AP PHYSICS C – MECHANICS SUMMER ASSIGNMENT 2020/2021

This course is equivalent to pre-engineering introductory Physics course for the university students. Comprehension of important physical concepts is emphasized in this course, together with development of significant proficiency in ability to understand and solve quantitative physical problems.

All students entering this course during the upcoming school year are required to complete this assignment.

This assignment is comprised of two components:

• Review of essential concepts

• Performance Task

There are four major topics included which will prepare students with stronger concept understanding and math skills for the upcoming school year. If you need assistance with any of the topics included in this assignment, you can use the following resource:

http://www.khanacademy.org/

If you would like additional practice/assistance with any topic in this assignment visit: Physics hyper textbook site or email me at anagpal@teaneckschools.org

Below are the POLICIES of the summer assignment:

•	The summer assignment is due the first day of class. On the first day of class, teacher will collect the summer assignment. Any student who does not have the assignment will be given one by the teacher.

•	Summer assignments will be graded basedoncompletionof theassignmentas pass/fail (pass-100%, fail-55%).

• Any student who registers as a new attendee of Teaneck High School after August 15thwill have an extra week to complete the summer assignment.

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Topic-1 Motion in 1 dimension

In analyzing the motion of objects, there are four basic parameters to keep track of. These are time, displacement, velocity, and acceleration. Time is a scalar, while the other three are vectors.

Displacement

The displacement represents the distance traveled, but it is a vector, so it also gives the direction. If you start in a particular spot and then move north 5 meters from where you started, your displacement is 5 m north. If you then turn around and go back, with a displacement of 5 m south, you would have traveled a total distance of 10 m, but your net displacement is zero, because you're back where you started. Displacement is the difference between your final position (x) and your starting point (xo):

Speed and velocity

Imagine that on your way to class one morning, you leave home on time, and you walk at 3 m/s east towards campus. After exactly one minute you realize that you've left your physics assignment at home, so you turn around and run, at 6 m/s, back to get it. You're running twice as fast as you walked, so it takes half as long (30 seconds) to get home again. There are several ways to analyze those 90 seconds between the time you left home and the time you arrived back again. One number to calculate is your average speed, which is defined as the total distance covered divided by the time. If you walked for 60 seconds at 3 m/s, you covered 180 m. You covered the same distance on the way back, so you went 360 m in 90 seconds.

Average speed = Total distance / elapsed time = 360 / 90 = 4 m/s. The average velocity, on the other hand, is given by: Average velocity = Total displacement / elapsed time. In this case, your average velocity for the round trip is zero, because you're back where you started so the displacement is zero

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Acceleration

An object accelerates whenever its velocity changes. Acceleration is the rate of change of velocity. Going back to the example we used above, let's say instead of instantly breaking into a run the moment you turned around, you steadily increased your velocity from 3m/s west to 6 m/s west in a 10 second period. If your velocity increased at a constant rate, you experienced a constant acceleration of 0.3 m/s/s (or, 0.3 m/s2).

Kinematics equations when the acceleration is constant

When the acceleration of an object is constant, calculations of the distance traveled by an object, the velocity it's traveling at a particular time, and/or the time it takes to reach a particular velocity or go a particular distance, are simplified. There are four equations that can be used to relate the different variables, so that knowing some of the variables allows the others to be determined.

Note that the equations apply under these conditions:

1. Acceleration is constant
2. Motion is measured from t = 0 , 3. Equations are:

V = V0+ a t (velocity = initial velocity + acceleration multiplied by time)

X = V0 t+ ½ a t2

V2 = V02+ 2ad (d = displacement)

The motion graphs can give the following information:

•	The slope of the velocity-time graph is the acceleration, while the area under the graph is the displacement.
•	The slope of position- time graph is velocity.( Refer to Hons Physics class notes for additional information)

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Example

A car travels up a hill at a constant speed of 37 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the average speed for the whole trip.

