Kinetic Energy and WorkKinetic Energy and Work

Chapter 7Chapter 7

Introduction to EnergyIntroduction to Energy

The concept of energy is one of the The concept of energy is one of the most important topics in sciencemost important topics in science

Every physical process that occurs in Every physical process that occurs in the Universe involves energy and the Universe involves energy and energy transfers or transformationsenergy transfers or transformations

Work and EnergyWork and Energy

Energy:Energy: scalar quantity associated with a state (or scalar quantity associated with a state (or condition) of one or more objects. condition) of one or more objects.

Work and energy are scalars, measured in NWork and energy are scalars, measured in N··m or m or

Joules, J, 1J=kgJoules, J, 1J=kg∙m∙m22/s/s22

Energy can exist in many forms - mechanical, Energy can exist in many forms - mechanical,

electrical, nuclear, thermal, chemical….electrical, nuclear, thermal, chemical….

Energy Approach to ProblemsEnergy Approach to Problems

The energy approach to describing The energy approach to describing motion is particularly useful when the motion is particularly useful when the force is not constantforce is not constant

An approach will involve An approach will involve Conservation Conservation of Energyof Energy• This could be extended to biological This could be extended to biological

organisms, technological systems and organisms, technological systems and engineering situationsengineering situations

Work and EnergyWork and Energy

Energy is a conserved quantity - the total Energy is a conserved quantity - the total amount of energy in the universe is amount of energy in the universe is constant.constant.

Energy can be converted from one type to Energy can be converted from one type to

another but never destroyed.another but never destroyed.

Work and energy concepts can simplify solutions of Work and energy concepts can simplify solutions of

mechanical problems - they can be used in an mechanical problems - they can be used in an

alternative analysisalternative analysis

SystemsSystems

A A systemsystem is a small portion of the is a small portion of the UniverseUniverse• We will ignore the details of the rest of the We will ignore the details of the rest of the

UniverseUniverse

A critical skill is to identify the systemA critical skill is to identify the system

Valid SystemValid System

A valid system mayA valid system may• be a single object or particlebe a single object or particle• be a collection of objects or particlesbe a collection of objects or particles• be a region of spacebe a region of space• vary in size and shapevary in size and shape

EnvironmentEnvironment

There is a There is a system boundarysystem boundary around the around the systemsystem• The boundary is an imaginary surfaceThe boundary is an imaginary surface• It does not necessarily correspond to a It does not necessarily correspond to a

physical boundaryphysical boundary

The boundary divides the system from The boundary divides the system from the the environmentenvironment• The environment is the rest of the UniverseThe environment is the rest of the Universe

WorkWork

The work,The work, WW, done on a system by an , done on a system by an agent exerting a constant force on the agent exerting a constant force on the system is the product of the magnitude,system is the product of the magnitude, FF, , of the force, the magnitudeof the force, the magnitude ΔΔrr of the of the displacement of the point of application of displacement of the point of application of the force, andthe force, and cos cos θθ,, where where θθ is the angle is the angle between the force and the displacement between the force and the displacement vectorsvectors

WorkWork

WW = = FF ΔΔrr cos cos θθ• The displacement is that of the point of The displacement is that of the point of

application of the forceapplication of the force• A force does no work on the object if the A force does no work on the object if the

force does not move through a displacementforce does not move through a displacement• The work done by a force on a moving object The work done by a force on a moving object

is zero when the force applied is is zero when the force applied is perpendicular to the displacement of its point perpendicular to the displacement of its point of applicationof application

Work ExampleWork Example

The normal force,The normal force, nn, , and the gravitational and the gravitational force,force, mgmg, do no work , do no work on the objecton the object

cos cos θθ = cos 90° = 0= cos 90° = 0 The forceThe force FF does do does do

work on the objectwork on the object

More About WorkMore About Work The system and the environment must be The system and the environment must be

determined when dealing with workdetermined when dealing with work The environment does work on the systemThe environment does work on the system

(Work (Work byby the environment the environment onon the system) the system) The sign of the work depends on the direction ofThe sign of the work depends on the direction of

FF relative torelative to ΔΔrr Work is positive when projection ofWork is positive when projection of FF ontoonto ΔΔrr

is in the same direction as the displacementis in the same direction as the displacement Work is negative when the projection is in the Work is negative when the projection is in the

opposite directionopposite direction

Work Is An Energy TransferWork Is An Energy Transfer

• This is important for a system approach This is important for a system approach to solving a problemto solving a problem

