homework #4

Homework #4

Due: 11:58pm on Monday, April 28, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

Video Solution Problem for Chapter 14 Sections 1-3 - Work-Energy and Particle Kinematics.

Watch the following video and answer the question.

Part A

The Skier starts from rest at point A. Determine the horizontal distance traveled after he leaves the ramp and the landing speed. Neglect friction.X

ANSWER:

Correct

Work-Energy Scaling

Learning Goal:

To develop a better understanding of the work needed to produce a given change in an objects speed as the objects mass or speed change.

A particle of mass moves along a straight line with initial speed . A force of magnitude pushes the particle a distance along the direction of its motion.

Part A

Find , the particles final speed after the particle has traveled a distance .

Express your answer in terms of , , , and .

X = 21.0 m, = 22.2 vc msX = 21.0 m, = 27.2 vc msX = 13.8 m, = 27.2 vc msX = 21.0 m, = 17.5 vc ms

M vi F s

vf s

vi M F s

You did not open hints for this part.

ANSWER:

Increase in mass

For the next two parts, assume that the particle's mass is increased to , while all other parameters in the problem introduction remain the same.

Part B

By what multiplicative factor, , does the initial kinetic energy increase, and by what multiplicative factor, , does the work done by the force increase (with

respect to the case when the particle had a mass )?

If one of the quantities doubles, for instance, it would increase by a factor of 2. If a quantity stays the same, then the multiplicative factor would be 1.

Express your answers numerically separated by a comma.


ANSWER:

Part C

The particle's change in speed over the distance will be _____ the change in speed when the particle had a mass equal to .

ANSWER:

= vf

3M

RT RU

M

, = RT RU

s M

Increase in initial speed

For the final two parts, assume that the initial speed of the particle is increased to , with the particle's mass once again equal to .

Part D

By what factor, , does the initial kinetic energy increase (with respect to the first scenario, with mass and speed ), and by what factor, , does the

work done by the force increase?

Express your answers numerically separated by a comma.

ANSWER:

Part E

The particle's change in speed over the distance will be _____ the change in speed when the particle had an initial speed equal to .


ANSWER:

greater than

equal to

less than

3vi M

RT M vi RU

, = RT RU

s vi

greater than

equal to

less than

Problem 14.9

Part A

If the 45 crate starts from rest and attains a speed of 5 when it has traveled a distance of 15 , determine the force acting on the crate. The

coefficient of kinetic friction between the crate and the ground is = 0.3.

Express your answer to three significant figures and include the appropriate units.

ANSWER:

Correct

Problem 14.18

The two blocks and have weights = 72 and = 13 .

kg m/s m P

k

= 170 P N

A B WA lb WB lb

Part A

If the kinetic coefficient of friction between the incline and block is = 0.2, determine the speed of after it moves 3 down the plane starting from rest.

Neglect the mass of the cord and pulleys.


ANSWER:

Correct

Problem 14.19

Part A

If the 10- block passes point on the smooth track with a speed of = 5 , determine the normal reaction on the block when it reaches point .


A k A ft

= 2.97 vAfts

lb A vA ft/s B

ANSWER:

Correct

Problem 14.24

The 0.3- ball of negligible size is fired up the smooth vertical circular track using the spring plunger.

Part A

The plunger keeps the spring compressed 0.04 when = 0.Determine how far it must be pulled back and released so that the ball will begin to leave the

track when = 135 .

Express your answer with the appropriate units.

= 20.5 NB lb

kg

m s s

ANSWER:

Correct

Bungee Jumping

Learning Goal:

To apply Newtons second law and the theorem of conservation of energy to solve kinetic problems.

A bungee jumper wants to jump off the edge of a bridge that spans a river below. The jumper has a mass , and the surface of the bridge is a height above the

water. The bungee cord, which has length when unstretched, will first straighten and then stretch as the jumper falls.

Assume the following:

The bungee cord behaves as an ideal spring once it begins to stretch and has spring constant .

The jumper does not actually jump but simply steps off the edge of the bridge and falls straight downward.

The jumper's height is negligible compared to the length of the bungee cord. Thus, the jumper can be treated as a point particle.

Use for the magnitude of the acceleration due to gravity.

= 0.155 s m

m hL

k

g

Part A - The height below the bridge at which the jumper hangs without oscillating

How far below the bridge, , will the jumper eventually be hanging, once the jumper stops oscillating and comes finally to rest? Assume that the jumper does not

touch the water.

