PC1431 Assignment 2 Answers

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Ass

Ass ignment 2: Newton'ignment 2: Newton's Laws of Motions Laws of Motion Due: 2:00am on Friday,

Due: 2:00am on Friday, FebruaFebruary 15, ry 15, 20132013 Note:

Note:_{To understand how points are awarded, read your instructor's}_{To understand how points are awarded, read your instructor's} _Gradin_Gradin_{g Polic}_{g Polic}_y_y_._.

Pulling Three Blocks

Thr

Three identical blocks connected by ideal strings are being pee identical blocks connected by ideal strings are being pulled alonulled along a horizontal fg a horizontal frictionless surface by a horizontal forictionless surface by a horizontal force rce . The magn. The magnitude of theitude of the tension

tension in thin the strie string ng betwbetween een blocks B blocks B and and C is C is = 3.0= 3.00 0 . Assume . Assume that that each each block block hashas m

maasss s = = 00..44000 0 ..

Part A Part A

What

What is is the the magnitude magnitude of of the the force?force?

Express your answer numerically in newtons. Express your answer numerically in newtons.

Hint 1.

Hint 1.Find the acceleration of block BFind the acceleration of block B What is

What is the the magnitude magnitude of of the the acceleration acceleration of of block block B?B?

Express your answer numerically in meter per second squared. Express your answer numerically in meter per second squared.

Hint 1.

Hint 1.Consider blocks A and B as a unitConsider blocks A and B as a unit

Since blocks A and B are not moving with respect to each other, you can treat them as one larger "object." This larger object has Since blocks A and B are not moving with respect to each other, you can treat them as one larger "object." This larger object has the same acc

the same acceleration as either block A eleration as either block A or block B or block B alone. Thalone. The adve advantageantageof of such an approach is that the larger object has only onesuch an approach is that the larger object has only one fo

force arce acting on cting on it (the teit (the tension nsion 3.00 3.00 in the in the roprope).e).

ANSWE ANSWER:R:

Hint 2.

Hint 2.Find the acceleration of all three Find the acceleration of all three blocksblocks Which of the f

Which of the following ollowing expressions givexpressions gives the magnes the magnitude itude of of the acceleration the acceleration of the of the three bthree blocks?locks?

Hint 1.

Hint 1.Consider all three blocks as a unitConsider all three blocks as a unit

Since the three blocks are not moving with respect to one another, you can treat them as one larger "object" of mass equal to the Since the three blocks are not moving with respect to one another, you can treat them as one larger "object" of mass equal to the sum of the ma

sum of the masses of all three bsses of all three blocks. locks. ThThe only hore only horizontal force aizontal force acting on this larger cting on this larger object is object is , so you can use New, so you can use Newton's 2nd laton's 2nd laww to determine an expression for its ac

to determine an expression for its acceleration.celeration.

ANSWE ANSWER:R:

PC1431AY1213SEM2

PC1431AY1213SEM2 Assig

Assignmenment 2: nt 2: Newton's LNewton's Laws of Mo...aws of Mo... ResourcesResources

= =

Signed in as Mikael Lemanza

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ANSWE ANSWER:R:

Correct

Part B Part B What

What is is the the tension tension in in the the string string between between block block A A and and block block B?B?

Express your answer numerically in newtons Express your answer numerically in newtons

Hint 1.

Hint 1.How to approach the questionHow to approach the question The

The tension tension is the ois the only hornly horizontal fizontal force aorce acting on cting on block A. Tblock A. Thus, you hus, you can fcan find the ind the acceleration acceleration of of block A anblock A and then d then apply Napply Newton'ewton's 2nds 2nd law. Note that all

law. Note that all three blocks three blocks havhave the se the same acceleration.ame acceleration.

ANSWE ANSWER:R:

Correct

± A Ride on the Ferris Wheel

A woman rides on a Ferris whee

A woman rides on a Ferris wheel of radius 16 l of radius 16 that maintains the sthat maintains the same speed througame speed throughout its mhout its motion. To better undotion. To better understand physics, erstand physics, she takes along ashe takes along a digital bathroom scale (with memory) and sits on it. When she gets off the ride, she uploads the scale readings to a computer and creates a graph of digital bathroom scale (with memory) and sits on it. When she gets off the ride, she uploads the scale readings to a computer and creates a graph of sc

scale reading ale reading vversus time. Note that the ersus time. Note that the graph graph has a has a minimminimum vum value of alue of 510 510 and and aa maxi

maximum mum vvalue alue of of 666 666 ..

Part A Part A

What is the woman's mass? What is the woman's mass?

Express your answer in kilograms. Express your answer in kilograms.

Hint 1.

Hint 1.How to approach the problemHow to approach the problem The w

The woman is moving in a circle with oman is moving in a circle with constant sconstant speed. To maintain this motion speed. To maintain this motion she must he must experienexperience a net ce a net acceleration (called centripetalacceleration (called centripetal acceleration) directed toward the center of t

acceleration) directed toward the center of the Ferris wheel.he Ferris wheel.

Draw and analyze the woman's free-body diagram at a wisely chosen point on the circular path and use Newton's 2nd law to determine her Draw and analyze the woman's free-body diagram at a wisely chosen point on the circular path and use Newton's 2nd law to determine her

= = _4.50_4.50

= = _1.50_1.50

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mass. mass.

Hint 2.