By definition the average speed is the ration of the total traveled distance and the total traveled time. Let us introduce the total traveled distance of the car as L. Then the time of the travel up the hill is

The time of the travel down the hill is

The total traveled time is

Then the average velocity is

Problems

a. b.

describestherelationshipbetweenthevelocityandtimeofthecart.
1. Intheaboveshowngraphalabcartmovesalongastraighthorizontaltrack.Thegraph

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the cart isdecreasing.
a.	Indicateeverytimeintervalforwhichspeed(magnitudeofthevelocity)of

b.	Indicateeverytimeatwhich the cartisatrest.
located at x = when t =0.
c.	Determinethehorizontalpositionxofthecartatt=4sifthecartis

0	0

10 s – 14 s, 14 s– 16 s, 16 s– 20 s.
d.	Determinethetraveleddistanceofthecartover10sfromthebeginning. e.	Findtheaccelerationofthecartduringtime:0s-4s,4s–8s,8s–10s,

2. The position-time graph below represents the motion of South's basketball coach during the last sixteen seconds of overtime during this past weekend's game.

Use the graph to answer the next several questions.

a. Determinethetotaldistancewalkedbythecoachduringthese16seconds. b. Determinetheresultingdisplacementofthecoachduringthese16seconds.
c.	Determinethedisplacementofthecoachafter12.0seconds.
position?
d. Atwhattimedidthecoachhavethegreatestdisplacementfromhisstarting

intervals for the last 16.0 seconds?
e. Whatwasthefastestspeedwhichthecoachwalkedduringanyofthetime

f.	Whatwastheaveragespeedofthecoachforthese16.0seconds?

After a long soccer practice down at the neighborhood soccer fields, Suzie begins
3. walkingupthesteephilltowardsherhome.Shegiveshersoccerballakickupthehill

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andcontinueswalkingtowardsit,meetingtheba lasitisrollingbackdown.The
following questions.
velocity-timegraphbelowdepictsthemotionofthelball.Usethegraphtoanswerthe

a. At what time did the ball change directions and begin rolling back downthe hill?

b. What is the acceleration of the ball as it rolls up the hill? Down the hill?

c. How far up the hill did the ball roll before it began to roll back down?

d. Determine the total distance traveled by the ball during the 5.00 seconds-both the distance up the hill and down the hill.

Youaredrivingalongthestreetatthespeedlimit(35mph/15.65m/s)and50meters
trafficlightwithinthe3secondsittakesforittoturnred.Whatisyourspeedasyou
4. beforereachingatrafficlightyounoticeitbecomingyellow.Youacceleratetomakethe

resistance.
crosstheintersection?Assumethattheaccelerationisconstantandthatthereisnoair

Acartravelsupahillataconstantspeedof24km/handreturnsdownthehillata
5. constantspeedof56km/h...Calculatetheaveragespeedforthewholetripinm/s.

Anengineerisdesigningtherunwayforanairport.Oftheplanesthatwillusethe
planewillbe65m/s.Assumingthisminimumacceeration,whatistheminimum
allowedlengthfortherunway.(Hint:initialvelocityiszero)
6. airport,thelowestaccelerationrateislikelytobe3lm/s2.Thetakeoffspeedforthis

Velocityatthetopiszeroanda=g=-9.8m/s2forverticalmotion)
7. Howhighcanahumanthrowaballifhecanthrowitwithinitialvelocity40m/s.(Hint

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Inacaraccidentinvolvingasportscar,skidmarksaslongas290.0mwereleftbyth
beforebrakingasbeing“inexcessof100mph”(161km/h).Supposethatittook10.0
8. carasitdeceleratedtoacompletestop.Thepolicereportcitedthespeedofthecare

secondsforthecartostop.Estimatethespeedofthecarbeforethebrakeswere
have to use system ofequations bysubstitution)
applied.(REMINDER:Answershouldread,“speedinexcessof...”)(Hint:youmay

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Topic-2 Vectors and Scalars

We'll move on from looking at motion in one dimension to motion in two or three dimensions. It's critical now to distinguish between two kinds of quantities, scalars and vectors. A scalar is something that's just a number with a unit, like mass ( 200 g ) or temperature ( -30°C). A vector has both a number and a direction, like velocity. For example a man is traveling 20 km/hr. east. Velocity is a combination of a scalar (speed, 20 km/hr.) and a direction (east).