• If the work is done on a system and it is If the work is done on a system and it is positive, energy is transferred to the positive, energy is transferred to the systemsystem

• If the work done on the system is If the work done on the system is negative, energy is transferred from the negative, energy is transferred from the systemsystem

Work done by a constant forceWork done by a constant force

Work done on an object by a constant force is Work done on an object by a constant force is defined to be the product of the magnitude of defined to be the product of the magnitude of the the displacementdisplacement and the component of the and the component of the forceforce parallel to the displacement parallel to the displacement

WhereWhere FFIIII is the component of the forceis the component of the force FF

parallel to the displacementparallel to the displacement dd

W = FW = FII II ·· d d

WorkWorkIn other words - In other words -

WhereWhere θθ is the angle betweenis the angle between FF and and dd

IfIf θθ is > 90is > 90oo, work is negative. A decelerating car has , work is negative. A decelerating car has negative work done on it by its engine.negative work done on it by its engine.

The unit of work is called The unit of work is called Joule Joule (J)(J), , 1 J = 1 N1 J = 1 N··mm

1J=kg1J=kg∙m∙m22/s/s22

d

F

W = F d cosW = F d cos

Scalar Product of Two VectorsScalar Product of Two Vectors

The scalar product The scalar product of two vectors is of two vectors is written aswritten as A A . . BB

It is also called It is also called the dot productthe dot product

A A . . B = B = A BA B cos cos θθ

θθ is the angle is the angle betweenbetween AA andand BB

Scalar ProductScalar Product

The scalar product is commutativeThe scalar product is commutativeA A . . B = B B = B . . AA

The scalar product obeys the distributive law The scalar product obeys the distributive law of multiplicationof multiplication

A A . . (B + C) (B + C) == A A . . B + A B + A . . CC

Dot Products of Unit VectorsDot Products of Unit Vectors

Using component form withUsing component form with A A andand BB::

0kjkiji

1kkjjii

zzyyxx

zyx

zyx

BABABABA

kBjBiBB

kAjAiAA

Work - on and byWork - on and by

A person pushes blockA person pushes block 30 m30 m along the along the ground by exerting force ofground by exerting force of 25 N25 N on the on the trolley. How much work does the person do trolley. How much work does the person do on the trolley?on the trolley?

W = F d = 25N x 30m = 750 NmW = F d = 25N x 30m = 750 Nm

Trolley does -Trolley does -750 Nm750 Nm work on the personwork on the person

FFpptFFtptp

d

A force acts on an object as A force acts on an object as the object moves in the the object moves in the xx direction from the origin direction from the origin to to xx = 5.00 m = 5.00 m. Find the work . Find the work W = W = FFddrr done on done on the object by the force.the object by the force.

F 4x i 3y j N

Mechanical EnergyMechanical Energy

Mechanical energy (energy associated Mechanical energy (energy associated

with with masses) can be thought of as having two masses) can be thought of as having two

components: components: kinetickinetic and and potentialpotential– Kinetic energy is energy of motionKinetic energy is energy of motion

– Potential energy is energy of positionPotential energy is energy of position

Kinetic EnergyKinetic Energy

In many situations, energy can be In many situations, energy can be considered asconsidered as “the ability to do work”“the ability to do work”

Energy can exist in different formsEnergy can exist in different forms

Kinetic energy is the energy associated Kinetic energy is the energy associated with the with the motionmotion of an object of an object

A moving object can do work on another A moving object can do work on another objectobject– E.g hammer does work on nail.E.g hammer does work on nail.

Kinetic EnergyKinetic Energy Consider an object with massConsider an object with mass mm moving in a moving in a straight line with initial velocitystraight line with initial velocity vvii. . To accelerate it To accelerate it

uniformly to a speeduniformly to a speed vvff a constant net forcea constant net force FF is is

exerted on it parallel to motion over a distanceexerted on it parallel to motion over a distance dd..

