PC1431 Assignment 2 Answers

PC1431AY1213SEM2

Signed in as Mikael Lemanza

 Assig nme nt 2: Newton's L aws of Mo...

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 Ass ignment 2: Newton's Laws of Motion Due: 2:00am on Friday, Februa February ry 15, 2013 Note:

To understand how points are awarded, read your instructor's Gradin Grading g Polic y .

Pulling Three Blocks Three Thr ee identical blocks connected by ideal strings are being pulled pulled along along a horizontal frictionless frictionless surface by a horizontal force force tension in the the string string betw between een blocks B and C is mas s

= 0. 400

= 3.00 3.00

. The magnitude magnitude of the

. Assume that each block has

.

Part A What is the magnitude

of the force?

Express your answer numerically in newtons.

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

of the acceleration of block B?

Express your answer numerically in meter per second squared.

Hint 1. Consider 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 the same acc eleration as either block A or block B alone. The The advantage advantage of  of such such an approach is that the larger object has only one force fo rce acting acting on it (the tension tension 3.00 in the rop rope). e).

 ANSWE R: =

Hint 2. Find the acceleration of all three blocks Which of the following following expressions gives gives the magnitude magnitude

of the acceleration of the three blocks? blocks?

Hint 1. Consider 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 sum of the masses masses of all three blocks. blocks. Th The e only horizontal horizontal force acting acting on this larger object is , so you can use Newton's Newton's 2nd law law to determine an expression for its ac celeration.

 ANSWE R:

Close

 ANSWE R: = 4.50

Correct

Part B What is the tension

in the string between block A and block B?

Express your answer numerically in newtons

Hint 1. How to approach the question The tension

is the only only horizontal horizontal force force acting acting on block A. Thus, Thus, you can find find the acceleration of block A and and then apply Newton' Newton's 2nd

law. Note that all three blocks hav have e the s ame acceleration.

 ANSWE R: = 1.50

Correct

± A Ride on the Ferris Wheel  A woman rides on a Ferris wheel wheel of radius 16

that maintains the s ame speed throughout throughout its m otion. To better understand understand physics, she 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  scale sc ale reading versus time. Note that the graph has a minim minimum um value of 510 and a maximum maxi mum value of 666

.

Part A What is the woman's mass? Express your answer in kilograms.

Hint 1. How to approach the problem The woman woman is moving in a circle with constant s peed. To maintain this motion s he must experien experience ce a net acceleration (called centripetal acceleration) directed toward the center of t 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 

mass.

Hint 2. Find 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 wheel. Instead, the scale records t he normal force force act ing on the woman, which can vary vary as she moves moves along the ci rcular path and experiences different different accelerations. Th This is normal force force is sometimes ref referred erred to as an apparent weight , because it mimics the feelings of  being heavier or lighter. Note that the normal force is equal in magnitude to the gravitational gravitational force on the flat surface of the earth, so t he apparent weight is weight is just called the weight  in this static situation. weight in  As the woman travels travels along the circular path, her apparent apparent weight fluctuates between a maximum value value and a minimum value. value. At what location (A - D) will the apparent apparent weight be a maxi mum? Where will it be a minimum?

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

Hint 1. Analyze 1. Analyze the free- body d iagram Draw a free-body diagram for the woman. Assume that the

following st atements y direction points vertically upward. Which of the following y direction

are true for every  for  every point point on the circle traveled by the woman? Check all that apply.  ANSWER: The gravitational force acting on the woman points in the

y direction. y  direction.

The gravitational force acting on the woman points in the

y direction. y direction.

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

y direction. y direction.

The normal force points in the

y direction. y  direction.

The magnitude magnitude of the normal force is constant.

Hint 2. Analyze 2. Analyze the accelera tion  At all t imes during the woman's woman's motion, she experiences a net ac celeration (called centripetal acceleration) acceleration) directed toward the center of the Ferris wheel. At what location (A - D) will t he acceleration be in only t he vertical vertical direction?

