potential energy and conservation of energyq3) a boy throws up a 0.15kg stone from the top of a 20m...

1

Potential Energy and Conservation of Energy Chap. 7 & 8

Level : AP Physics

Potential Energy of a System see p.191 in the textbook

- Potential energy is the energy associated with the arrangement of a system of objects that exert forces on

each other

- Potential energy can be thought of as energy of position

- Potential energy can be thought as stored energy that can either do work or be converted to kinetic

energy

Gravitational Potential Energy & Conservation of Mechanical Energy

- Gravitational Potential Energy(Ug) is defined as

Ug = mgy

- Conservation of Mechanical Energy is described as

The total mechanical energy(E) of a system remains constant in any isolated system if the force

acting on the object is a conservative force *~conservative force is explained in details on p.9 in the worksheet

Etot = Ki + Ui = Kf + Uf or ΔK + ΔU =0 where ΔEtot = 0

- An object is falling towards the Earth where the initial height at some instantaneous point is yi and the

final height is yf. (air resistance is ignored)

y=yi-yf

yi

yf

- The gravitational force Fg= mg does work on the object.

=> Ki + ΣW = Kf. (Work.Kinetic Energy Theorem )

=> Ki +Wg = Kf. (The only force doing work on the object is force of gravity)

- If y= yf - yi, and since the force of gravity(Fg) is doing work on the object over

a distance of ‘y’,

Wg = mgyi – mgyf = Ui – Uf = –∆Ug

Wg = –∆Ug : work done by gravity = the negative change in Ug

=> Ki +Wg = Kf

=> Ki – ∆Ug = Kf

=> ∆K + ∆Ug = 0 or Etot = Ki + Ui = Kf + Uf

2

If you drop a ball from a certain height, the gravitational potential energy(Ug ) decreases and the

kinetic energy(K) increases, but the total mechanical energy is conserved.

*~ Ug = mgh K = 𝟏

𝟐𝒎𝒗𝟐 ~*

m=10kg, h=10.2m, g=9.8m/s2

Ug =1000J K=0J Total Energy = Ug + K = 1000J

h Ug =500J K=500J Total Energy = Ug + K = 1000J



Q1) An object of mass 10kg is lifted at a height of 10m.

i) What is the gravitational potential energy of the object at that height? What is the kinetic energy at that

height? What is the total mechanical energy?

ii) When the object falls 8m, what is the gravitational potential energy and kinetic energy at that height?

What is the speed at that height? What is the total mechanical energy?

Q2) A 0.04kg body starting from rest falls through a vertical distance of 0.25m to the ground.

(a) What is the kinetic energy of the body just before it hits the ground?

(b) What is the speed of the body just before it hits the ground?

W=Fdcosφ f = µFN K =

1

2mv² Ug = mgh Ki +ΣW=Kf Ki + Ui = Kf + Uf

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1


Q3) A boy throws up a 0.15kg stone from the top of a 20m cliff with a speed of 15m/s. Find its

kinetic energy and speed when it lands in a river below.

Ug,max, K=0

Ug + K

Ug=0, Kmax

Ans) 46.3J, 24.8m/s

Q4) A block of mass 2kg is sliding down

a frictionless inclined surface. The block

begins at rest 2m above the plane

surface. The angle of the incline is 30°.

i) Find the speed when the block reaches

the bottom using Work.Kinetic Energy

Theorem.

FN

h=2m φ

30º

Fg

Solution) Ki + ΣW = Kf

Only force doing work on block is gravity Fg => Ki + Wg = Kf

Since Work is defined as W=Fcosφd => Wg=Fgcos(90º-30º)d, so Ki + Fgcos60ºd = Kf

Since the block starts at rest Ki =0 => 0 + mgcos60ºd = 1

2𝑚𝑣𝑓

2

Finally, solve for vf (remember that h=dsin30º) ans) 6.26m/s

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ii) Find the speed when the block reaches the bottom using Conservation of Mechanical Energy Theorem.

Solution) Since gravitational force Fg is conservative, mechanical energy is conserved throughout the motion

From Ki + Ugi = Kf + Ugf , Ki=Ugf =0, so Ugi = Kf !! => mgh = 1

2𝑚𝑣𝑓

2,then solve for vf

Energy is Path Independent

Mechanical Energy is conserved when a conservative force does work on an object in an isolated system.

If there is an applied force from outside of the system or frictional force, then mechanical energy is not

conserved.

A block is released at height of h=5m for all three cases. Using Ki +ΣW=Kf, find the speed of the

block at the bottom for each case. ‘d ’ will be the distance traveled.

θ=30°

case-1

θ=60°

case-2

case-3

Solution) Ki + ΣW = Kf where Ki=0

Only force doing work on block is gravity Fg => Wg = Kf

=> (mgcosφ)d= 1

2𝑚𝑣𝑓

2 since φ=90°‒ θ, cosφ=sinθ and sinθ=h/d

=> mgh= 1

2𝑚𝑣𝑓

2 . So vf =√2𝑔ℎ = √2 × 9.8 × 5 = 9.9m/s for all cases

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Q5) A child rides on a smooth slide of a height of 2m. The child starts from rest at the top.

