 \mathrm{N},\) causing a \(2.00-\mathrm{m}\) sideways displacement of the rope at its midpoint, what force will the rope exert on the poplar tree? Compare this with pulling the rope directly away from the poplar with a force of \(360.0 \mathrm{N}\) and explain why the values are different. [Hint: Until the poplar is cut through enough to start falling, the rope is in equilibrium.] (IMAGE CAN'T COPY)
Find and compare the weight of a \(65-\mathrm{kg}\) man on Earth with the weight of the same man on (a) Mars, where \(g=3.7 \mathrm{N} / \mathrm{kg}.\) (b) Venus, where \(g=8.9 \mathrm{N} / \mathrm{kg} ;\) and (c) Earth's Moon, where \(g=1.6 \mathrm{N} / \mathrm{kg}.\)
The forces on a small airplane (mass \(1160 \mathrm{kg}\) ) in horizontal flight heading eastward are as follows: gravity \(=16.000 \mathrm{kN}\) downward, lift \(=16.000 \mathrm{kN}\) upward, thrust \(=1.800 \mathrm{kN}\) eastward, and \(\mathrm{drag}=1.400 \mathrm{kN}\) westward. At \(t=0,\) the plane's speed is \(60.0 \mathrm{m} / \mathrm{s} .\) If the forces remain constant, how far does the plane travel in the next \(60.0 \mathrm{s} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Scientific Method Physics

Read Explanation

Oscillations

Read Explanation

Nuclear Physics

Read Explanation

Quantum Physics

Read Explanation

Fluids

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.