quiz samples for chapter 5 general physics i force and motion -...

General Physics IQuiz Samples for Chapter 5

Force and Motion - IMarch 30, 2020

Name: Department: Student ID #:

Notice

� +2 (−1) points per correct (incorrect) answer.

� No penalty for an unanswered question.

� Fill the blank ( ) with � (8) if the statement iscorrect (incorrect).

� Textbook: Walker, Halliday, Resnick, Principlesof Physics, Tenth Edition, John Wiley & Sons(2014).

5-1 Newton’s First and Second Laws

Newtonian Mechanics

1. (�) The relation between a force and theacceleration caused by that force was firstunderstood by Isaac Newton (1642-1727) and thecorresponding theory is called Newtonianmechanics.

2. (�) Newtonian mechanics breaks down in the limitwhen a particle’s motion is very fast comparable tothe speed of light. In this regime, we must takeinto account a more fundamental theory called thespecial theory of relativity whose non-relativisticlimit reproduces Newtonian mechanics.

3. (�) Newtonian mechanics also breaks down in anatomic scale or smaller scales, where we mustrequire quantum mechanics to provide a sensibledescription.

Newton’s First Law

1. (�) An inertial reference frame is a referenceframe in which the law of inertia is valid. Thelaw of inertia is the Newton’s first law of motion:

� If there is no net force applied to a particle,then the particle’s velocity is a constant ofmotion.

Fnet = 0 → v = constant.

� The net force is the vector sum of all of theforces.

Force

1. (�) The dimensions of force are given by

[F ] = [M ][L][T ]−2.

2. (�) The SI unit of force is

1 N = kg ·m/s2.

1 Newton is the force that would accelerate aone-kilogram object by 1 m/s2.

3. (�) The CGS unit of force is

1 dyne = g · cm/s2.

Thus,

1 N = 105 dyne.

1 N is the force that would accelerate aone-kilogram object by 1 m/s2.

4. (�) Force is a vector. If two or more forcesapplies to a body, then the net force on that bodyis the vector sum of the forces. The vector sum isequivalent to all of the applied forces. This iscalled the principle of superposition for forces.

5. (�) A contact force is any force that occurs onlythrough contact such as punching, kicking,pressing, pulling, and scratching. A normal force,that sustains an object on a surface by cancellingthe pressing force onto the surface, is also a contactforce. In a tug of war, two teams pull at oppositeends of a rope. This involves a contact force socalled tension. Most of the contact forces that weexperience are not fundamental but reducible intomore basic interactions. Most of the familiarcontact forces can be decomposed into an intricatecombination of the electromagnetic interactions.

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6. (�) In nature, there are fundamental forces.These are not to be reducible to more basic forces.Gravity, electromagnetic force, and subatomicforces. Because the gravity and electromagneticforces reach to long ranges, we were able toexperience these forces for a very long time. Thesubatomic interactions reveal its effect only atsubatomic distances. They are the weakinteraction and the strong interaction. The specialtheory of relativity has unified the electric forceand the magnetic force. The standard model ofparticle physics was successful to explain theelectromagnetic, weak, and strong interaction in aunified way. For more details, refer to Wikipedia.In this course, we mainly focus on the contact forceand elementary treatments of specific long-rangeforces, gravity and electromagnetism.

Newton’s Second Law

1. (�) The following strategy is a systematic way ofsolving problems involving Newton’s second law.

� Identify the object to consider the motion.

� List all of the acting forces to that object.

� Introduce a convenient frame of axes.

� Read off the components.

2. (�) A free-body diagram is a convenient tool.

� Draw the body only and draw a dot.

� Draw each force vector placing its tail at thedot.

� Place the origin of the frame to that dot.

� Read off the components and carry out theanalysis.

3. Consider two forces applying to a particle whosenet force is vanishing.

(a) (�) The sum of two forces F1 and F2 is thezero vector:

Fnet = F1 + F2 = 0.

(b) (�) The two forces are antiparallel (opposite)and are of equal magnitude.