Determine the speed at the bottom.

Ans) v=6.3m/s *~If the path is not linear, it is impossible to use W-K theorem because φ is a variable!

Q6) A roller coaster starts from rest at the top of

an 18m hill. The car travels to the bottom of the

hill and continues up the next hill that is 10m high.

How fast is the car moving at the top of the hill if

the friction is ignored?

h=18m

h=10m

Ans) v=12.5m/s

Q7) A roller coaster starts from rest at position 1 shown below and rolls without friction along the

loop. The roller coaster first pass A, goes around the loop and passes B and then passes C. The

height h=1m and the radius of the loop is r=0.4m. Find the speed of the roller coaster at position

A, B and C.

B

r

h

A

C

Ans) A&C is 4.43m/s, B is 1.98m/s

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Q8) A 0.8kg pendulum bob on a 2m cord is pulled sideways until the cord makes an angle of 36.9° with

the vertical. Find the speed of the bob as it passes through the position B after being released at rest.

Position A is the maximum height and the position B will the low point of the bob.

36.9°

A

h

B

a) 2.8m/s b) 3.6m/s c) 4.5m/s d) 6.1m/s

Q9) A skier leaves the top of a slope with an

initial speed of 5m/s. Her speed at the bottom

of the slope is 13m/s. What is the height of the

slope? (*~remember that the direction of the

speed is irrelevant when using 'Conservation

of Mechanical Energy' equation)

vi=5m/s

h=? vf =13m/s

a) 1.14m b) 4.61m c) 6.45m d) 7.35m

Q10) A projectile of mass 2kg is fired with initial speed of 50m/s at an angle of θ=53.1° from the

horizontal surface. *~vx=vcosθ, vy=vsinθ~*

i) What is the initial kinetic energy?

a) 900J b) 1600J c) 2500J d) 0J

ii) What is the kinetic energy when the projectile is at the highest point in its trajectory?

a) 900J b) 1600J c) 2500J d) 0J (*'0J ' is not the answer!!!)

iii) What is the gravitational potential energy at the highest point in its trajectory?

a) 900J b) 1600J c) 2500J d) 0J

iv) Find the maximum height of the trajectory.

a) 81.6m b) 62.9m c) 49.9m d) 35.4m

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Elastic Potential Energy see p.194

The work done by a spring force (Fs = - kx) on a block connected to the spring is given

Ws = 𝟏

𝟐kxi

2 – 𝟏

𝟐kxf

2

Since the elastic potential energy is expressed as Us = 𝟏

𝟐kx2 where x=xf -xi

Ws = – ∆Us

Note that the spring force is a conservative force, so mechanical energy is conserved

Q11) A 1.6kg block on a frictionless surface is pressed against a spring with a spring constant of

k=1000N/m. The pressure causes the spring to compress 0.02m and then released.

What is the kinetic energy of the moment ball passes the equilibrium position? What is the speed

at that position? see solution below for your reference

i) Solution using W-E method

Ki + ΣW = Kf

Only force doing work is the spring force Fs

=> Ki + Ws = Kf

If we choose the initial state when the moment the block is

released, then Ki =0, Ws=1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2 and xi=0.02m, xf=0

=> 1

2𝑘𝑥𝑖

2= Kf then find Kf and solve for vf

ii) Solution using Conservation of Mech. E method

Ki + Usi = Kf + Usf , here Ki= Usf =0 => Usi = Kf

=> 1

2𝑘𝑥𝑖

2= Kf

Ans)0.2J, 0.5m/s

W=Fcosφ×d f =µFN K=

1

2mv² Ug=mgh Ki+ΣW=Kf Ws =

1

2kxi

2– 1

2kxf

2 Us = 1

2kx2 ∆K+∆U =0

8

Q12) The horizontal surface on which the block slides is frictionless. The speed of the block before it

touches the spring is 6m/s. How fast is the block moving at the instant the spring has been compressed

0.15m? k=2000N/m, mass is 2kg.

If we choose the initial state when the moment block comes in

contact with spring, then xi=0 and xf=0.15m.

Ans) v=3.67m/s

xf =0.15m xi=0

vf =? vi =6m/s

Q13) revisit! A crate of mass 12kg slides from rest

down a frictionless 35° incline and is stopped by a

strong spring with k=3×104N/m. The block slides

3m from the point of release to the point where it

comes to rest against the spring. When the block

comes to rest, how far has the spring been

compressed?