Hint 2.Find the extreme points on the circular pathFind the extreme points on the circular path

The bathroom scale does not record the gravitational force acting on the woman. If it did, the reading would not vary as she rides the Ferris The bathroom scale does not record the gravitational force acting on the woman. If it did, the reading would not vary as she rides the Ferris wheel. Instead, the scale records t

wheel. Instead, the scale records the normal forhe normal force actce acting on the woman, which can varing on the woman, which can vary as y as she movshe moves along the cies along the circular path andrcular path and experiences diffe

experiences different accelerations. rent accelerations. ThThis is normal fornormal force is ce is sometimes refsometimes referred to as erred to as ananapparent weight apparent weight , because it mimics the feelings of , because it mimics the feelings of being heavier or lighter.

being heavier or lighter. Note that the normal force is

Note that the normal force is equal in magnitude to the gravequal in magnitude to the gravitational force on the flat surface of the earth, so titational force on the flat surface of the earth, so thehe apparent weight apparent weight is justis just called the

called theweight weight in this static situation.in this static situation. As

As the woman travthe woman travels along the circular path, her apparels along the circular path, her apparent weight fluctuates between a maximum vent weight fluctuates between a maximum value and a minimum value and a minimum value. At whatalue. At what location (A -

location (A - D) will the apparenD) will the apparent weight be a maxit weight be a maximum? Where will it mum? Where will it be a minimum?be a minimum?

Enter the letters that correspond to the correct positions separated by a comma. Enter the letters that correspond to the correct positions separated by a comma.

Hint 1.

Hint 1. Analyze the free- Analyze the free-body dbody diagramiagram

Draw a free-body diagram for the woman. Assume that the

Draw a free-body diagram for the woman. Assume that the y y direction points vertically upward. Which direction points vertically upward. Which of the followinof the following stg statementsatements are true for

are true for every every point on the circle traveled by the woman?point on the circle traveled by the woman? Check all that apply.

Check all that apply. ANSWER:

ANSWER:

Hint 2.

Hint 2. Analyze the accelera Analyze the accelerationtion At all t

At all times during the womanimes during the woman's motion, she experiences a net ac's motion, she experiences a net acceleration (called centripetal accelerationceleration (called centripetal acceleration) directed toward the) directed toward the center of the Ferris wheel. At

center of the Ferris wheel. At what location (A - D) will twhat location (A - D) will the acceleration be in only the acceleration be in only the vhe vertical direction?ertical direction?

Check all that apply. Check all that apply. ANSWER:

ANSWER:

The gravitational force acting on the woman points in the

The gravitational force acting on the woman points in the y y direction.direction. The gravitational force acting on the woman points in the

The gravitational force acting on the woman points in the y y direction.direction. The magnitude of the gravitational force acting on the woman is constant. The magnitude of the gravitational force acting on the woman is constant. The normal force points in the

The normal force points in the y y direction.direction. The normal force points in the

The normal force points in the y y direction.direction. The magn

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Hint 3.

Hint 3. Applying New Applying Newton's 2ton's 2nd lawnd law

Since the normal force and graviational force are the only forces acting on the woman in the vertical (

Since the normal force and graviational force are the only forces acting on the woman in the vertical (y y ) direction,) direction, ,

, where

where and and are are the the magnitudes magnitudes of of the the normal normal force force and and the the gravgravitatitational ional force, force, respectrespectively.ively.

Apply what you know about the acceleration of the wo

Apply what you know about the acceleration of the woman in theman in the y y direction at certain points along the circular path to determinedirection at certain points along the circular path to determine when

when will will be be a a maximum maximum or or minimum.minimum.

ANSWE ANSWER:R:

Hint 3.

Hint 3.Find the acceleration of the womanFind the acceleration of the woman What

What is is the the acceleration acceleration of of the the woman?woman?

Express your answer in meters per second squared. Express your answer in meters per second squared.

Hint 1.

Hint 1.How to approach the problemHow to approach the problem

You can use the information from the problem statement and the graph to determine the woman's speed. This can be used to find You can use the information from the problem statement and the graph to determine the woman's speed. This can be used to find her centripetal acc

her centripetal acceleration:eleration:

. .

Hint 2.

Hint 2.Find the woman's speedFind the woman's speed What is

What is the the speed speed of of the the woman?woman?

Express your answer in meters per second. Express your answer in meters per second.

Hint 1.

Hint 1.Determining the speedDetermining the speed Since the Ferris wh

Since the Ferris wheel turns at constant speed, the distance eel turns at constant speed, the distance the womathe woman travn travels during els during some time intervsome time interval is gival is given byen by , whe

, where re is the speis the speed aed and nd is the time. is the time. For For a complete a complete cyclcycle, the e, the distance travdistance traveled eled is the circumfis the circumference erence of of the Ferthe Ferrisris wheel

wheel and and the time the time required required is one is one period period . T. Thus fhus for or a a Ferris wheel Ferris wheel of of radius radius ,,

. .

Hint 2.

Hint 2.Find the periodFind the period The ea

The easiest siest way to determine the speed of the woman is way to determine the speed of the woman is to calculate to calculate the distance the distance she travshe travels during one complete cycels during one complete cycle of le of motion and

motion and divdivide this by the time ide this by the time that it takes to complete this cycthat it takes to complete this cycle. Thle. The time fe time for a or a complete cycle is called the percomplete cycle is called the period iod .. What is

What is the period of the Ferris wheethe period of the Ferris wheel?l? Express y

Express your aour ansnswewer in r in secondsseconds.. ANSWE ANSWER:R: ANSWER: ANSWER: ANSWE ANSWER:R: A A B B C C D D = = = = = =

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ANSWE ANSWER:R:

Correct

The Window Washer

A window wa

A window washer of mass sher of mass is sis sitting on a platform suspended by a systitting on a platform suspended by a system of cables and pulleys as shown . He is em of cables and pulleys as shown . He is pulling on the cable with a fopulling on the cable with a force of rce of magnitude

magnitude . The . The cables and cables and pulleys are pulleys are ideal (ideal (massless and frictionless), and massless and frictionless), and the platfothe platformrm has negligible mass.

has negligible mass.

Part A Part A

Find the

Find the magnitude of magnitude of the minimum fthe minimum force orce that allows the that allows the window window washer washer to movto move e upward.upward.

Express

Express your your answer in answer in terms terms of of the the mass mass and tand the magnitude of he magnitude of the the acceleacceleration due ration due to to gravity gravity ..

Hint 1.