Check all that apply.  ANSWER:

 A B C D

Hint 3. Applying 3. Applying Newton's Newton's 2 nd law Since the normal force and graviational force are the only forces acting on the woman in the vertical ( y ) direction, , where

and

are the magnitudes of the normal force and the grav gravitat itational ional force, respect ively.

 Apply what you know about the acceleration of the woman woman in the y  direction at certain points along the circular path to determine y direction when will be a maximum or minimum.

 ANSWE R:

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

of the woman?

Express your answer in meters per second squared.

Hint 1. How 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 her centripetal acc eleration: .

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

of the woman?

Express your answer in meters per second.

Hint 1. Determining the speed Since the Ferris wheel wheel turns at constant speed, the distance , where where

is the speed speed and and

the woman woman travels travels during some time interval interval is given given by

is the time. For a complete cycl e, the distance traveled traveled is the circumference circumference of the Ferris Ferris

wheel and the time required is one period

. Thus Thus for for a Ferris wheel of radius

,

.

Hint 2. Find the period The easiest easiest way to determine the speed of the woman is to calculate the distance she travels travels during one complete cyc le of  motion and div divide ide this by the time that it takes to complete this cyc le. The The time for for a complete cycle is called the period period . What is the period of the Ferris wheel? wheel? Express your your a ns nswe we r in seconds seconds..  ANSWE R: =

 ANSWER: =

 ANSWE R: =

 ANSWE R: = 60

Correct

The Window Washer   A window washer washer of mass magnitude

is s itting on a platform suspended by a syst em of cables and pulleys as shown . He is pulling on the cable with a force force of 

. The cables and pulleys are ideal (massless (massless and frictionless), and the platform platform

has negligible mass.

Part A Find the magnitude of the minimum force force

that allows the window washer to move move upward.

Express your answer in terms of the mass

and the the magnitude of the accele ration due to gravity

.

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

in the cable on which the man is pulling. pulling.

Express your answer in terms of some or all of the varia bles

,

, and

.

 ANSWE R: =

Hint 2. Upward force on the platform The tension along along the cable exerted on the platfor platform m

is equal equal to the force force

. The The 3 sections of the cable (strands) are seperated by the pulleys. What is the for force ce

, by the pulley to the left left of the diagram diagram in the upwa upward rd direction? direction? Remember Remember that the pulley to the left left is supported

by two strands of the cable. Express your answer in terms terms of

, the tension in the cable .

 ANSWE R: =

Hint 3. Upw Upward ard forces on window washer  What forces pull the window washer upward?  ANSWE R:  A force equal to

and a force force equal to

 A force equal to

and the normal force force acting on the window washer washer from from the platform equal equal to

The normal force force acting on the window washer from from the platform equal to  A force equal to

.

and a force force equal to

and the normal force force acting on the window washer washer from from the platform equal equal to

and a force force equal to

Hint 4. All 4. All forces on window washer  washer  What objects exert for forces ces on the window washer? washer?  ANSWE R: Only the platform and the string being pulled Only the platform and the earth Only the earth and the cable supporting the platform Only the cable being pulled and the cable s uppor upporting ting the platfor platform m Only the platform; the earth; and the cable.

 ANSWE R: =

Correct

Skydiving  A s ky diver of mass 80.0

(including parachute) parachute) jumps off a plane plane and begins begins her descent.

Throu Th rougho ghout ut this pro problem blem use 9.80 9.80

forr the magn fo magnitude itude of the acceler acceleration ation due to grav gravity.

Part A  At t he beginning beginning of her fall, fall, does the s ky diver hav have an acceleration?

Hint 1. Free fall The speed of an an object in fre free e fall through a medium (liquid or gas) increases as t he object initially begins to descend. If the resist 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, 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 terminal speed .

 ANSWE R: No; the sky diver falls at constant speed. Yes and her acceleration is directed upward. upward. Yes and her acceleration is directed downward.

Correct This applet shows the sk y diver (not (not to scale) with her position, speed, and acceleration graphed graphed as functions of tim e. You c an see how her  acceleration drops to zero over time, giving constant speed after a long time.