Express your answer in terms of , , , and .


ANSWER:

Part B

This question will be shown after you complete previous question(s).

Problem 14.70

The 2- ball of negligible size is fired from point with an initial velocity of 10 up the smooth inclined plane.

d

m L g k

= d

kg A m/s

Part A

Determine the distance from point to where it hits the horizontal surface at .


ANSWER:

Correct

Part B

Also, what is its velocity when it strikes the surface?


ANSWER:

Correct

Principle of Linear Impulse and Momentum

Learning Goal:

To use the principle of linear impulse and momentum to relate a force on an object to the resulting velocity of the object at different times.

The equation of motion for a particle of mass can be written as

By rearranging the terms and integrating, this equation becomes the principle of linear impulse and momentum:

C D

= 8.53 d m

= 10.0 vms

m

F = ma = m dvdt

For problem-solving purposes, this principle is often rewritten as

The integral is called the linear impulse, , and the vector is called the particle's linear momentum.

Part A

A jetliner of mass 8.61104

is in level flight when it encounters a downdraft (a downward wind) that lasts for 1.15 . The vertical component of the jetliner's

velocity is 82.5 after the downdraft subsides. What is the downdraft's average force, , on the jetliner?

Express your answer numerically in newtons to three significant figures.


ANSWER:

Part B


Part C


F dt = m dv = m m t2t1

v2v1

v2 v1

m + F dt = mv1 t2

t1v2

F dt I mv

kg s

m/s F

= F N

Problem 15.28

The winch delivers a horizontal towing force to its cable at which varies as shown in the graph.

Part A

Determine the speed of the 80 bucket when . Originally the bucket is released from rest.

Express your answer with the appropriate units.

ANSWER:

Correct

Problem 15.32

Block weighs 10 and block weighs 3 . If is moving downward with a velocity = 3 at = 0. The coefficient of kinetic friction between the

horizontal plane and block is = 0.15.

F A

kg t = 24 s

= 16.6 v2ms

A lb B lb B ( )vB 1 ft/s tA A

Part A

Determine the velocity of when = 1 .


ANSWER:

Correct

Video Solution Problem for Chapter 15 Sections 1,2 - Linear Impulse and Momentum.


A t s

= 6.00 ( )vA 2fts

Part A

A golf ball having a mass of is struck such that it has an initial velocity of as shown. Determine the horizontal and vertical components of the

impulse given to the ball.

ANSWER:

Video Solution Problem for Chapter 15 Section 3 - Conservation of Linear Momentum.

40 g 200 m/s

Im = 6.93 N s, Im = 4.00 N spx py

Im = 4.00 N s, Im = 7.32 N spx py

Im = 4.00 N s, Im = 6.93 N spx py

Im = 6.93 N s, Im = 4.39 N spx py


Part A

A 0.6-kg brick is thrown into a 25-kg wagon which is initially at rest. If, upon entering, the brick has a velocity of as shown, determine the final velocity of

the wagon.

ANSWER:

Problem 15.61

Block has a mass of 2 and is sliding on a rough horizontal surface with a velocity = 1 when it makes a direct collision with block , which has a

mass of 1 and is originally at rest. The coefficient of kinetic friction between the blocks and the plane is = 0.3.

10 m/s

= 0.240 m/svwagon

= 0.208 m/svwagon

= 0.234 m/svwagon

= 0.203 m/svwagon

A kg ( )vA 1 m/s Bkg k

Part A

If the collision is perfectly elastic , determine the velocity of each block just after collision. Assume rightward direction is positive.

Express your answers using three significant figures separated by a comma.

ANSWER:

Correct

Part B

Also, determine the distance between the blocks when they stop sliding.

Express your answer using three significant figures and include the appropriate units.

ANSWER:

Correct

Ballistic Pendulum

Learning Goal:

(e = 1)

= 0.333,1.33 ( , (vA)2 vB)2 m/s

= 0.283 d m

To apply conservation of energy and momentum.

As shown, an object of mass is fired with an initial speed at a ballistic pendulums bob. The bob has mass and is suspended by a massless rod of length .

After the collision, the pendulums bob and object stick together and swing to a maximum angular displacement .

Part A - Expression for the initial velocity of the object

Find an expression for , the initial speed of the object fired.

Express your answer in terms of some or all of the variables , , , and and , the acceleration due to gravity.