Work done on objectWork done on object

W = F d = m a dW = F d = m a d (NII)

SoSo

da

ada

if

iifif

2

2)(222

222

vv

vxxvv

dd

mW if

2

22 vv

Kinetic EnergyKinetic Energy

If we rearrange this we obtain

We define the quantity ½mv2 to be the translational kinetic energy (KE) of the object

This is the ‘Work-Energy Theorem’:

“The net work done on an object is equal to its change in kinetic energy”

22

2

1

2

1if mmW vv

W = KE

The Work-Energy TheoremThe Work-Energy Theorem

• NoteNote– The net work done on an object is the work The net work done on an object is the work

done by the net force.done by the net force.– Units of energy and work must be the Units of energy and work must be the

same (J)same (J)

W = W = KEKE

I.I. Kinetic energyKinetic energy

Energy associated with the state of motion of an object.Energy associated with the state of motion of an object.

2

2

1mvK

Units:Units: 1 Joule = 1J = 1 kgm1 Joule = 1J = 1 kgm22/s/s22

II. WorkII. Work

Energy transferred “to” or “from” an Energy transferred “to” or “from” an object by means of a force acting on the object by means of a force acting on the object. object.

To To +W +WFrom From -W -W

- Constant force- Constant force: xx maF

dFWdFKKvvm

dvvmmaF

d

vvadavv

xxiff

fxxf

xxf

)(2

1

1)(

2

1

22

20

2

20

220

220

2

Work done by the force = Energy transfer due to the force.

-To calculate the work done on an object by a force during To calculate the work done on an object by a force during a displacement, we use only the force component alonga displacement, we use only the force component along the object’s displacement. The force component perpendicularthe object’s displacement. The force component perpendicular to the displacement does zero work.to the displacement does zero work.

dFFddFW x

cos

Assumptions:Assumptions: 1) F = constant force 1) F = constant force 2) Object is particle-like (rigid object, all 2) Object is particle-like (rigid object, all parts of the object must move together).parts of the object must move together).

090

90180

90

W

W

A force doesA force does +W+W when it has a vector component in the when it has a vector component in the same direction as the displacement, andsame direction as the displacement, and –W–W when it has when it has a vector component in the opposite direction.a vector component in the opposite direction. W=0W=0 when it when it has no such vector component.has no such vector component.

Net work done by several forces = Sum of works done by Net work done by several forces = Sum of works done by individual forces.individual forces. Calculation:Calculation: 1) W1) Wnetnet= W= W11+W+W22+W+W33+…+…

2) F2) Fnet net W Wnetnet=F=Fnetnet d d

II. Work-Kinetic Energy TheoremII. Work-Kinetic Energy Theorem

WKKK if

Change in the kinetic energy of the particle = Net work done on the particle

III. Work done by a constant forceIII. Work done by a constant force

- Gravitational force:- Gravitational force:

cosmgdW

Rising objectRising object:: W= mgd cos180W= mgd cos180º º = -mgd= -mgd FFgg transfers transfers mgdmgd energyenergy from from the object’s kinetic energy. the object’s kinetic energy.

Falling down object:Falling down object: W= mgd cos 0W= mgd cos 0º = +mgdº = +mgd Fg transfers transfers mgd energyenergy to to the object’s kinetic energy the object’s kinetic energy.

External applied force + Gravitational force: External applied force + Gravitational force:

gaif WWKKK

Object stationary before and Object stationary before and after the lift:after the lift: Wa+Wg=0

The applied force transfers the same amount of The applied force transfers the same amount of energy to the object as the gravitational force energy to the object as the gravitational force transfers from the object.transfers from the object.

IV. Work done by a variable forceIV. Work done by a variable force

- Spring force:- Spring force:

xkF

Hooke’s lawHooke’s law

k = spring constant = spring constant measures spring’s stiffness. measures spring’s stiffness. Units: N/m

Hooke’s LawHooke’s Law

When When xx is positive is positive (spring is stretched),(spring is stretched), FF is is negativenegative

WhenWhen xx is is 00 (at the (at the equilibrium position),equilibrium position), FF is is 00

When When xx is negative is negative (spring is compressed),(spring is compressed), FF is positiveis positive

Hooke’s LawHooke’s Law

The force exerted by the spring is The force exerted by the spring is always directed opposite to the always directed opposite to the displacement from equilibriumdisplacement from equilibrium

FF is called the is called the restoring forcerestoring force If the block is released it will oscillate If the block is released it will oscillate

back and forth between back and forth between ––xx and and xx

kxFD x 1

Work done by a spring force:Work done by a spring force:

Hooke’s law:Hooke’s law:

Assumptions:Assumptions:

• Spring is massless Spring is massless mspring << mblock

• Ideal spring Ideal spring obeys Hooke’s law exactly. obeys Hooke’s law exactly.• Contact between the block and floor is frictionless.Contact between the block and floor is frictionless.• Block is particle-like.Block is particle-like.

xxi Δx

1)1) The block displacement must be divided The block displacement must be divided into many segments of infinitesimal width,into many segments of infinitesimal width, Δx.

