Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 10, Problem 39AP

(a)

To determine

The distance d of the car coasts upward as a function of n .

(a)

Expert Solution
Check Mark

Answer to Problem 39AP

The distance d of the car coasts upward as a function of n is (1890+80n)(0.459m80n150) .

Explanation of Solution

Given information: The mass of car is 800kg , mass of counterweight is 950kg , mass of pulley is 280kg , radius of pulley is 0.700m , mass of each people is 80.0kg , speed of elevator car is 3.00m/s .

From the law of energy conservation,

Ei=Ef (1)

Here,

Ei is the total initial energy of the system.

Ef is the total final energy of the system.

Formula to calculate the total initial energy of the system is,

Ei=Kei+Kci+Ksi (2)

Here,

Kei is the initial translational kinetic energy of the elevator.

Kti is the initial translational kinetic energy of the counterweight.

Ksi is the initial rotational kinetic energy of the sheave pulley.

Write the expression for the initial translational kinetic energy of the elevator is,

Kei=12mev2

Here,

me is the mass of the elevator.

v is the initial speed of the system.

Write the expression for the initial translational kinetic energy of the counterweight is,

Kci=12mcv2

Here,

mc is the mass of the counterweight.

Write the expression for the initial rotational kinetic energy of the sheave is,

Ksi=12Isω2 (3)

Here,

Is is the moment of inertia of the sheave pulley.

ω is the angular speed of the sheave pulley.

Write the expression for the moment of inertia of the pulley is,

I=12msr2

Here,

ms is the mass of the sheave pulley.

Write the expression for the initial angular speed of the pulley is,

ω=vr

Here,

r is the radius of the sheave pulley.

Substitute 12msr2 for I and vr for ω in equation (3).

Ksi=12(12msr2)(vr)2=14msv2

Substitute 12mev2 for Kei , 12mcv2 for Kci and 14msv2 for Ksi in equation (2).

Ei=12mev2+12mcv2+14msv2=12(me+mc+12ms)v2 (4)

Since at the end the system comes to rest hence all the kinetic energies will be zero only potential energy remains in the system.

Formula to calculate the total final energy of the system is,

Ef=Uef+Uec+Ues (5)

Here,

Uef is the final potential energy of the elevator.

Kti is the final potential energy of the counterweight.

Ksi is the final potential energy of the sheave pulley.

Write the expression for the final potential energy of the elevator is,

Uef=megd

Here,

g is the acceleration due to gravity.

d is the distance cover by the system.

Write the expression for the final potential energy of the counterweight is,

Ucf=mcgd

Here,

g is the acceleration due to gravity.

d is the distance cover by the system.

Since the sheave pulley remains at its position so its final potential energy is zero.

Usf=0

Substitute megd for Uef , mcgd for Ucf and 0 for Usf in equation

Ef=megdmcgd+0=(memc)gd

Substitute 12(me+mc+12ms)v2 for Ei and (memc)gd for Ef in equation (1).

12(me+mc+12ms)v2=(memc)gd (6)

Formula to calculate the mass of the elevator is,

me=mcar+nm

Here,

mcar is the mass of the car.

n is the number of people enter in the elevator.

m is the mass of each person.

Substitute mcar+nm for me in equation ()

12((mcar+nm)+mc+12ms)v2=((mcar+nm)mc)gd

Substitute 800kg for mcar , 80kg for m , 950kg for mc , 280kg for ms , 3.00m/s for v and 9.8m/s2 for g in above equation to find the expression for d in terms of n .

12[(800kg+n(80kg))+950kg+12×280kg](3.00m/s)2=((800kg+n(80kg))950kg)(9.8m/s2)dd=[1890+n(80kg)](4.5m2/s2)[n(80kg)150kg](9.8m/s2)=(1890+80n)(0.459m80n150)

Conclusion:

Therefore, the distance d of the car coasts upward as a function of n is (1890+80n)(0.459m80n150) .

(b)

To determine

The distance d for n=2 .

(b)

Expert Solution
Check Mark

Answer to Problem 39AP

The distance d for n=2 is 94.1m .

Explanation of Solution

Given information: The mass of car is 800kg , mass of counterweight is 950kg , mass of pulley is 280kg , radius of pulley is 0.700m , mass of each people is 80.0kg , speed of elevator car is 3.00m/s .