ANSWER:

Part B - The ratio of initial speeds of bullets fired from 9-mm and 0.44-caliber guns

An experiment compares the initial speed of bullets fired from two handguns: a 9 and a 0.44 caliber. The guns are fired into a 10.0- pendulum bob that is

m v0 M L

v0

m M L g

= v0

mm kg

attached to an arm of length . Assume that the 9- bullet has a mass of 6.00 and the 0.44-caliber bullet has a mass of 12.0 . If the 9- bullet causes

the pendulum to swing to a maximum angular displacement of 4.30 and the 0.44-caliber bullet causes a maximum displacement of 10.1 , find the ratio of the

initial speed of the 9- bullet to the speed of the 0.44-caliber bullet, .

Express your answer numerically to three significant figures.


ANSWER:

Video Solution Problem for Chapter 15 Sections 5, 6 - Angular Momentum of a Particle.


L mm g g mm

mm ( /(v0)9 v0)0.44

= ( /(v0)9 v0)0.44

Part A

Two identical spheres with each are rigidly attached to the rotating light rigid structure. The velocity of the spheres is . Determine the

angular momentum about the axis of rotation.

m = 3 kg v = 2 m/s

ANSWER:

Video Solution Problem for Chapter 15 Section 7 - Conservation of Angular Momentum.


= 7.20 kg Ho m2

s

= 3.60 kg Ho m2

s

= 1.80 kg Ho m2

s

= 6.00 kg Ho m2

s

Part A

The projectile having a mass of is fired from a cannon with a muzzle velocity of . Determine the projectile's angular momentum about

point at the instant it is at the maximum height of its trajectory.

ANSWER:

Angular Impulse and Momentum Principles

Learning Goal:

To apply the principle of angular impulse and momentum to describe a particle's motion.

The moment of a force about a point O, fixed in an inertial coordinate system, , and the angular momentum about the same point, , are related as follows:

where is the time derivative of the angular momentum, . Integrating this equation with respect to time yields the following equation:

m = 3 kg = 500 m/svo

O

= 6.76( ) kg /s CWH0 106 m2

= 6.76( ) kg /s CCWH0 106 m2

= 15.11( ) kg /s CWH0 106 m2

= 15.11( ) kg /s CCWH0 106 m2

MO HO

=MO HOHO = r mvHO

This equation is the principle of angular impulse and momentum, and it is often rearranged to its more familiar form

Part A

A centrifugal governor consists of a central rotating shaft that has two thin, pin-connected rods attached to it; a heavy sphere caps the end of each rod. A

centrifugal governor mechanically limits an engine's speed. A part of the engine turns the centrifugal

governor, and if the speed exceeds a set amount, the height of the spheres decreases the driving force

of the engine by reducing the fuel flow. The two rods move freely about the pin. If the whole apparatus

is rotating about the central shaft and the spheres have a tangential velocity, , the thin rods will create

an angle, , between each rod and the central shaft. Develop an equation for the tangential velocity, ,

in terms of some or all of the following: , the angle between the thin rods and the central shaft; , the

length from the pin to each sphere's center; , each sphere's mass; and , the acceleration due to

gravity. Neglect the mass of the thin rods.

Express your answer in terms of some or all of the variables , , , and .


ANSWER:

Part B

dt = ( ( t2t1

MO HO)2 HO)1

( + dt = (HO)1 t2

t1MO HO)2

v

v

l

m g

l m g

= v


Part C


Problem 15.100

The small cylinder has a mass of 10 and is attached to the end of a rod whose mass may be neglected. The frame is subjected to a couple

, where is in seconds, and the cylinder is subjected to a force of 60 , which is always directed as shown. The cylinder has a speed =

2 when = 0.

Part A

Determine the speed of the cylinder when = 2 .


ANSWER:

C kgM = (8 + 5)N mt2 t N v0

m/s t

t s

Correct

Problem 15.105

The four 9- spheres are rigidly attached to the crossbar frame having a negligible weight. A couple moment , where is in seconds, is

applied as shown.

Part A

Determine the speed of each of the spheres in 5 seconds starting from rest. Neglect the size of the spheres.


ANSWER:

Correct

= 13.4 vms

lb M = (0.5t + 0.8) lb ft t

= 15.3 v fts

Score Summary:

Your score on this assignment is 99.7%.

You received 99.7 out of a possible total of 100 points.

homework #4

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