2)2) F(x) ≈ cte within each shortwithin each short Δx segment.segment.

- Calculation:- Calculation:

xx

FFxx

xxii xxffΔΔxx

FFjj

)(2

1

2

1)(

0

222if

x

x

x

x

x

x

x

x

js

xxkxkdxxkdxkxdxF

xxFW

f

i

f

i

f

i

f

i

22

2

1

2

1fis xkxkW

WWss>0 >0 Block ends up closer to the Block ends up closer to the

relaxed position (x=0) than it was initially.relaxed position (x=0) than it was initially.WWss<0 <0 Block ends up further away from Block ends up further away from

x=0x=0.WWss=0 =0 Block ends up at Block ends up at x=0x=0.

02

1 2 ifs xifxkW

Work done by an applied force + spring Work done by an applied force + spring force:force:

saif WWKKK

Block stationary before and after the displacementBlock stationary before and after the displacement:: WWaa= -W= -Wss The work done by the The work done by the applied forceapplied force displacing displacing

the block is the negative of the work done by the the block is the negative of the work done by the spring forcespring force..

If it takesIf it takes 4.00 J4.00 J of work to stretch a Hooke's-of work to stretch a Hooke's-law springlaw spring 10.0 cm10.0 cm from its unstressed from its unstressed length, determine the extra work required to length, determine the extra work required to stretch it an additionalstretch it an additional 10.0 cm10.0 cm..

AA 2.00-kg2.00-kg block is attached to a spring of force block is attached to a spring of force constantconstant 500 N/m.500 N/m. The block is pulledThe block is pulled 5.00 cm5.00 cm to the to the right of equilibrium and released from rest. Find the right of equilibrium and released from rest. Find the speed of the block as it passes through equilibrium if speed of the block as it passes through equilibrium if (a) the horizontal surface is frictionless and (b) the (a) the horizontal surface is frictionless and (b) the coefficient of friction between block and surface iscoefficient of friction between block and surface is 0.3500.350..

Work done by a general variable force:Work done by a general variable force:

• Assume that during a Assume that during a very small displacement,very small displacement, ΔΔxx,, FF is constantis constant

• For that displacement,For that displacement, WW ~ ~ FF ΔΔxx

• For all of the intervals,For all of the intervals,

f

i

x

xx

W F x

Work done by a general variable force:Work done by a general variable force:

•

• Therefore,Therefore,

• The work done is The work done is equal to the area equal to the area under the curveunder the curve

lim0

ff

ii

xx

x x xxx

F x F dx

f

i

x

xxW F dx

3D-Analysis3D-Analysis

f

i

f

i

f

i

f

i

z

z

z

y

y

y

x

x

x

r

r

zyx

zyxzyx

dzFdyFdxFdWWdzFdyFdxFrdFdW

kdzjdyidxrd

zFFyFFxFFkFjFiFF

ˆˆˆ

)(),(),(;ˆˆˆ

Work-Kinetic Energy Theorem - Variable forceWork-Kinetic Energy Theorem - Variable force

KKKmvmvdvvmdvmvW

vdx

dv

dt

dx

dx

dv

dt

dv

mvdvdxvdx

dvmdx

dt

dvmdxma

dxmadxxFW

ifif

v

v

v

v

x

x

x

x

f

i

f

i

f

i

f

i

22

2

1

2

1

)(

V. PowerV. Power Time rate at which the applied force does work.Time rate at which the applied force does work.

- Average power:- Average power: amount of work doneamount of work donein an amount of time in an amount of time ΔΔtt by a force. by a force.