Hint 1.Find a simple expression for the tensionFind a simple expression for the tension Find an

Find an expression fexpression for the or the tension tension in the in the cable on cable on which the which the man is pman is pulling.ulling.

Express

Express your your answer answer in in terms terms of of ssome ome or or all all of of the the variavariables bles , , , , and and ..

ANSWE ANSWER:R:

Hint 2.

Hint 2.Upward force on the platformUpward force on the platform The

The tension atension along long the cable the cable is equais equal to l to the fthe force orce . The . The 3 sections of 3 sections of the cable the cable (strands) (strands) are are seperated seperated by the by the pulleys. What is the pulleys. What is the forforcece exerted on the

exerted on the platforplatform m , by the pulley to the lef, by the pulley to the left of the diat of the diagram in the gram in the upwaupward direrd direction? Remembection? Remember that the pulley to the ler that the pulley to the left is supportedft is supported by

bytwotwostrands of the cable.strands of the cable. Express

Express your your answer in tanswer in terms erms of of , the , the tension tension in the in the cablecable..

ANSWE ANSWER:R:

Hint 3.

Hint 3.UpwUpward ard forces forces on window washer on window washer What forces pull the window washer upward? What forces pull the window washer upward? ANSWE ANSWER:R: = = ₆₀₆₀ = = = =

A force equal to

A force equal to and a forand a force equal toce equal to A force equal to

A force equal to and the normal fand the normal force acting on the window worce acting on the window washer frasher from the platform equaom the platform equal to l to .. The

The normal fornormal force acting on ce acting on the window the window washer fwasher from the rom the platform equal platform equal to to and and a fa force equal orce equal toto A force equal to

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Hint 4.

Hint 4. All forces on window w All forces on window washer asher What

What objects exert objects exert forforces on ces on the window wthe window washer?asher? ANSWE ANSWER:R: ANSWE ANSWER:R:

Correct

Skydiving

A s

A sky diver of ky diver of mass 80.0 mass 80.0 (including pa(including parachute) jumps off a prachute) jumps off a plane and belane and begins her descent.gins her descent.

Th

Throuroughoghout this ut this proproblem blem use 9use 9.80 .80 fofor the r the magnmagnitude itude of of the the acceleracceleration ation due due to gto gravravity.ity.

Part A Part A At t

At the beginnhe beginning of her fing of her fall, does the sall, does the sky diver haky diver havve an acceleration?e an acceleration?

Hint 1.

Hint 1.Free fallFree fall The speed of a

The speed of an object in n object in frefree fall through a medium (liquid or gas) increases as te fall through a medium (liquid or gas) increases as the object initially he object initially begins to descend. If the resistbegins to descend. If the resistance of ance of the medium is neglected, the speed of the object increases at a constant rate. If one takes into account the drag force due to resistance of the medium is neglected, the speed of the object increases at a constant rate. If one takes into account the drag force due to resistance of the medium, the acceleration of the object is no longer constant. The speed of the object increases as the object falls down, but only up to the medium, the acceleration of the object is no longer constant. The speed of the object increases as the object falls down, but only up to a certain value, called the

a certain value, called the terminal speed terminal speed ..

ANSWE ANSWER:R:

Correct

This

This appletappletshows the skshows the sk y diver (ny diver (not to ot to scale) with her position, scale) with her position, speed, and acceleration graphespeed, and acceleration graphed as functions of timd as functions of tim e. You ce. You c an see how her an see how her acceleration drops to zero over time, giving constant speed after a long time.

acceleration drops to zero over time, giving constant speed after a long time.

Part B Part B At s

At some point during home point during her freer free fall, the sky e fall, the sky divdiver reaches her terminaer reaches her terminal speed. What l speed. What is the magnitude of the drag fis the magnitude of the drag force orce due to air resistance thatdue to air resistance that acts

acts on the skon the sky diver wy diver when she has reached terminal speed?hen she has reached terminal speed? Express your answer in newtons.

Express your answer in newtons.

Hint 1.

Hint 1.Dynamic equilibriumDynamic equilibrium A body moving a

A body moving at constt constant vant velocity ielocity is in dynamic s in dynamic equilibriuequilibrium: The net form: The net force acting on the body is ce acting on the body is zero. Since tzero. Since the only forces acting on thehe only forces acting on the sky

sky divdiver are her weer are her weight and the drag force, when she falls at constight and the drag force, when she falls at constant speed these tant speed these two forwo forces mces must balance.ust balance.

ANSWE ANSWER:R:

Correct

Only the platform and the string being pulled Only the platform and the string being pulled Only the platform and the earth

Only the platform and the earth

Only the earth and the cable supporting the platform Only the earth and the cable supporting the platform Only the

Only the cable being pulled and the cable scable being pulled and the cable supporupporting the ting the platforplatformm Only the platform; the earth;

Only the platform; the earth; and the cable.and the cable.

= =

No; the sky diver falls at constant speed. No; the sky diver falls at constant speed. Yes

Yes and her acceleration is and her acceleration is directed upwardirected upward.d. Yes and her acceleration is directed downward. Yes and her acceleration is directed downward.

= = ₇₈₄₇₈₄

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Part C Part C

For an ob

For an object falling througject falling through air at a high speeh air at a high speed d , the drag fo, the drag force acting on it due to air resistance can be expressed asrce acting on it due to air resistance can be expressed as

, , where

where the coefficient the coefficient depends on depends on the shape and the shape and sizsize of e of the falling othe falling object and on bject and on the density of air. For the density of air. For a human a human body, the numerical vbody, the numerical value falue for or i

is s aabboouut t 00..22550 0 ..

Using

Using this this vvalue alue for for , , what what is is the terminal the terminal speed speed of of the the sky sky diverdiver??

Express you answer in meters per second. Express you answer in meters per second.

Hint 1.