Examples of scalars: mass, temperature, speed, distance

Examples of vectors: displacement, velocity, acceleration, force

One crucial difference between scalars and vectors involves the use of plus and minus signs. A scalar with a negative sign means something very different from a scalar with a plus sign; +30°C feels an awful lot different than -30°C, for example. With a vector the sign simply tells you the direction of the vector. If you're traveling with a velocity of 20 km/hr. east, it means you're traveling east, and your speed is 20 km/hr. A velocity of -20 km/hr. east also means that you're traveling at a speed of 20 km/hr., but in the direction opposite to east: 20 km/hr. west, in other words. With a vector, the negative sign can always be incorporated into the direction. Note that a vector will normally be written in bold, like this: A. A scalar, like the magnitude of the vector, will not be in bold face (e.g., A).

Components of a vector

A vector pointing in a random direction in the x-y plane has x and y components: it can be split into two vectors, one entirely in the x-direction (the x-component) and one entirely in the y-direction (the y-component). Added together, the two components give the original vector. The easiest way to add or subtract vectors, which is often required in physics, is to add or subtract components. Splitting a vector into its components involves nothing more complicated than the trigonometry associated with a right-angled triangle.

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Example

A vector, which we will call A, has a length of 5.00 cm and points at an angle of 25.0° above the negative x-axis, as shown in the diagram. The x and y components of A, Ax and Ay are found by drawing right-angled triangles, as shown. Only one right-angled triangle is actually necessary; the two shown in the diagram are identical.

Knowing the length of A, and the angle of 25.0°, Ax and Ay can be found by re-arranging the expressions for sin and cos.

Note that this analysis, using trigonometry, produces just the magnitudes of the vectors Ax and Ay. The directions can be found by looking on the diagram. Usually, positive x is to the right of the origin; Ax points left, so it is negative:

Ax = -4.53 cm in the x-direction (or, 4.53 cm in the negative x-direction)

Positive y is generally up; Ay is directed up, so it is positive:

Ay = 2.11 cm in the y-direction

Adding vectors

It's fairly easy to add vectors that are parallel to each other, or at right angles. In general, however, the angle between vectors being added (or subtracted) will be something other than 0, 90, or 180°. Consider the following example, where the vector C equals A + B.

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A has a length of 5.00 cm and points at an angle of 25.0° above the negative x-axis. B has a length of 7.00 cm and points at an angle of 40.0° above the positive x-axis. If C = A + B, what is the magnitude and direction of C? There are basically two ways to answer this. One way is to draw a vector diagram, moving the tail of B to the head of A, or vice versa. The vector C will then extend from the origin to wherever the tip of the second vector is.

The second way to find the magnitude and direction of C we'll use a lot in this course, because we'll often have vector equations of the form C =A + B. The simplest way to solve any vector equation is to split it up into one-dimensional equations, one equation for each dimension of the vector. In this case we're working in two dimensions, so the one vector equation can be replaced by two equations:

In the x-direction: Cx = Ax + Bx

In the y-direction: Cy = Ay + By

In other words, to find the magnitude and direction of C, the vectors A and B are split into components. The components are:

Ax = -4.532 cm in the x-direction

Ay = 2.113 cm in the y-direction

Bx = 7.00 cos40 = 5.362cm

By = 7.00 sin40 = 4.500 cm

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Bx = 5.362 cm in thex-direction

By = 4.500 cm in they-direction

The components of C are found by adding the components of A and B:

Cx = Ax + Bx = (-4.532 + 5.362) cm in the x-direction = 0.83 cm in the x-direction

Cy = Ay + By= (2.113 + 4.500) cm in the y-direction = 6.61 cm in the y-direction

The magnitude of C can be found from its components using the Pythagorean Theorem:

The direction of C can be found by taking the inverse tangent of Cy/Cx:

Inverse tan of 6.613 / 0.830 = 82.8°.