Part B  At s ome point during her her free free fall, the sky div diver er reaches her terminal terminal speed. What is the magnitude of the drag force force

due to air resistance that

acts on the sk y diver when when she has reached terminal speed? Express your answer in newtons.

Hint 1. Dynamic equilibrium  A body moving at at const ant velocity velocity i s in dynamic equilibriu equilibrium: m: The net force force acting on the body is zero. Since t he only forces acting on the sky div diver er are her weight weight and the drag force, when she falls at const ant speed these t wo forces forces m ust balance.

 ANSWE R: = 784

Correct

Part C For an object object falling through through air at a high speed speed

, the drag force force acting on it due to air resistance can be expressed as ,

where the coefficient is about 0. 250

depends on the shape and siz e of the falling object object and on the density of air. For a human body, the numerical value value for  for  .

Using this value for

, what is the terminal speed

of the sky diver diver? ?

Express you answer in meters per second.

Hint 1. Terminal 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 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 equilibriu equilibrium m mathematically and solve solve for the s peed.

 ANSWE R: = 56.0

Correct Recreational Recreational sky divers divers can control t heir terminal speed to some ext ent by c hanging hanging their body posture. W hen oriented in a headfirst headfirst dive, a sky diver diver can reach speeds of about 54 meters per sec ond (120 miles per hour). hour). For maxi mum drag and stability, sky divers divers often will orient themselves themselves " belly-first." belly-first." In this position, their terminal speed is typically around around 45 meters per sec ond (100 miles per hour). hour).

Part D When the s ky div diver er descends to a certain height from the ground, she deploys her parachute to ensure a safe landing. (Usually the parachute is deployed when the sky diver diver reaches an altit ude of about 900 --3000 .) Immediat Immediat ely after deploying the parachute, does the sk skydiver ydiver hav have ea nonzero acceleration?

Hint 1. How to approach the problem The parachute parachute acts to decrease the s peed of the sky div diver. er. A decrease in speed corresponds corresponds t o an acceleration opposite to t he direction of  motion.

 ANSWE R: No; the sk y diver keeps falling at constant s peed. Yes and her acceleration is directed downward. Yes and her acceleration is directed upward. upward.

Correct

Part E When the par parachu achute te is fully fully open, open, the eff effectiv ective e drag drag coeff coefficient icient of of the sky diver diver plus plus parach parachute ute increa increases ses to 60.0

. What is the drag drag force force

acting on the s ky div diver er immediately after she has opened the parachute? Express your answer in newtons.

Hint 1. How to approach the problem The drag force depends of the speed of the falling object object and the coeffici coefficient ent

. You simply need to determine the speed of the sky diver 

when she opens the parachute.

Hint 2. Find the speed of the sky diver when when the parachute is deployed When the parachu parachute te is deployed, what is the speed Express your answer in meters per second.

Hint 1. How to approach the question

of the sky div diver? er?

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

 ANSWE R: =

 ANSWE R: = 1.88×105

Correct

Part F What is the terminal speed

of the sky div diver er when the parachute is opened opened? ?

Express your answer in meters per second.

Hint 1. How to approach the problem Calculate the terminal terminal speed of the sky diver as you did in Part Part C, but take into account the new value of the coefficient coefficient

.

 ANSWE R: = 3.61

Correct  A ty pical "s tudent" parachute for recreational recreational skydiving skydiving has a drag coefficient coefficient that gives gives a terminal speed for landing landing of about about 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 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).

Exercise 4.30  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 and and 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 right.)

Part A 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 not be graded but the relative length of one to the other will be graded.  ANSWE R:

Correct

Part B Draw clearly labeled free-body diagram for the truck. (The bed of the truck is not  frictionless.) not 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 not be graded but the relative length of one to the other will be graded.  ANSWE R:

Correct

Block on an Incline Adjacent to a Wall  A wedge with an inclination of angle angle

rests next to a wall. A block of mass

wedge and the block or between the wedge and the horizontal surface.

is s liding down down the plane, as shown. There There is no friction between the

Part A Find the magnitude, Express

, of of the sum of of all forces acting on the block.

in terms of

and

, alo ng with any nece ssary cons constants. tants.