Hint 1.Terminal speedTerminal speed

When an object in free fall has reached terminal speed, it has no acceleration and the net force acting on it is zero. Since the only forces When an object in free fall has reached terminal speed, it has no acceleration and the net force acting on it is zero. Since the only forces acting on the skydiver are the drag force and her weight, they must balance each other. To find the terminal speed of the skydiver, express acting on the skydiver are the drag force and her weight, they must balance each other. To find the terminal speed of the skydiver, express the condition of dynamic

the condition of dynamic equilibriuequilibrium mathematically m mathematically and solvand solve for the se for the speed.peed.

ANSWE ANSWER:R:

Correct

Recrea

Recreational sky tional sky divdivers can control ters can control t heir terminal speed to some extheir terminal speed to some ext ent by cent by c hanginhanging their body posture. Wg their body posture. W hen oriented in a headfhen oriented in a headfirst dive, airst dive, a sky

sky divdiver can reach speeds of about 54 meters per secer can reach speeds of about 54 meters per sec ond (120 miles per hour)ond (120 miles per hour). For maxi. For maxi mum drag and stability, mum drag and stability, sky sky divdivers often will orienters often will orient themselv

themselves "es " belly-firbelly-first." st." In this position, In this position, their terminal speed is their terminal speed is typically typically arounaround 45 meters per secd 45 meters per sec ond (100 miles per hour)ond (100 miles per hour)..

Part D Part D

When the s

When the sky ky divdiver descends to er descends to a certain height from the ground, she deploys her parachute to ensure a safe landing. (Usually the parachute isa certain height from the ground, she deploys her parachute to ensure a safe landing. (Usually the parachute is deployed when

deployed when the sky divthe sky diver er reaches an reaches an altitaltitude ude of of about 900 about 900 --3000 --3000 .) I.) Immediatmmediately after ely after deploying the deploying the parachute, does the parachute, does the skskydiver ydiver havhave e aa nonzero acceleration?

nonzero acceleration?

Hint 1.

Hint 1.How to approach the problemHow to approach the problem The pa

The parachute acts rachute acts to decrease the sto decrease the speed of the sky peed of the sky divdiver. A er. A decrease in speed corresponddecrease in speed corresponds ts to an acceleration opposite to to an acceleration opposite to the direction of he direction of motion. motion. ANSWE ANSWER:R:

Correct

Part E Part E When the

When the parparachuachute is fute is fully opelly open, the en, the effffectivective dre drag coeag coefffficient oicient of the f the sky divsky diver per plus palus parachrachute inute increacreases to 60.0 ses to 60.0 . What is the dr. What is the drag fag forceorce acting on the s

acting on the sky ky divdiver immediately after she has opened the parachute?er immediately after she has opened the parachute?

Express your answer in newtons. Express your answer in newtons.

Hint 1.

Hint 1.How to approach the problemHow to approach the problem The

The drag drag force depends force depends of of the speed of the speed of the falling obthe falling object and the coeffject and the coefficiicient ent . You simply need to . You simply need to determine the speed of determine the speed of the sky diver the sky diver when she opens the parachute.

when she opens the parachute.

Hint 2.

Hint 2.Find the speed of the Find the speed of the sky diver wsky diver when the parachute hen the parachute is deployedis deployed When

When the the parachuparachute te is is deployed, deployed, what what is is the the speed speed of of the the sky sky divdiver?er? Express your answer in meters per second.

Express your answer in meters per second.

Hint 1.

Hint 1.How to approach the questionHow to approach the question =

= _56.0_56.0

No; the sk

No; the sky diver keeps falling at constant sy diver keeps falling at constant speed.peed. Yes and her acceleration is directed downward. Yes and her acceleration is directed downward. Yes

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When she opens the parachute, she is still falling at a constant speed equal to the terminal speed calculated in Part C. When she opens the parachute, she is still falling at a constant speed equal to the terminal speed calculated in Part C.

ANSWE ANSWER:R: ANSWE ANSWER:R:

Correct

Part F Part F What

What is is the the terminal terminal speed speed of of the the sky sky divdiver er when when the the parachute parachute is is openedopened?? Express your answer in meters per second.

Express your answer in meters per second.

Hint 1.

Hint 1.How to approach the problemHow to approach the problem Calculate the termin

Calculate the terminal speed of al speed of the sky diver the sky diver as you did in Paras you did in Part C, but take into account the new t C, but take into account the new vvalue of alue of the coeffthe coefficient icient ..

ANSWE ANSWER:R:

Correct

A ty

A ty pical "spical "s tudent" parachute for rtudent" parachute for recreational skydivecreational skydiving has a drag coeffing has a drag coefficient that icient that givgives a terminal speed for landes a terminal speed for landing of abing of about 2 meters per out 2 meters per second (5 miles per hour). If this seems slow based on video or real-life sky divers you have seen, that may be because the sky divers you second (5 miles per hour). If this seems slow based on video or real-life sky divers you have seen, that may be because the sky divers you saw were using high-performance parachutes; these offer the sky divers more maneuverability in the air but increase the terminal speed up to saw were using high-performance parachutes; these offer the sky divers more maneuverability in the air but increase the terminal speed up to 4 meters per second (10 miles per hour).

4 meters per second (10 miles per hour).

Exercise 4.30

A large box containing your new computer sits on the bed of your pickup truck.

A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. The light turns green aYou are stopped at a red light. The light turns green and you stomp onnd you stomp on the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. (Assume that the truck is accelerating to the the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. (Assume that the truck is accelerating to the right.)

right.)

Part A Part A

Draw clearly labeled free-body diagram for the box. (The bed of the truck is

Draw clearly labeled free-body diagram for the box. (The bed of the truck is not not frictionless.)frictionless.)

Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded.

not be graded but the relative length of one to the other will be graded. ANSWE ANSWER:R: = = = = _1.88×10_1.88×1055 = = _3.61_3.61

(9)

Correct

Part B Part B

Draw clearly labeled free-body diagram for the truck. (The bed of the truck is

Draw clearly labeled free-body diagram for the truck. (The bed of the truck is not not frictionless.)frictionless.)

Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded.

not be graded but the relative length of one to the other will be graded. ANSWE

ANSWER:R:

Correct

Block on an Incline Adjacent to a Wall

A wedge with an inclination of a

A wedge with an inclination of angle ngle rests next rests next to a wall. A to a wall. A block of mass block of mass is sis sliding dowliding down the plane, as shown. Thern the plane, as shown. There is no friction between thee is no friction between the wedge and the block or between the wedge and the horizontal surface.

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Part A Part A

Find

Find the the magnitude, magnitude, , o, of f the the sum osum of f all all forces forces acting acting on on the the block.block.

Express

Express in in terms terms of of and and , , aloalong ng with with any any necenecesssary sary consconstants.tants.

Hint 1.

Hint 1.DirectiDirection of the non of the net force on the et force on the blockblock The ne

The net force on the block t force on the block must be tmust be the force in the direction of motion, which is he force in the direction of motion, which is down the incline.down the incline.

Hint 2.

Hint 2.DetermiDetermine the ne the forces acting on the forces acting on the blockblock

What forces act on the block? Keep in mind that there is no friction between the block and the wedge. What forces act on the block? Keep in mind that there is no friction between the block and the wedge. ANSWE

ANSWER:R:

Correct

Hint 3.

Hint 3.Find the magnitude of the Find the magnitude of the force acting along the force acting along the direction of motiondirection of motion Consider a coordinate system with the

Consider a coordinate system with the x x direction pointing down the incline and tdirection pointing down the incline and thehey y direction perpedirection perpendicular to the incline. ndicular to the incline. In theseIn these coordinates,

coordinates, what what is is , the , the component component of of the the block's wblock's weight eight in in thethe x x direction?direction? Express

Express in in terms terms of of , , , , and and ..

ANSWE ANSWER:R:

Correct

"Normal," in this context, is a synonym for "perpendicular." The normal force has no component in the direction of the block's motion "Normal," in this context, is a synonym for "perpendicular." The normal force has no component in the direction of the block's motion (down the incline).

(down the incline).

ANSWE ANSWER:R:

Correct

Part B Part B Find

Find the the magnitude, magnitude, , o, of f the the forforce that the ce that the wall wall exerts oexerts on n the the wedgwedge.e.

Express

Express in in terms terms of of and and , along , along with with any any necessnecessary ary consconstants.tants. The weight of the block and friction

The weight of the block and friction

The weight of the block and the normal (contact) force The weight of the block and the normal (contact) force The weight of the block and the weight of the wedge The weight of the block and the weight of the wedge

The weight of the block and the force that the wall exerts on the wedge The weight of the block and the force that the wall exerts on the wedge

= =

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Hint 1.

Hint 1.The force between the wall and the wedgeThe force between the wall and the wedge

There is no friction between the wedge and the horizontal surface, so for the wedge to remain stationary, the net horizontal force on the There is no friction between the wedge and the horizontal surface, so for the wedge to remain stationary, the net horizontal force on the wedge must be zero. If the block

wedge must be zero. If the block exerts a exerts a force with a horizontal component on the wedgeforce with a horizontal component on the wedge, s, some other horizontal forome other horizontal force must ce must act on tact on thehe wedge so that the net force is zero.

wedge so that the net force is zero.

Hint 2.

Hint 2.Find the normal force between the block and Find the normal force between the block and the wedgethe wedge What is

What is the magnitude, the magnitude, , of the norma, of the normal (contact) force betwel (contact) force between the block and the ween the block and the wedge? (dge? (You might havYou might have computed this already whe computed this already whileile answering Part A.)

answering Part A.) Express

Express in in terms terms of of , , , , and and ..

ANSWE ANSWER:R:

Correct

Hint 3.

Hint 3.Find the horizontal comFind the horizontal component of ponent of the normal forcethe normal force In

In the prevthe previous hint you found ious hint you found the magnitude of the magnitude of the normal fthe normal force between orce between the block and the the block and the wedge. What is the magnitude, wedge. What is the magnitude, , of , of thethe horizontal

horizontal component of this normal force?component of this normal force? Express

Express in in terms terms of of and and ..

ANSWE ANSWER:R:

Correct

ANSWE ANSWER:R:

Correct

Your

Your answer answer to to Part Part B B could could be be expressed expressed as as either either or or . . In In either either form, form, we we see see that that as as gets gets vvery ery small small or or asas app

approaroaches 90 deches 90 degregrees ( es ( radradians)ians), the contact for, the contact force betwce between theen the wall ae wall and the wnd the wedgedge goee goes to zero. Ths to zero. This is what we shouis is what we should expeld expect; inct; in the f

the first limit ( irst limit ( small), the small), the block is accelerating block is accelerating vvery slowery slowly, and ly, and all hoall horizontal frizontal forces arorces are small. Ie small. In the n the second limit ( second limit ( about about 90 90 degredegrees), thees), the block s

block s imply falls vertically and exerts imply falls vertically and exerts no horizontal forno horizontal force on the ce on the wedgwedge.e.

Velocity from Force Diagram Ranking Task

Below a

Below are bre birds-eye virds-eye views of iews of six identical toy cars movsix identical toy cars moving to ing to the righthe right at 2 t at 2 . Various f. Various forces act on orces act on the cars wthe cars with magnitudeith magnitudes and s and directions indicateddirections indicated below. All forces act

below. All forces act in the horizontal plane and are either parain the horizontal plane and are either parallel or at 45 llel or at 45 or 90 degreor 90 degrees to es to the car's motion.the car's motion.

Part A Part A

Rank these cars

Rank these cars on the basis on the basis of their speed a short timof their speed a short time (ie. before any car's speed can reach zero) after the forces are appliede (ie. before any car's speed can reach zero) after the forces are applied.. Rank from largest to smallest. To rank items as equivalent, overlap them.

Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1.