Combined, this gives C = 6.66 cm at an angle of 82.8° above the positive x-axis.

Note how the calculations and the diagrams go hand-in-hand. This will often be the case; it is always a good idea to draw diagrams as you resolve the vectors.

Problems

9. A busmakesthree displacements Ithefollowing order:

1) 58 mi, 38 east of north

2) 69 mi, 46 west of north; and 3) 75 mi,south-east

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horizontalpoints(north,east, south,andwest).
a. Drawacleardiagramshowingallthreedisplacementvectorswithrespectto

b. FindtheXandYcomponentsofdisplacementD1. c. FindtheXandYcomponentsofdisplacementD2.
d. Find the Xand Y componentsofdisplacementD3. e. Find the magnitude ofthe resultant vector.

f. Find the direction ofthe resultant vector.

10.Which of the following statements are true of scalars and vectors? (State the corrected statement if it is False.)

a. A vector quantity always has a direction associated with it. b. A scalar quantity can have a direction associated with it.
c. Vectors can be added together; scalar quantities cannot.
d. Vectors can be represented by an arrow on a scaled diagram; the length of the arrow represents the vector's magnitude and the direction it points represents the vector's direction.

11. Whichofthefollowingquantitiesarevectors?Includeallthatapply.

a. distance traveled b. displacement
c. average speed d. average velocity
e. instantaneous velocity f. acceleration

vectoradditiondiagrams?(StatethecorrectedstatementisitisFalse.)
12. Whichofthefollowingstatementsaretrueofvectoraddition,vectorsubtraction,and

a. Vectors A, B, and C are added together as A + B + C. If the order in which they are added is changed to C + B + A, then the result would be different.
b. Vectors A, B, and C are added together as A + B + C. If the order in which they are added is reversed to C + B + A, then the result would be a vector with the same magnitude but the opposite direction.
c. When constructing a vector diagram for A + B + C, it is not absolutely necessary that vectors B and C use the same scale that is used by vectorA.
d. The resultant in a vector addition diagram always extends from the head of the last vector to the tail of the first vector.
e. If vectors A and B are added at right angles to each other, then one can be sure that the resultant will have a magnitude that is greater than the magnitudes of either one of the individual vectors A and B.
f. If vectors A and B are added at right angles to each other, then one can be sure that the resultant will have a magnitude that is less than the arithmetic sum of the magnitudes of A and B.

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Topic-3 Projectile Motion

A projectile is shot at an initial velocity v0 at an angle  above the horizon. How far will it go? How high will it rise?

Denote the projectile's range R and the height of its trajectory h. Let us choose the origin of the coordinate system at the launch point, the x axis in the horizontal direction and the y axis in the vertical direction pointing up. Then the components of v0, the initial velocity vector, are

v0x = v0 cos  v0y = v0 sin 	(1)

Each vector equation is actually a set of two independent equations (this is a twodimensional motion) for the x and y components of the corresponding vectors.
So in reality we are dealing with two sets of independent equations for the x and y directions and can envision the problem as a superposition of two completely independent motions that together are equivalent to the projectile motion.

The Horizontal Component

Neglecting air resistance, the velocity x component does not change throughout the entire motion. Therefore, vx = v0x and

x = v0x t	(2) That's all for the horizontal motion.

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The Vertical Component

The y component of the projectile's motion is subject to the gravitational acceleration which points downward. So vy as a function of time is

vy = v0y - g t	(3)

For our choice of axes. The y motion can be envisioned as an object thrown straight up with velocity v0y. Such an object will rise until its velocity is zero, so from the last equation the time to reach maximum height is

tmax = v0y/g	(4)

The object will then fall to the ground and the downward trip is the mirror image of the trip up Therefore, the total time the object stays in the air/time of flight is

ttot = 2tmax = 2v0y/g	(5)

Putting it Together:

That's it. We have derived the total duration of the projectile's motion. To find the range R, all that is left is to plug ttot into equation (2) which gives us the distance covered at any time t. The result is

R = v0xttot

= 2v0xv0y/g

= 2(v02/g) sin  cos 	(6)

Where in the last step we inserted v0x and v0y from equation (1).