Hint 1. Directi Direction on of the n et force on the block The net net force on the block must be t he force in the direction of motion, which is down the incline.

Hint 2. Determi Determine ne the forces acting on the block What forces act on the block? Keep in mind that there is no friction between the block and the wedge.  ANSWE R: The weight of the block and friction 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 force that the wall exerts on the wedge

Correct

Hint 3. Find the magnitude of the force acting along the direction of motion Consider a coordinate system with the  x   x direction direction pointing down the incline and t he y  y direction direction perpendicular perpendicular to the incline. In these coordinates, what is , the component of the block's weight weight in the  x  direction?  x direction? Express

in terms of

,

, and

.

 ANSWE 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 (down the incline).

 ANSWE R: =

Correct

Part B Find the magnitude, Express

, of of the for force ce that the wall exerts on on the wedg wedge. e.

in terms of

and

, along with any necess necessary ary cons constants. tants.

Hint 1. The 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 wedge must be zero. If the block exerts a force with a horizontal component on the wedge, wedge, s ome other horizontal force force must act on t he wedge so that the net force is zero.

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

, of the normal normal (contact) force between between the block and the wedge? wedge? (You (You might have have computed this already while while

answering Part A.) Express

in terms of

,

, and

.

 ANSWE R: =

Correct

Hint 3. Find the horizontal component component of the normal force In the previous previous hint you found the magnitude of the normal force force between the block and the wedge. What is the magnitude,

, of the

horizontal component of this normal force? horizontal component Express

in terms of

and

.

 ANSWE R: =

Correct

 ANSWE R: =

Correct Your answer to Part B could be expressed as either approa approaches ches 90 degre degrees es ( the first first limit (

or

. In either form, we see that as

gets very small or as

radians) radians),, the contact force force between between the the wall and and the wedg wedge e goes goes to zero. This This is what we should should expect; expect; in

small), the block is accelerating very slowly, slowly, and all horizontal horizontal forces forces are are small. In In the second limit (

about 90 degrees), degrees), the

block s imply falls vertically and exerts no horizontal force force on the wedge. wedge.

Velocity from Force Diagram Ranking Task Below are are birds-eye birds-eye views of six identical toy cars moving moving to the right right at 2

. Various forces forces act on the cars with with magnitudes magnitudes and directions indicated

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

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

Hint 1. How to approach the problem First, added up the force force vectors to find the net force acting on the car. S ince you are ask ed about the speed a short t ime after the forces are applied, the speeds of all cars will be close to 2 . For each car, the small diff difference erence from 2 will be due to the acceleration caused by the net force force acting on the car. A net force force acting to the right right will result in a speed greater than 2 to the left left will result result in a speed less than 2

, while a net for force ce acting

. Recall that the acceleration acceleration may be be found found from the net net force force by using New Newton's ton's 2nd 2nd law, law,

.

Hint 2. Summing force vectors Forces are vectors vectors and s um in t he same way that velocity or acceleration vectors vectors s um. Be carefu carefull when resolving resolving vectors vectors into components, if you need to. Once the force vectors are are summed into a single total, or or net for force ce , the acceleration of the car of mass can be

determined by Newton’s 2nd law: .

 ANSWE R:

Correct

± Mass on Turntable  A s mall metal cy linder rests on a circular turntable that is rotating at a constant s peed as illustrated in the diagram . The small metal cylinder has has a mass of 0.20

, the coeff coefficient icient of static friction between between the

cylinder and and the turntable is 0.080, and the cylinder is located 0.15

from the center of the

turntable. Take the magnitude of the accel acceleration eration due to grav gravity ity to be 9.81

.

Part A What is the maximum speed speed

that the cylinder can mov move along its circular path without slipping off the turntable?