Hint 1.How to approach the problemHow to approach the problem First,

First, added up the foradded up the force vectors to find the net force acting ce vectors to find the net force acting on the car. Son the car. Since you are askince you are asked about the speed a short ted about the speed a short time after the forcesime after the forces are

are applied, applied, the the speeds speeds of of all all cars cars will will be be close close to to 2 2 . . For For each each car, car, the the small small diffdifference erence from from 2 2 will will be be due due to to the the accelerationacceleration caused by the

caused by the net fnet force acting orce acting on the on the car. A net fcar. A net force acting orce acting to the rto the right will ight will result in result in a speed a speed greater greater than 2 than 2 , while , while a net a net forforce actingce acting to the lef

to the left will rt will result in a esult in a speed speed less than 2 less than 2 . Recall . Recall that the accelerathat the acceleration may btion may be fe found ound from from the nethe net fort force by using ce by using NewNewton's 2nton's 2nd lawd law,, .

.

Hint 2.

Hint 2.Summing force vectorsSumming force vectors Forces are ve

Forces are vectors and sctors and sum in tum in the same way that he same way that vvelocity elocity or acceleration vor acceleration vectors sectors sum. Be um. Be carefucareful when resolvl when resolving ving vectors into ectors into components,components, if

if you you need need to. to. Once Once the the force force vvectors aectors are re summed summed into into a a single single total, ototal, or r net net forforce ce , , the the acceleration acceleration of of the the car car of of mass mass can can bebe = = = = = =

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determined by Newton’s 2nd law: determined by Newton’s 2nd law:

. . ANSWE ANSWER:R:

Correct

± Mass on Turntable

A s

A small metal cymall metal cylinder rests on a circular turntable that is linder rests on a circular turntable that is rotating at a constant srotating at a constant speed as illustrated in the diagram .peed as illustrated in the diagram . The

The small metal small metal cylinder hacylinder has a s a mass of mass of 0.20 0.20 , the , the coeffcoefficient of icient of static friction bestatic friction between tween thethe cylinder an

cylinder and the d the turntable turntable is 0.080, is 0.080, and and the cylinder the cylinder is located is located 0.15 0.15 from from the center the center of of thethe turntable.

turntable. Take

Take the the magnitude magnitude of of the the accelacceleration eration due due to to gravgravity to ity to be be 9.81 9.81 ..

Part A Part A

What is

What is the the maximum speemaximum speed d that the that the cylinder cylinder can mocan movve e along along its circular its circular path path without without slipping slipping off off the the turntable?turntable? Express your answer numerically in meters per second to two significant figures.

Express your answer numerically in meters per second to two significant figures.

Hint 1.

Hint 1.Centripetal aCentripetal acceleraticcelerationon If

If you know you know a boa body is in dy is in unifouniform circular rm circular motion, you motion, you know know what what its acceleration its acceleration must be. Imust be. If f a boa body of dy of mass mass is travis traveling eling with spewith speed ed in ain a circle

circle of of radius radius , , what what is is the the magnitude magnitude of of its its centripetal centripetal acceleration?acceleration?

ANSWE ANSWER:R:

(13)

Correct

Hint 2.

Hint 2.DetermiDetermine the ne the force causing accelerationforce causing acceleration Whenev

Whenever you see uniform circular motion, ter you see uniform circular motion, there is a real force that chere is a real force that causes the assauses the associated centripetal acceleration. In this ociated centripetal acceleration. In this probleproblem, whatm, what force causes the c

force causes the centripetal acceleration?entripetal acceleration? ANSWE

ANSWER:R:

Correct

Hint 3.

Hint 3.Find the maximum possible friction forceFind the maximum possible friction force The

The magnitude magnitude of of the the force force due due to to static static friction satisfies friction satisfies . . What What is is in in this this probleproblem?m?

Express your answer numerically in newtons to three significant figures. Express your answer numerically in newtons to three significant figures. ANSWE

ANSWER:R:

Correct

Hint 4.

Hint 4.Newton's 2nd lawNewton's 2nd law

To solve this problem, relate the answers to the previous two hints using Newton's 2nd law: To solve this problem, relate the answers to the previous two hints using Newton's 2nd law:

. . ANSWE ANSWER:R:

Correct

Two Blocks and Two Pulleys

A block

A block of mass of mass is attis attached to a masslessached to a massless, ideal string. This st, ideal string. This string wrring wraps around a massless pulley and then wraps arouaps around a massless pulley and then wraps around a second pulley that isnd a second pulley that is attached to a

attached to a block of block of mass mass that is frthat is free to ee to slide on slide on a fa frictionless table. The string rictionless table. The string is firmly anis firmly anchored chored to a wto a wall and all and the whthe whole system is frictionless.ole system is frictionless.

Use the coordinate sys

Use the coordinate system indicated in tem indicated in the figure wthe figure when solvhen solving this ing this probleproblem.m. normal force

normal force static friction static friction weigh

weight t of cylinder of cylinder

a force other than those above a force other than those above

= = _0.157_0.157

= = _0.34_0.34

(14)

Part A Part A

Assuming t

Assuming that hat is the magnitude of the horizontal acceleration of the blois the magnitude of the horizontal acceleration of the block of mass ck of mass , what is , what is , the t, the tension in the string?ension in the string?

Express

Express the the tension tension in in terms terms of of and and ..

Hint 1.

Hint 1.Which physical principle to useWhich physical principle to use You should use Newton's 2nd law: You should use Newton's 2nd law:

, ,

wher

where e , , , , ... ... are are forforces ces acting acting on on the the block block of of mass mass . . Keep Keep in in mind mind that that the the whole whole systsystem em is is frictionless.frictionless.

Hint 2.

Hint 2.Force diagram for the Force diagram for the block of massblock of mass Which f

Which figure igure correctly illustrates correctly illustrates the the forces forces acting acting on on the the block oblock of f mass mass ??

The

The vvectors ectors , , , , , , and and denote denote the the normal normal force, force, the the gravgravitatitational ional force, force, thethe tension in the

tension in the string, and the friction force, respectively.string, and the friction force, respectively.