. This result can be simplified when we recall the identity from trigonometry sin2 = sin cos. With this identity, the range becomes simply

R = (v02/g) sin2	(7)

The maximum value of the sine function is 1, obtained when its angle is  (°). Therefore, for a given v0, the maximal range is v02/g, obtained for a launch angle  = °. The functionsin is symmetric about its peak at °; therefore the function sin2 is symmetric about °. For example, a projectile launched at 30° will have the same range as one launched at 60°.

The Trajectory Height

2	2
can directly use	h = v /2g = (v 2g) sin 
You can take half of the2total time/of flight and substitute in the second equation of motion or 0y	0
.

Horizontal Projectiles (Simple Projectile Motion)

When an object is launched or thrown completely horizontally, such as a rock thrown horizontally off a cliff, the initial velocity of the object is its initial horizontal velocity. Because horizontal velocity doesn’t change, this velocity is also the object’s final horizontal velocity, as well as its average horizontal velocity. Further, the initial vertical velocity of the projectile is

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zero. This means that you could hurl an object 1000 m/s horizontally off a cliff, and simultaneously drop an object off the cliff from the same height, and they will both reach the ground at the same time (even though the hurled object has traveled a greater distance).

Example

Fred throws a baseball 42 m/s horizontally from a height of 2m. How far will the ball travel before it reaches the ground?
To solve this problem, you must first find how long the ball will remain in the air. This is a vertical motion problem.

•	v0=0 m/s •	v=?
•	Δy=2m
•	a= g =9.8 m/s2 •	t=?

Now that you know the ball is in the air for 0.639 seconds, you can find how far it travels horizontally before reaching the ground. This is a horizontal motion problem, in which the acceleration is 0 (nothing is causing the ball to accelerate horizontally.) Because the ball doesn’t accelerate, its initial velocity is also its final velocity, which is equal to its average velocity.

•	v0=42 m/s •	v=42 m/s •	Δx=?
•	a=0
•	t=0.639s

You can therefore conclude that the baseball travels 26.8 meters horizontally before reaching the ground.

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Problems

anangle?(StatethecorrectedstatementifyoufindastatementFalse.)
13. Whichofthefollowingstatementsaretrueoftheverticalmotionofprojectilelaunched at

a. The vertical component of a projectile's velocity is a constant value of 9.8 m/s. b. The vertical component of a projectile's velocity is changing at a constantrate. c. The vertical velocity of a projectile is 0 m/s at the peak of its trajectory.
d. The vertical velocity of a projectile is unaffected by the horizontal velocity; thesetwo components of motion are independent of each other.
e. The final vertical velocity of a projectile is always equal to the initialvertical velocity.
f. The vertical acceleration of a projectile is 0 m/s/s when it is at the peak of its trajectory.
g. As a projectile rises towards the peak of its trajectory, the vertical acceleration will decrease; as it falls from the peak of its trajectory, its vertical acceleration will decrease.
h. As a projectile rises towards the peak of its trajectory, the vertical acceleration is directed upward; as it falls from the peak of its trajectory, its vertical acceleration is directed downward.
i. The peak height to which a projectile rises above the launch location isdependent upon the initial vertical velocity.
j. As a projectile rises towards the peak of its trajectory, the vertical velocity will decrease; as it falls from the peak of its trajectory, its vertical velocity willincrease.
k. Consider a projectile launched from ground level at a fixed launch speed and a variable angle and landing at ground level. The vertical displacement of the projectile during the first half of its trajectory (i.e., the peak height) will always increase as the angle of launch is increased from 0 degrees to 90 degrees.
l. Consider a projectile launched from ground level at a fixed launch angle and a variable launch speed and landing at ground level. The vertical displacement of the projectile during the first half of its trajectory (i.e., the peak height) will always increase as the launch speed is increased.