The expression for the distance d for the n from part (a) is given by,

d=(1890+80n)(0.459m80n150)

Substitute 2 for n in above equation to find d .

d=(1890+80×2)(0.459m80×2150)=2050×0.459m10=94.1m

Conclusion:

Therefore, the distance d for n=2 is 94.1m .

(c)

To determine

The distance d for n=12 .

(c)

Expert Solution
Check Mark

Answer to Problem 39AP

The distance d for n=12 is 1.62m .

Explanation of Solution

Given information: The mass of car is 800kg , mass of counterweight is 950kg , mass of pulley is 280kg , radius of pulley is 0.700m , mass of each people is 80.0kg , speed of elevator car is 3.00m/s .

The expression for the distance d for the n from part (a) is given by,

d=(1890+80n)(0.459m80n150)

Substitute 12 for n in above equation to find d .

d=(1890+80×12)(0.459m80×12150)=2850×0.459m810=1.62m

Conclusion:

Therefore, the distance d for n=12 is 1.62m .

(d)

To determine

The distance d for n=0 .

(d)

Expert Solution
Check Mark

Answer to Problem 39AP

The distance d for n=0 is 5.79m .

Explanation of Solution

Given information: The mass of car is 800kg , mass of counterweight is 950kg , mass of pulley is 280kg , radius of pulley is 0.700m , mass of each people is 80.0kg , speed of elevator car is 3.00m/s .

The expression for the distance d for the n from part (a) is given by,

d=(1890+80n)(0.459m80n150)

Substitute 0 for n in above equation to find d .

d=(1890+80×0)(0.459m80×0150)=1890×0.459m150=5.79m

Conclusion:

Therefore, the distance d for n=0 is 5.79m .

(e)

To determine

The integral values of n for which the expression in part (a) apply.

(e)

Expert Solution
Check Mark

Answer to Problem 39AP

The expression in part (a) is valid only when n2 .

Explanation of Solution

Given information: The mass of car is 800kg , mass of counterweight is 950kg , mass of pulley is 280kg , radius of pulley is 0.700m , mass of each people is 80.0kg , speed of elevator car is 3.00m/s .

The expression for the distance d for the n from part (a) is given by,

d=(1890+80n)(0.459m80n150)

From the above expression, the distance d is a function of n . The distance can never be negative so, the minimum integral values of n should be greater than 2 .

Conclusion:

Therefore, the expression in part (a) is valid only when n2 .

(f)

To determine

The explanation for the answer in part (e).

(f)

Expert Solution
Check Mark

Answer to Problem 39AP

The mass of the elevator is less than the mass of the counterweight for the value of n=0 and for n=1 , so If it released the car would accelerate in upward direction.

Explanation of Solution

Given information: The mass of car is 800kg , mass of counterweight is 950kg , mass of pulley is 280kg , radius of pulley is 0.700m , mass of each people is 80.0kg , speed of elevator car is 3.00m/s .

The expression for the distance d for the n from part (a) is given by,

d=(1890+80n)(0.459m80n150)

Substitute 1 for n in above equation to find d .

d=(1890+80×1)(0.459m80×1150)=904.23m70=12.9m

Since the value of distance d is negative for n=0 or n=1 that indicates car would accelerate in upward direction because the mass of the counterweight becomes greater than the mass of the elevator which is undesirable.

Conclusion:

Therefore, the mass of the elevator is less than the mass of the counterweight for the value of n=0 and for n=1 , so If it released the car would accelerate in upward direction.

(g)

To determine

The value of d for infinite number of people enter into the elevator.

(g)

Expert Solution
Check Mark

Answer to Problem 39AP

The value of d for which the infinite number of people enter into the elevator is 0.459m .

Explanation of Solution

Given information: The mass of car is 800kg , mass of counterweight is 950kg , mass of pulley is 280kg , radius of pulley is 0.700m , mass of each people is 80.0kg , speed of elevator car is 3.00m/s .

The expression for the distance d for the n from part (a) is given by,

d=(1890+80n)(0.459m80n150)

Rearrange the above equation.

d=(1890n+80)(0.459m80150n)

Substitute for n in above equation to find d .

d=(1890+80)(0.459m80150)=(0+80)(0.459m800)=0.459m

Since the value of distance d is negative for n=0 or n=1 but it becomes positive after the value of n become greater than 2 because distance cannot be negative. So, the expression for the part (a) is valid only for n2 .