-Instantaneous power:Instantaneous power: -instantaneous time rate of doing work.instantaneous time rate of doing work.

t

WPavg

dt

dWP

Units:Units: 1 watt= 1 W = 1J/s1 watt= 1 W = 1J/s 1 kilowatt-hour = 1000W1 kilowatt-hour = 1000W··h = 1000J/s x 3600s = 3.6 x 10h = 1000J/s x 3600s = 3.6 x 1066 J J = 3.6 MJ= 3.6 MJ

vFFvdt

dxF

dt

dxF

dt

dWP

coscos

cos

xx

FF

φφ

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor. Find an expression for the speedacross a frictionless floor. Find an expression for the speed vvff at at

the end of that distance if the block’s initial velocity is: (a)the end of that distance if the block’s initial velocity is: (a) 00 and and (b)(b) 1m/s1m/s to the right.to the right.

N

mg

Fx

Fy

N

mg

Fx

Fy

(a)

In the figure below aIn the figure below a 2N 2N force is applied to aforce is applied to a 4kg4kg block at a block at a downward angledownward angle θθ as the block moves right-ward throughas the block moves right-ward through 1m1m across a frictionless floor. Find an expression for the speedacross a frictionless floor. Find an expression for the speed vfvf at the end of that distance if the block’s initial velocity is: (a)at the end of that distance if the block’s initial velocity is: (a) 00 and (b)and (b) 1m/s1m/s to the right.to the right.

N

mg

Fx

Fy

smv

vkgmN

mvKva

f

f

f

/cos

)4(5.0cos)1)(2(

5.00)(2

20

smv

JvkgN

JmvKsmvb

f

f

f

/cos1

2)4(5.0cos)2(

25.0/1)(

2

20

N

mg

Fx

Fy

dFdFW )cos(

)(5.0 20

2 vvmKW f

(a)

(c) The situation in fig.(b) is similar in that the block is initially (c) The situation in fig.(b) is similar in that the block is initially moving atmoving at 1m/s1m/s to the right, but now theto the right, but now the 2N2N force is directed force is directed downward to the left. Find an expression for the speed of the downward to the left. Find an expression for the speed of the block at the end of the 1m distance.block at the end of the 1m distance.

N

mg

Fx

Fy

N

mg

Fx

Fy

A small particle of mass A small particle of mass mm is pulled to the top of a is pulled to the top of a frictionless half-cylinder (of radius frictionless half-cylinder (of radius RR) by a cord that passes ) by a cord that passes over the top of the cylinder, as illustrated in Figure. (a) If the over the top of the cylinder, as illustrated in Figure. (a) If the particle moves at a constant speed, show that particle moves at a constant speed, show that FF = = mgmgcoscos. . ((NoteNote: If the particle moves at constant speed, the : If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must component of its acceleration tangent to the cylinder must be zero at all times.) (b) By directly integrating be zero at all times.) (b) By directly integrating W = W = FFddrr, , find the work done in moving the particle at constant speed find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder. from the bottom to the top of the half-cylinder.

Two springs with negligible masses, one with spring Two springs with negligible masses, one with spring constant constant kk11 and the other with spring constant and the other with spring constant kk22, are , are attached to the endstops of a level air track as in Figure. attached to the endstops of a level air track as in Figure. A glider attached to both springs is located between them. A glider attached to both springs is located between them. When the glider is in equilibrium, spring When the glider is in equilibrium, spring 11 is stretched by is stretched by extension extension xxii11 to the right of its unstretched length and spring to the right of its unstretched length and spring 22 is stretched by is stretched by xxii22 to the left. Now a horizontal force to the left. Now a horizontal force FFappapp is is applied to the glider to move it a distance applied to the glider to move it a distance xxaa to the right from to the right from its equilibrium position. Show that in this process (a) the its equilibrium position. Show that in this process (a) the work done on spring work done on spring 11 is is kk11((xxaa

22+2+2xxaaxxii11)) , (b) the work done , (b) the work done on spring on spring 2 2 is is kk22((xxaa

22 – 2 – 2xxaaxxii22)) , (c) , (c) xxii22 is related to is related to xxii11 by by xxii2 2 = = kk11xxii11//kk22, and (d) the total work done by the force , and (d) the total work done by the force FFappapp is ( is (kk11 + + kk22))xxaa

22..