m/s. (Hint: simple Projectile Motion)
14. Aballisthrownhorizontallyfromtheroofofabuilding75mtallwithaspeedof4.6

a. Howmuch laterdoesthe ballhit the ground? b. How far from the building will it land?
c. Whatisthevelocityoftheballjustbefore ithitstheground

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horizontal.
15. A projectile is fired with an initial speed of 150 m/s at an angle of 47 above the

a. Determine the total time in the air.

b. Determinethemaximumheightreachedbytheprojectile.

c. Determinethemaximumhorizontaldistancecoveredbytheprojectile.

anangleof40.0degreesabovethehorizontal.Whatistheprojectile'slaunchspeed?
16. Thehighestbarrierthataprojectilecanclearis20m,whentheprojectileislaunchedat

degreesandtheotherat60degrees.Theprojectiletohitthegroundfirstwillbethe
17. Twoprojectilesarefiredatequalspeedsbutdifferentangles.Oneisfiredatangleof30

onefiredat(neglectairresistance) a)60 degrees

b) 30 degrees

c)bothhit me at the sametime

projectile.Onthediagram,drawvectorarrowsrepresentingthevxandvyvelocity
themagnitudeofthevelocitycomponents.Labeleachcomponent.(Notethatthe
18. Consider the trajectory diagram shown below for a non-horizontally launched componentsduringthecourseofthemotion.Thelengthofthearrowsshouldrepresent velocitycomponentsarealreadyshownforthefirstposition.)

faultybarricadeonabridge(quiteunfortunately),andlandedinapileofhay(quite
19. Anunfortunateaccidentoccurredonthetollway.Adriveraccidentallypassedthrough a

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fortunately).Measurementsattheaccidentscenerevealthatthedriverplungeda
fromthelocationwhereitleftthebridge.Ifthedriverwasina65mi/hr.speedzone,
thendeterminetheamountbywhichthedriverwasexceedingthespeedlimtatthe
down.(1.00m/s=2.24mi/hr.)(Hint-simpleprojectilemotion)
verticaldistanceof8.26meters.Thecarcarriedahorizontaldistanceof42.1meters timeoftheaccident.Assumethatthecontactwiththebarricadedidnotslowithecar

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Topic-4 Forces

Aforceisaninteractionbetweenobjectsthattendstoproduceacceleratonoftheobjec s.
Acceerationoccurswhenthereisanetforceonanobject;noaccelerationoccurswhenthene
force,butnoaccelerationoccurswhentheforcesarebalanced.Rememberthatacceleration
the velocity of anobject.
forcel(thesumofalltheforces)iszero.Inotherwords,accelerationoccursiwhenthereisatnett producesachangeinvelocity(magnitudeand/ordirection),soanunbalancedforcewillchange

Isaac Newton (1642-1727) studied forces and noticed three things in particular about them.

Newton's first law

Newton's first law states that an object at rest tends to remain at rest, and an object in motion tends to remain in motion with a constant velocity (constant speed and direction of motion), unless it is acted on by a nonzero net force.

Newton's second law

If there is a net force acting on an object, the object will have an acceleration and the object's velocity will change. How much acceleration will be produced by a given force? Newton's second law states that for a particular force, the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object. This can be expressed in the form of an equation:

In the MKS system of units, the unit of force is the Newton (N). In terms of kilograms, meters, and seconds, 1 N = 1 kg m / s2.

Newton's third law

A force is an interaction between objects, and forces exist in equal-and-opposite pairs. Newton's third law summarizes this as follows: when one object exerts a force on a second object, the

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second object exerts an equal-and-opposite force on the first object. Note that "equal-and-opposite" is the shortened form of "equal in magnitude but opposite in direction".