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

Hint 1. Centripetal a ccelerati cceleration on If you know a body body is in unifo uniform rm circular motion, you know what its acceleration must be. IfIf a body body of mass circle of radius  ANSWE R:

, what is the magnitude

of its centripetal acceleration?

is traveling traveling with speed speed

in a

Correct

Hint 2. Determi Determine ne the force causing acceleration Whenever you see uniform circular motion, t here is a real force that c auses the ass ociated centripetal acceleration. In this proble Whenever problem, m, what force causes the c entripetal acceleration?  ANSWE R: normal force static friction weightt of cylinder  weigh a force other than those above

Correct

Hint 3. Find the maximum possible friction force The magnitude

of the force due to static friction satisfies

. What is

in this proble problem? m?

Express your answer numerically in newtons to three significant figures.  ANSWE R: = 0.157

Correct

Hint 4. Newton's 2nd law To solve this problem, relate the answers to the previous two hints using Newton's 2nd law: .

 ANSWE R: = 0.34

Correct

Two Blocks and Two Pulleys  A block of mass

is att ached to a massless , ideal string. This st ring wraps wraps around a massless pulley and then wraps around around a second pulley that is

attached to a block of mass

that is free free to slide on a frictionless frictionless table. The The string is firmly anchored anchored to a wall wall and the whole whole system is frictionless.

Use the coordinate sys tem indicated in the figure when when solving solving this proble problem. m.

Part A  Assuming t hat

is the magnitude of the horizontal acceleration of the block block of mass

Express the tension in terms of

and

, what is

, the t ension in the string?

.

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

,

where wher e

,

, ... are for forces ces acting on the block of mass

. Keep in mind that the whole syst em is frictionless.

Hint 2. Force diagram for the block of mass Which figure figure correctly illustrates the forces acting on the block of of mass The vectors

,

,

, and

denote the normal force, the grav gravitat itational ional force, the

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

 ANSWE R: Figure 1 Figure 2 Figure 3 None of the above

Correct

 ANSWE R: =

Correct

Part B

?

Given Giv en

, the tension in the string, calculate

Express the accel era tion magni tude

, the magnitude of the vertical acceleration of the block of of mass

in terms of

,

, and

.

.

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

,

where wher e

,

, ... are for forces ces acting on the block of mass

.

Hint 2. Force diagram for the block of mass Which figure figure correctly illustrates the forces acting on the block of of mass The vectors

,

, and

?

denote the grav gravitat itat ional force, the tensi tension on in the stri string, ng,

and the inertial force, respect iv ively. ely.

 ANSWE R: Figure 1 Figure 2 Figure 3 None of the above

Correct

 ANSWE R:

=

Correct

Part C Given Giv en the magnitude Express

of the acceleration of the block of mass

in terms of

, find

, the magnitude of the horizontal acceleration of the block of mass

.

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

and

as the vertical coordinate of the block of mass

respectively. respectiv ely. It is clear that

and the horizontal coordinate of the block of mass

,

, the length of the string, is

, where wher e

is a constant that accounts for the wound wound portions portions of the string and and the length length of string between between the y  axis and the wall. Do not y axis

worry worr y about the value value of

, as it will vanish vanish in the next next st ep. By differentiating differentiating this equation twice with respect to time, you should obtain obtain a

relation between

. The variables

and

and

will vanish upon differentiation.

.

Hint 2. Method 2: Intuition (does not involve calculus) You should notice that, while while the block of of mass , where

and

descends a height

are the speeds of the block s of mass es

, the other will mov move e only half of and

. Hence, Hence, at each instant,

, respect ively. The formula for

versus

should be

obvious.

 ANSWE R: =

Correct

Part D Using the result of Part C in the formu formula la for Express your answer in terms of

that you prev previously iously obtained obtained in Part A, express express

and

as a function of

.

.

 ANSWE R: =

Correct

Part E Having Hav ing solved the prev previous parts, you hav have e all the pieces needed to calculate calculat e an express ion for

, the magnitude of the acc acceleration eleration of the block of mass

. Write

.

Express Expres s the acceleration magnitude

in terms of

,

, and

.