ANSWE ANSWER:R:

Correct

ANSWE ANSWER:R:

Correct

Part B Part B Figure 1 Figure 1 Figure 2 Figure 2 Figure 3 Figure 3

None of the above None of the above

= =

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Giv

Given en , , the the tension tension in in the the string, string, calculate calculate , , thethemagnitudemagnitudeof of the the vvertical ertical acceleration acceleration of of the the block oblock of f mass mass ..

Express

Express the the accelacceleraeration tion magnimagnitude tude in in terms terms of of , , , , and and ..

Hint 1.

Hint 1.Which physical principle to useWhich physical principle to use Apply Newton's 2nd law

Apply Newton's 2nd law::

, ,

wher

where e , , , , ... ... are are forforces ces acting acting on on the the block block of of mass mass ..

Hint 2.

Hint 2.Force diagram for the Force diagram for the block of massblock of mass Which f

Which figure igure correctly illustrates correctly illustrates the the forces forces acting acting on on the the block oblock of f mass mass ??

The

The vvectors ectors , , , , and and denote denote the the gravgravitatitational ional force, force, the the tensitension on in in the the stristring,ng, and the inertial force, respect

and the inertial force, respectivively.ely.

ANSWE ANSWER:R:

Correct

ANSWE ANSWER:R:

Correct

Part C Part C Giv

Given en the the magnitude magnitude of of the the acceleration acceleration of of the the block block of of mass mass , , find find , , the the magnitude magnitude of of the the horizontal horizontal acceleration acceleration of of the the block block of of mass mass .. Express

Express in in terms terms of of ..

Hint 1.

Hint 1.Method 1: String constraint (uses calculus)Method 1: String constraint (uses calculus) Def

Define ine and and as as the the vvertical ertical coordinate coordinate of of the the block block of of mass mass and and the the horizontal horizontal coordinate coordinate of of the the block block of of mass mass ,, respectiv

respectively. It is ely. It is clear that clear that , the , the length length of of the string, the string, isis

, , wher

where e is a constant that accounts for is a constant that accounts for the wothe wound pund portions of ortions of the string anthe string and the lend the length of gth of string betwestring between theen they y axis and the wall. Do notaxis and the wall. Do not worr

worry about the vy about the value of alue of , as it will v, as it will vanish in the neanish in the next stxt step. By differep. By differentiating this equation entiating this equation twice with respect to time, you should obtwice with respect to time, you should obtain atain a relation

relation between between and and . . The The vvariables ariables and and will will vvanish anish upon upon differentiation.differentiation. Figure 1 Figure 1 Figure 2 Figure 2 Figure 3 Figure 3

None of the above None of the above

= =

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Hint 2.

Hint 2.Method 2: Intuition (does not involve calculus)Method 2: Intuition (does not involve calculus) You

You should should notice notice that, wthat, while hile the the block oblock of f mass mass descends descends a a height height , the other , the other will will movmove e only only half half of of . H. Hence, ence, at at each each instant,instant, ,

, where where and and are are the the speeds speeds of of the the blockblocks s of of massmasses es and and , , respectrespectively. ively. The The formula formula for for vversus ersus should should bebe obvious. obvious. ANSWE ANSWER:R:

Correct

Part D Part D Using

Using the the result result of of Part Part C C in in the the formuformula la for for that that you you prevpreviously oiously obtained btained in in Part Part A, expreA, express ss as as a a function function of of ..

Express

Express your your answer answer in in terms terms of of and and .. ANSWE ANSWER:R:

Correct

Part E Part E Hav

Having ing solved solved the prethe prevvious parts, you ious parts, you havhave e all the all the pieces needed pieces needed to calculatto calculate e , the , the magnitude of magnitude of the the accacceleration of eleration of the block the block of of mass mass . Write. Write an

an expressexpression ion for for .. Expres

Express s the the acceleration acceleration magnitude magnitude in in terms terms of of , , , , and and ..

Hint 1.

Hint 1.How to approach this problemHow to approach this problem In

In Part B, Part B, using Nusing Newton's ewton's 2nd 2nd law, law, you you derived derived a a relation between relation between and and the the tensitension in on in the the stristring, ng, . . In In Part D Part D you fyou found ound as as a a function of function of .

. Now Now eliminateliminate e from from this systthis system em of of two two linear linear equations equations and and solve solve for for ..

ANSWE ANSWER:R:

Correct

Hanging Chandelier

A c

A chandehandelier with mass lier with mass is attis attached to the ceiling of a large concert hall by two cables. Bached to the ceiling of a large concert hall by two cables. Because the ceiling is ecause the ceiling is covcovered wered with intricate architectith intricate architecturalural decorations (not indicated in the figure, which uses

decorations (not indicated in the figure, which uses a humbler depiction), the workers who huna humbler depiction), the workers who hung the cg the chandehandelier couldn't attach tlier couldn't attach the cables to he cables to the ceilingthe ceiling directly abov

directly above e the chathe chandelier. ndelier. Instead, Instead, they attached they attached the cabthe cables to les to the ceiling the ceiling near near the wthe walls. Caballs. Cable 1 le 1 has tension has tension and and makes an makes an angle angle of of with with thethe ceiling.

ceiling. Cable Cable 2 2 has has tension tension and and makes makes an an angle angle of of with with the the ceiling.ceiling. = = = = = =

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Part A Part A

Find

Find an an expression expression for for , the , the tension tension in in cable cable 1, 1, that dthat does oes not not dependepend d on on ..

Expres

Express s your your answer in answer in terms terms of of ssome or ome or all of all of the vthe variableariables s , , , and , and , as , as well as well as the magnituthe magnitude of de of the the acceleration due acceleration due to to gravitygravity .

.

Hint 1.