Conclusion:

Therefore, the value of d for which the infinite number of people enter into the elevator is 0.459m .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
An elevator system in a tall building consists of a 800-kg car and a 950-kg counterweight joined by a light cable of constant length that passes over a pulley of mass 280 kg. The pulley, called a sheave, is a solid cylinder of radius 0.700 m turning on a horizontal axle. The cable does not slip on the sheave. A number n of people, each of mass 80.0 kg, are riding in the elevator car, moving upward at 3.00 m/s and approaching the floor where the car should stop. As an energy-conservation measure, a computer disconnects the elevator motor at just the right moment so that the sheave–car– counterweight system then coasts freely without friction and comes to rest at the floor desired. There it is caught by a simple latch rather than by a massive brake. (a) Determine the distance d the car coasts upward as a function of n. Evaluate the distance for (b) n = 2, (c) n = 12, and (d) n = 0. (e) For what integer values of n does the expression in part (a) apply? (f) Explain your answer to part…
A wheel of radius 0.269 m, which we can model as a thin disk, is mounted on a frictionless horizontal axis. The mass of the wheel is 2.44 kg. A massless cord wrapped around the wheel is attached to a block of 4.16 kg that slides on a horizontal frictionless surface. If a horizontal force P with a magnitude of 10.1 N is applied to the block as shown below. If the wheel and block start at rest, and the block is moved through a displacement of 2.60 m what is the final angular velocity of the wheel in rad/s, assuming the cord does not slip. P
Two blocks are connected by a cord (of negligible mass) that passes over a pulley of negligible friction. The pulley is a solid disk of radius 14.6 cm and mass 12.3 kg. The surface on which mass M2 sits is frictionless. The two blocks have masses M1 = 3.23 kg and M2 = 7.99 kg. Determine the speed (in m/s) of block M2 after it has fallen a distance of 1.04 m.

Chapter 10 Solutions

Physics for Scientists and Engineers with Modern Physics

Ch. 10 - A machine part rotates at an angular speed of...Ch. 10 - A dentists drill starts from rest. After 3.20 s of...Ch. 10 - Why is the following situation impossible?...Ch. 10 - Review. Consider a tall building located on the...Ch. 10 - Prob. 8PCh. 10 - A discus thrower (Fig. P10.9) accelerates a discus...Ch. 10 - Prob. 10PCh. 10 - A car accelerates uniformly from rest and reaches...Ch. 10 - Review. A small object with mass 4.00 kg moves...Ch. 10 - Prob. 13PCh. 10 - Find the net torque on the wheel in Figure P10.14...Ch. 10 - A grinding wheel is in the form of a uniform solid...Ch. 10 - Review. A block of mass m1 = 2.00 kg and a block...Ch. 10 - Prob. 17PCh. 10 - Prob. 18PCh. 10 - Your grandmother enjoys creating pottery as a...Ch. 10 - Prob. 20PCh. 10 - You have just bought a new bicycle. On your first...Ch. 10 - Imagine that you stand tall and turn about a...Ch. 10 - Following the procedure used in Example 10.7,...Ch. 10 - Two balls with masses M and m are connected by a...Ch. 10 - Rigid rods of negligible mass lying along the y...Ch. 10 - A war-wolf or trebuchet is a device used during...Ch. 10 - Big Ben, the nickname for the clock in Elizabeth...Ch. 10 - Consider two objects with m1 m2 connected by a...Ch. 10 - Review. An object with a mass of m = 5.10 kg is...Ch. 10 - Prob. 30PCh. 10 - A uniform solid disk of radius R and mass M is...Ch. 10 - This problem describes one experimental method for...Ch. 10 - A tennis ball is a hollow sphere with a thin wall....Ch. 10 - A smooth cube of mass m and edge length r slides...Ch. 10 - Prob. 35PCh. 10 - Prob. 36APCh. 10 - Prob. 37APCh. 10 - Prob. 38APCh. 10 - Prob. 39APCh. 10 - Prob. 40APCh. 10 - Review. A string is wound around a uniform disk of...Ch. 10 - Review. A spool of wire of mass M and radius R is...Ch. 10 - Review. A clown balances a small spherical grape...Ch. 10 - Prob. 44CPCh. 10 - A spool of thread consists of a cylinder of radius...Ch. 10 - Prob. 46CPCh. 10 - A uniform, hollow, cylindrical spool has inside...Ch. 10 - A cord is wrapped around a pulley that is shaped...
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
Moment of Inertia; Author: Physics with Professor Matt Anderson;https://www.youtube.com/watch?v=ZrGhUTeIlWs;License: Standard Youtube License