(c) (�) One dimension is enough to analyze theproblem.

4. Consider three forces applying to a particle whosenet force is vanishing.

(a) (�) The sum of three forces F1, F2, and F3 isthe zero vector:

Fnet = F1 + F2 + F3 = 0.

(b) (�) Applying the associative law of addition,we can find that the sum of two vectorscancels the remaining vector.

(c) (�) All of the three vectors must be coplanar.Thus the analysis can be carried out in twodimensions.

(d) (�) The three vectors can be translated tomake a triangle. After that, every vector’shead meets another vector’s tail.


https://en.wikipedia.org/wiki/Fundamental_interaction



5. Consider the particle A that makes a projectilemotion.

(a) (�)

In an inertial reference frame S, the particle’sposition at time t is expressed as

rA = x0 + v0t+1

2gt2,

where x0 and v0 are the initial position andvelocity of the particle.

(b) (�)

The position of the origin O of S relative tothe particle is

rO/A = −x0 − v0t−1

2gt2.

(c) (�) The velocity of the origin O of S relativeto the particle is

vO/A = −v0 − gt.

(d) (�) The acceleration of the origin O of Srelative to the particle is

aO/A = −g.

(e) (�) According to the result, vO/A = −v0 − gt,the velocity of the origin of an inertialreference frame relative to A is not constant.Thus one might guess that the origin is underthe non-vanishing net force. However, thisacceleration is not originated from a realforce. This acceleration is solely due to thechoice of a wrong frame of reference. Theframe attached to A that is accelerating is notan inertial reference frame. Therefore, theacceleration is not due to a real force but dueto a fictitious force. A fictitious is observedin an accelerating frame (non-inertialframe).

6. (�) In Newtonian mechanics, the origin of theacceleration is the force. The Newton’s secondlaw of motion states that the acceleration of aparticle is related to the mass and force as

F = ma,

where F is the force and m is the mass of theparticle.

7. (�) For a given net force, F , the accelerationdepends on the mass of the particle:

a =F

m.

5-2 Some Particular Forces

1. (�) The gravitational force Fg on an object onthe ground is dominated by the gravitationalattractive force between the object and the Earth.The value is the mass times the gravitationalacceleration:

Fg = mg.

2. The weight W of an object is the magnitude ofthe net force that prevents the object from fallingfreely, as measured by someone on the ground.




(a) (�) In an inertial reference frame,

W = Fg.

It is because

Fnet = ma = Fg −W,

and a = 0 if the weight exactly cancels thegravitational force.

(b) (�) If the object is at rest hang on a rope,then the weight is equivalent to the tension ofthe rope.

(c) (�) If the object is at rest on the floor, thenthe weight is equivalent to the normal force ofthe floor.

3. (�)

The normal force FN (N) acts along the normal(perpendicular) direction from a surface. Thenormal force is against another force that ispressing down the surface. The normal force is notgreater than the applied pressing force and themagnitude is only to cancel the perpendicularcomponent of the pressing force.

4. Consider an object at rest on the floor of anelevator.

(a) (�) In general, the frame of reference S? fixedto the floor of an elevator is a non-inertialreference frame. We describe the motion in aninertial reference frame S fixed to the ground.The net force applied to the object is the sumof the gravitational force Fg and the weight−W (normal force of the elevator floor):

Fnet = ma = Fg −W,

where a is the acceleration of the elevator.Note that we have chosen the positivedirection parallel to the gravitational force.

(b) (�) The weight W is then expressed as

W = Fg −ma.

(c) (�) If the acceleration is the same direction asthe gravitational acceleration, then Wdecreases.

(d) (�) If the acceleration is opposite to thegravitational acceleration, then W increases.

(e) (�) If the elevator falls freely, then a = g.Then the object becomes weightless.

5. Consider an object hanging on a rope that ishanging from a ceiling.




(a) (�) The weight W is the tension T of therope.

W = Fg −ma,where a is the acceleration of the elevatorrelative to the ground. a > 0 if theacceleration is parallel to the gravitationalacceleration.