Hint 1. How to approach this problem In Part B, using Newton's Newton's 2nd law, you derived a relation between . Now eliminat e

and the tensi tension on in the stri string, ng,

from this system syst em of two linear equations and solve for

. In Part D you found found

as a function of 

.

 ANSWE R:

=

Correct

Hanging Chandelier   A c hande handelier lier with mass

is att ached to the ceiling of a large concert hall by two cables. B ecause the ceiling is cov covered ered with with intricate architect ural

decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung hung the c hande handelier lier couldn't attach t he cables to the ceiling directly above the chandelier. chandelier. Instead, they attached the cables cables to the ceiling near the walls. walls. Cable Cable 1 has tension and makes an angle of with the ceiling. Cable 2 has tension

and makes an angle of

with the ceiling.

Part A Find an expression for

, the tension in cable 1, that does does not depen depend d on

Express Expres s your answer in terms of some or all of the variable variable s

,

.

, and

, as well as the magnitude magnitude of the acceleration due to gravity

.

Hint 1. Find the sum of forces in the x  the x direction direction The chandelier is static; hence the vector forces on it sum to zero. Type in the sum of the  x  components of the forces acting on t he  x components chandelier, using the coordinate system shown. Express your answer in terms of some or all of the varia bles and

,

,

,

,

.

 ANSWE R: =

Correct

Hint 2. Find the sum of forces in the y  direction y direction Now type the corresponding equation relating the y  components of the forces acting on t he chandelier, chandelier, again using the c oordin oordinate ate sy stem y components shown. Express Expres s your answer in terms of some or all of the variables due to gravity

,

,

,

, and

, as well as the magnitude of the acceleration

.

 ANSWE R: =

Correct

Hint 3. Putting it all together  There are two unknowns in this problem, Eliminate Eli minate

 ANSWE R:

and

. Each of the prev previous ious two hints leads you to an an equation involving involving these two unknowns.

from this pair of equations and solve for

.

=

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 coefficient coeff icient of friction. friction. The racetrack consists of a flat, flat, circular road with a radius of 45 . The applet shows the result of driving the car around the track at various speeds.

Part A What is

, the coeff coefficient icient of static friction between the tires and the track? track?

Express your answer to two significant figures.

Hint 1. How 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 express ion for for the frictional force equal to the centripetal force needed to keep the car going in a circle. Solving this equation for gives the answer. Use

for the mass of the car in your calculations.

Hint 2. Use the applet to find 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  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 the maximum possible possible static friction. friction. What is the lowest speed at which which the car leav leaves es the track in the applet? Express your answer in meters per second as an integer.  ANSWE R: = 21

Correct If the lowest lowest speed at at which the car leaves the track is frictional force must be between between

, then the speed at at which the needed needed centripetal force force equals equals the maximum

and the closest speed below

in integer multiples of meters per second, so you would have have than

that you can measure. In this applet, you can only measure . You can use any speed greater than

and less

to calculate the centripetal force force that equals the maximum static friction.

Hint 3. Find an expres expression sion for  Find an expression for

, the coeff coefficient icient of of static friction between the car's tires and and the road. Use

magnitude of the acceleration due to grav gravity,

for the mass of the car,

for the speed at which which the car is just about to leav leave e the track, and

for the

for the radius radius of the

track. Express your answer in terms of 

,

,

, and

.

Hint 1. Expression for centripetal acceleration The centri centripetal petal accel acceleration eration

required to move an object in a circ ular path of radius

at speed

is

.

Hint 2. Expression for the force of static friction Recall that that static friction will equal the force force attempting to move move an object unless the magnitude of that force force exceeds is the magnitude of the normal force and magnitude to the weight of the car, giv giving ing

 ANSWE R:

=

Correct

, where where

is the coefficient coefficient of static friction. Since the track track is flat, the normal for force ce is equal in . Thu Thus, s, the magnitude of the maximum force of static friction is

.

 ANSWE R: = 0.91

Correct Score Summary: Your score on this assignment is 100%. You received 50 out of a possible total of 50 points.