N6. A 2kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x axis along the surface. Beginning at t=0, a steady wind pushes on the lunchbox in the negative direction of x so it’s position is changed as x=t–0.1t2. Estimate the kinetic energy of the lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force from the wind do on the lunch box from t=1s to t=5s?

N12. In the figure below a horizontal forceIn the figure below a horizontal force FFaa of magnitudeof magnitude

20N 20N is applied to ais applied to a 3kg3kg book, as the book slides a distance ofbook, as the book slides a distance of d=0.5m d=0.5m up a frictionless ramp. (a) During the displacement, up a frictionless ramp. (a) During the displacement, what is the net force done on the book bywhat is the net force done on the book by FFaa, the gravitational , the gravitational

force on the book and the normal force on the book? (b) If the force on the book and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, book has zero kinetic energy at the start of the displacement, what is the speed at the end of the displacement?what is the speed at the end of the displacement?

x

y

mg

N

Fgy

Fgx

N15. (a) Estimate the work done represented by the graph below (a) Estimate the work done represented by the graph below in displacing the particle from in displacing the particle from x=1x=1 to to x=3mx=3m. (b) The curve is given. (b) The curve is givenby by F=a/xF=a/x22, with , with a=9Nma=9Nm22. Calculate the work using integration. Calculate the work using integration

N19.N19. An elevator has a mass of An elevator has a mass of 4500kg 4500kg and can carry a maximum and can carry a maximum load of load of 1800kg1800kg. If the cab is moving upward at full load at constant . If the cab is moving upward at full load at constant speed speed 3.8m/s3.8m/s, what power is required of the force moving the cab , what power is required of the force moving the cab to maintain that speed?to maintain that speed?

mgmg

FFaa

N17. A single force acts on a body that moves along an x-axis. A single force acts on a body that moves along an x-axis. The figure below shows the velocity component versus time for The figure below shows the velocity component versus time for the body. For each of the intervals the body. For each of the intervals AB, BC, CD, and DE, give the signAB, BC, CD, and DE, give the sign (plus or minus) of the work done by (plus or minus) of the work done by the force, or state that the work is the force, or state that the work is zero.zero.

vv

ttAA

BB CC

DD

EE

15E. In the In the figure below, a cord runs around two massless, figure below, a cord runs around two massless, frictionless pulleys; a canister with mass frictionless pulleys; a canister with mass m=20kgm=20kg hangs from one hangs from one pulley; and you exert a force pulley; and you exert a force FF on the free end of the cord. on the free end of the cord. (a) What must be the magnitude of (a) What must be the magnitude of FF if if you are to lift the canister at a constant you are to lift the canister at a constant speed? (b) To lift the canister by speed? (b) To lift the canister by 2cm2cm, , how far must you pull the free end of how far must you pull the free end of the cord? During that lift, what is the the cord? During that lift, what is the work done on the canister by (c) your work done on the canister by (c) your force (via the cord) and (d) the force (via the cord) and (d) the gravitational force on the canister?gravitational force on the canister?mg

TT T

P1

P2

Challenging problems – Chapter 7

Two trolleys of masses mTwo trolleys of masses m11=400 kg and m=400 kg and m22=200 kg are connected by a =200 kg are connected by a

rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push them with a force of 1200N. to push them with a force of 1200N. From the point of view of kinetics, From the point of view of kinetics, does the relative position of the does the relative position of the trolleys matter? If the rod can trolleys matter? If the rod can only stand an applied force of 500N, only stand an applied force of 500N, which trolley should be up front? which trolley should be up front?

Situation 1

m1g m2g

N1 N2

F

F1r F2r

2.2. A car with a weight of A car with a weight of 2500N2500N working with a power of working with a power of 130kW 130kW develops a velocity of develops a velocity of 112 km/hour112 km/hour when traveling along a when traveling along a horizontal straight highway. Assuming that the frictional forces horizontal straight highway. Assuming that the frictional forces (from the ground and air) acting on the car are constant but not (from the ground and air) acting on the car are constant but not negligible: What is the value of the frictional forces? negligible: What is the value of the frictional forces? (a) What is the car’s maximum velocity on a (a) What is the car’s maximum velocity on a 5500 incline hill? (b) incline hill? (b) What is the power if the car is traveling on a What is the power if the car is traveling on a 101000 inclined hill at inclined hill at 36km/h36km/h??