Although the forces between two objects are equal-and-opposite, the effect of the forces may or may not be similar on the two; it depends on their masses. Remember that the acceleration depends on both force and mass, and let's look at the force exerted by the Earth on a falling object. If we drop a 100 g (0.1 kg) ball, it experiences a downward acceleration of 9.8 m/s2, and a force of about 1 N, because it is attracted towards the Earth. The ball exerts an equal-and-opposite force on the Earth, so why doesn't the Earth accelerate upwards towards the ball? The answer is that it does, but because the mass of the Earth is so large (6.0 x 1024 kg) the acceleration of the Earth is much too small (about 1.67 x 10-25 m/s2) for us to notice.

In cases where objects of similar mass exert forces on each other, the fact that forces come in equal-and-opposite pairs is much easier to see.

Inclined planes and resolution of Forces

N (Normal Force is Normal to surface) so N= Wcos(a)

Wsin(a)= Force of friction when the object just starts sliding

Force of friction = μS N

Wsin(a) = μSWcos(a)

μS = Wsin(a)/Wcos(a)

Thus:

μS = tan(a)

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Example

A rightward force is applied to a 5-kg object to move it across a rough surface with a rightward acceleration of 2 m/s/s (or m/s2). The coefficient of friction between the object and the surface is 0.1. Use the diagram to determine the gravitational force, normal force, applied force,frictional force, and net force. (Neglect air resistance.)

Fnet =10 N, right; Fgrav = 49 N; Fnorm = 49 N; Ffrict = 4.9 N; Fapp = 14.9 N

Fnet can be found using Fnet = m • a = (5 kg) • (2 m/s/s) = 10 N, right.

Since the mass is known, Fgrav can be found: Fgrav = m • g = 5 kg • 9.8 m/s/s = 49 N

Since there is no vertical acceleration, the normal force equals the gravity force.

Once Fnorm is known, Ffrict can be found using Ffrict = "mu" • Fnorm = (0.1) • (49 N) = 4.9 N.

Since the Fnet = 10 N, right, the rightward force (Fapp) must be 10 N more than the leftward force (Ffrict); thus, Fapp must be 14.9 N.

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Problems

hat apply.
20. tWhichofthefollowingarealwaystrueofanobjectthatisatequilibrium?Includeall

a. All the forces acting upon the object are equal. b. The object is at rest.
c. The object is moving and moving with a constant velocity. d. The object has an acceleration of zero.
e. There is no change in the object's velocity. f. The sum of all the forces is 0 N.
g. All the forces acting upon an object are balanced.

constantspeed?Writeallpossibleanswers(ifany)intheblank:
21. Whichone(s)ofthefollowingforcediagramsdepictanobjectmovingtotherightwith

Explain your answer(s) to this question.

vel city of2m/s?	Explain or show your work.
22.oHowmuchnetforceisrequiredtokeepa5-kgobjectmovingrightwardwithaconstant

23. Usetheapproximationthatg=~10m/s2tofillintheblanksinthefollowingdiagram (Fgrav =weight)

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24. Usetheapproximationthatg=~10m/s2tofillintheblanksinthefollowingdiagrams.
force of friction= μN
Fnorm istheNormalforce(N)exertedbythesurface.FtensistheappliedForce.Ffrictisthe

25. A945-kgcartravelingrightwardat22.6m/sslamsonthebrakesandskidstoastop(w
distance required tostop.
lockedwheels).Ifthecoefficientoffrictionbetweentiresandroadis0.972,determinethe ith

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Performance Task

The task is to design an experiment to measure the static sliding coefficient of friction between two surfaces by using a ramp and measuring its inclination.

The idea is that if you put a solid object on a ramp and start to tilt the ramp upward, there is a point where the object will start to slide. That is the angle where the force of gravity is strong enough to overcome the static friction.

By simply knowing the angle or the inclination with a protractor, you can then calculate the static sliding coefficient of friction between the two materials. You can cover the ramp with various materials to determine different coefficients.

(Also see: Sliding Friction on an Inclined Surface)

Questions you will include:

•	What materials are needed? •	What are the steps to take?
•	How are the calculations made?

You need to answer the above questions and find coefficients of friction for three different surfaces. (You can either keep ramp surface same and change the solid object or keep change the ramp surface and keep the same object).

Have Fun!!

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