(b) (�) If the acceleration is the same direction asthe gravitational acceleration, then T = Wdecreases.

(c) (�) If the acceleration is opposite to thegravitational acceleration, then T = Wincreases.

(d) (�) If a� g, then the object and the rope areattached to the ceiling.

6. Consider the frictional force.

(a) (�) When an object standing on a surface isresisted by the bonding between the objectand the surface, the object experiences thestatic frictional force.

(b) (�) When an object slides over a surface, theobject experiences the kinetic frictionalforce.

(c) (�) The direction of a frictional force isparallel to the surface and opposite to theintended motion.

(d) (�) The static frictional force on an object isopposite to the force that attempts theacceleration of that object. The magnitude ofthe static frictional force cannot be greaterthan that intended force.

(e) (�) The kinetic frictional force is opposite tothe velocity of the object under the friction.

7. Consider the tension.

(a) (�) Tension is a pulling force transmittedalong a line made of a one-dimensionalcontinuous object like string, cord, cable,chain, or rope. This force is irrelevant to anycompression.

(b) (�) If A and B are interacting through astring, then the magnitude of the tension Tequally applies to A and B. However, thedirection of the tension for B is opposite tothat for A.

(c) (�) If the chord that exerts tension ismassless, then the tension is uniform. If themass of the chord is not negligible, then thetension varies depending on the position. Themassless approximation is valid in the limitthat the mass that experiences the tension ismuch greater than the mass of the chord.

(d) (�) The assumption of the unstretchablechord is usually necessary if one can neglectthe restoring (spring) force. In addition, thisassumption provides a constraint completelydetermines the displacement of one object atone end in terms of the displacement of theobject at the other end.

5-3 Applying Newton’s Laws

Newton’s Third Law

1. Newton’s Third Law : The forces between twointeracting bodies are the same in magnitude butopposite in direction.

F2→1 = −F1→2,

where Fi→j is the force applied to object j byobject i.




(a) (�) There is no self interaction:

F1→2 + F2→1 = 0.

Substituting 2 = 1, we find that

F1→1 + F1→1 = 0,

which leads to

F1→1 = 0.

(b) (�) We shall find that the conservation oflinear momentum is guaranteed by Newton’sthird law.

2. Three connected blocks with masses m1, m2, andm3 are pulled to the right on a horizontalfrictionless table by a force of magnitude T3 .

(a) (�) The magnitude of the system’sacceleration. The net force on the wholesystem is T3 and the total mass ism1 +m2 +m3. Therefore,

T3 = (m1 +m2 +m3)a.

The acceleration a is

a =T3

m1 +m2 +m3.

(b) (�) The system m1 +m2 are pulled by thenet force T2 and they have a commonacceleration a. Therefore,

T2 = (m1 +m2)a =m1 +m2

m1 +m2 +m3T3.

(c) (�) The system m1 is pulled by the net forceT1 and its acceleration is a. Therefore,

T1 = m1a =m1

m1 +m2T2 =

m1

m1 +m2 +m3T3.

3. A block of mass M is pushed by a horizontal forceFh with an acceleration a (up to the hill) along thefrictionless ramp that makes an angle of θ with thehorizontal plane. We introduce the coordinate axessuch that x axis is along the ramp down the slope.y axis is normal to the ramp.

(a) (�) The gravitational force Fg is

Fg = Mg(i sin θ − j cos θ)

(b) (�) The horizontal force Fh is

Fh = Fh(−i cos θ − j sin θ)

(c) (�) The normal force FN is

FN = jFN .

(d) (�) The net external force is

Fnet = (Mg sin θ − Fh cos θ)i

+(−Mg cos θ − Fh sin θ + FN )j.

(e) (�) Because the particle is accelerating with−ai, the net force must be the same as

Fnet = −Mai.

(f) (�) The magnitude of the horizontal force is

Fh =M(g sin θ + a)

cos θ.


quiz samples for chapter 5 general physics i force and motion -...

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