Hint 1.Find the sum of forces in theFind the sum of forces in the x x directiondirection

The chandelier is static; hence the vector forces on it sum to zero. Type in the sum of the

The chandelier is static; hence the vector forces on it sum to zero. Type in the sum of the x x components of the forces acting on tcomponents of the forces acting on thehe chandelier, using the coordinate system shown.

chandelier, using the coordinate system shown. Express

Express your your answer answer in in terms terms of of ssome ome or or all all of of the the variavariables bles , , , , , , ,, and . and . ANSWE ANSWER:R:

Correct

Hint 2.

Hint 2.Find the sum of forces in theFind the sum of forces in they y directiondirection Now type the corresponding equation relating the

Now type the corresponding equation relating the y y components of the forces acting on tcomponents of the forces acting on the chandelierhe chandelier, again using the c, again using the coordinoordinate syate systemstem shown.

shown. Expres

Express s your your answer answer in in terms terms of of ssome ome or or all all of of the the variables variables , , , , , , , , and and , , as as well as well as the the magnitude magnitude of of the the accelerationacceleration due

due to to gravity gravity .. ANSWE

ANSWER:R:

Correct

Hint 3.

Hint 3.Putting it all together Putting it all together There

There are are two two unknowns in unknowns in this problem, this problem, and and . Each . Each of of the prthe prevevious two ious two hints leads you hints leads you to ato an n equation invequation involving olving these two these two unknowns.unknowns. Eli

Eliminate minate from from this this pair pair of of equations equations and and solve solve for for ..

ANSWE ANSWER:R: = = = =

(18)

Correct

At the Test Track

You want to test the grip of the tires on your new race car. You decide to take the race car to a small test track to experimentally determine the You want to test the grip of the tires on your new race car. You decide to take the race car to a small test track to experimentally determine the coeff

coefficient of ficient of friction. The riction. The racetrack consists of a fracetrack consists of a flat, circular road lat, circular road with a with a radius of radius of 45 45 .. The appletThe appletshows the result of driving the car around the trackshows the result of driving the car around the track at various speeds.

at various speeds.

Part A Part A

What is

What is , the , the coeffcoefficient of icient of static friction static friction between between the tires the tires and and the trathe track?ck? Express your answer to two significant figures.

Express your answer to two significant figures.

Hint 1.

Hint 1.How to approach the problemHow to approach the problem

You need to find the point at which the force of friction is just strong enough to keep the car on the circular track. Then, you can set the You need to find the point at which the force of friction is just strong enough to keep the car on the circular track. Then, you can set the express

expression fion for or the frictional force the frictional force equal to equal to the centripetal force needed the centripetal force needed to keep to keep the car the car going in going in a circle. Solving a circle. Solving this equation for this equation for gives gives thethe answer.

answer. Use Use for for the the mass of mass of the the car car in in your your calculations.calculations.

Hint 2.

Hint 2.UUse the applet to se the applet to find the speedfind the speed

When the car successfully goes around the track, without leaving the track at all, then the needed centripetal force must be less than or When the car successfully goes around the track, without leaving the track at all, then the needed centripetal force must be less than or equal to the maximum possible static friction. When the car leaves the track during a lap, then the needed centripetal force must be greater equal to the maximum possible static friction. When the car leaves the track during a lap, then the needed centripetal force must be greater than

than the the maximum pomaximum possible static fssible static friction. What riction. What is the is the lowest lowest speed speed at wat which hich the the car car leavleaves the es the track in track in the the applet?applet? Express your answer in meters per second as an integer.

Express your answer in meters per second as an integer. ANSWE

ANSWER:R:

Correct

If

If the lowthe lowest speed aest speed at which the t which the car leavcar leaves the track is es the track is , then the , then the speed aspeed at which t which the nethe needed eded centripetal fcentripetal force eqorce equals the uals the maximummaximum frictional

frictional force force must must be be betweebetween n and and the the closest closest speed speed below below that that you you can can measure. measure. In In this this applet, applet, you you can can only only measuremeasure in

in integer integer multiples multiples of of meters meters per per second, second, so so you you would would havhave e . . You You can can use use any any speed speed greater greater than than and and lessless than

than to to calculate calculate the the centripetal centripetal forforce ce that that equals equals the the maximum maximum static static friction.friction.

Hint 3.

Hint 3.Find an Find an expresexpression for sion for Find

Find an an expression expression for for , the , the coeffcoefficient oicient of f static friction static friction between between the the car's car's tires atires and nd the the road. road. Use Use for for the the mass of mass of the the car, car, for for thethe magnitude

magnitude of of the acceleration the acceleration due due to grato gravvity, ity, for for the speed the speed at which the car at which the car is just about to is just about to leavleave the e the track, and track, and for for the rathe radius of dius of thethe track.

track.

Express your answer in terms of

Express your answer in terms of , , , , , , and and ..

Hint 1.

Hint 1.Expression for centripetal accelerationExpression for centripetal acceleration The

The centricentripetal petal accelacceleration eration required required to to move move an an object object in a in a circcircular ular path path of of radius radius at at speed speed is is ..

Hint 2.

Hint 2.Expression for the force of static frictionExpression for the force of static friction Recall tha

Recall that static friction will t static friction will equal equal the fthe force attempting orce attempting to movto move an object unless the e an object unless the magnitude magnitude of of that fothat force exceeds rce exceeds , whe, wherere is the magnitude

is the magnitude of the of the normal normal force force and and is the coeffis the coefficient of static friction. Since the traicient of static friction. Since the track is flat, the normal ck is flat, the normal forforce is equal ce is equal inin magnitude

magnitude to to the the weight weight of of the the car, car, givgiving ing . . ThuThus, the s, the magnitude magnitude of of the the maximum maximum force force of of static static friction friction is is ..

ANSWE ANSWER:R:

Correct

= = = = ₂₁₂₁ = =

(19)

ANSWE ANSWER:R:

Correct

Score Summary: Score Summary:

Your score on this assignment is 100%. Your score on this assignment is 100%.

You received 50 out of a possible total of 50 points. You received 50 out of a possible total of 50 points.

= = 0.910.91