Ans) 0.116m

3m


1


1

2kxi

2– 1

2kxf

2 Us = 1

2kx2 ∆K+∆U =0

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Conservative and Nonconservative Forces (see p.196 for more details)

Conservative Forces

1. A force is conservative if the work it does on a particle moving between any two points is independent

of the path taken by the particle ex) Fg is a conservative force because it is path independent

2. The work done by a conservative force on a particle moving through any closed path is zero.

( A closed path is one in which the beginning and end points are identical)

- Examples of conservative forces are gravitational force (Fg) and elastic force(Fs ).

- For an object free-falling from a certain height, the total mechanical energy(E) of the system consists of

kinetic and gravitational potential energy.

- The total mechanical energy(E) of a falling object remains constant in any isolated system if the

force acting on the object is a conservative force

- If an object is free-falling, then Ki + Wg = Kf

Since the force of gravity is conservative, Wg = ‒∆Ug

=> ∆K + ∆U = 0 for conservative force (∆Etot = 0)

=> E = Ki + Ugi = Kf + Ugf

**Non-conservative Forces

- If the forces acting on objects within a system are conservative, then the total mechanical energy of the

system remains constant. However, if some of the forces acting on objects within the system are not

conservative, then the total mechanical energy does not remain constant

- A force is non-conservative if it causes a change in mechanical energy E, which is the sum of kinetic

and potential energies.

Examples of non-conservative forces are an applied force outside the isolated system and frictional force

i) Work Done by an Applied Force ( = an external force exerted outside the system)

- If you lift a book over a certain distance(Wapp), then

Ki + Wapp + Wg = Kf

Since, work done by gravity(Wg) is conservative, Wg= - ∆Ug, so

Ki + Wapp ‒ ∆Ug = Kf

=> Wapp = ∆K + ∆Ug and Wapp = ∆Etot for non-conserative force (∆Etot ≠ 0)

- We can see that the total mechanical energy is not conserved because the Wapp changes ∆Etot.

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ii) Work Done by Kinetic Friction

- If an object moves a distance d on a flat surface, the only force that does work is the force of kinetic

friction

Ki + Wfriction = Kf

Ki – fk d = Kf

=> ∆K = - fkd

- If the book moves on an incline that has friction, a change in the gravitational potential energy of the

book-Earth system also occurs

∆E = ∆K + ∆Ug = - fkd

- Since frictional force causes a change in mechanical energy, it is non-conservative force.

Summary


1

2mv² Ug=mgh Us=

1

2kx² Ki+ΣW=Kf Ws =

1

2kxi

2– 1

2kxf

2

If there is no friction or an applied force outside the system and only conservative force exists

=> ∆K + ∆U = 0

If there is friction acting in the system =>∆K + ∆U = - fkd

If there is an applied force from outside the system on a frictionless surface => ∆K + ∆U = Wapp

If there is an applied force from outside the system on a rough surface => ∆K + ∆U = Wapp ‒ fkd

Q14) compare with Q4!

A block of mass 2kg is sliding down a rough

inclined surface. The block begins at rest 2m above

the plane surface. The friction acting on the block

is 4N. The angle of the incline is 30°. Find the

speed when the block reaches the bottom using

Conservation of Energy Theorem. That is, use

∆E = ∆K + ∆Ug = - fkd.

(solution is on the next page)

FN

fk

φ

h=2m 30º

Fg

11

Solution) ∆K + ∆Ug = - fkd.

=>Kf - Ki + Ugf - Ugi = - fkd , here Ki=Ugf =0 => Kf - Ugi = - fkd

So rearrange according to Kf

=> 1

2𝑚𝑣𝑓

2=mghi - fkd , then solve for vf ans) 4.82m/s


1


1

2kxi

2– 1

2kxf

2 Us =

1

2kx2

∆K + ∆U =0: for conservative forces ∆K + ∆U = Wapp ‒ fkd: for non-conservative forces

Q15) compare with Q11!

A 1.6kg block on a rough surface is pressed against a spring with a spring constant of k =1000N/m. The

pressure causes the spring to compress 0.02m and then released. A constant frictional force of 4N retards

its motion from the moment it is released

What is the speed the moment the block passes equilibrium position?

Ans) 0.387m/s

Q16) A 10kg block on a horizontal frictionless surface is attached to a light spring(k=200N/m). The block

is initially at rest at its equilibrium position when a horizontal force Fp=80N pulls on the block. What is

the speed of the block when it is 0.13m from its equilibrium position?

Fp

Ans) 1.32m/s

12


1

2mv² Ug=mgh Ki+ΣW=Kf Ws=

1

2kxi

2– 1

2kxf

2 Us =

1

2kx2

∆K + ∆U =0: for conservative forces ∆K + ∆U = Wapp - fkd: for non-conservative forces

Q17) Two boxes are connected to each other on a

frictionless surface as shown. The system is

released from rest and the m2=1kg box falls

through a vertical distance of h=1m. What is the

kinetic energy of the m2 just before it reaches the

floor? Solve using

i) Newton’s 2nd law, ii) Work.Kinetic Energy

method and iii) Conservation of Energy method

Solution is on the next page for your reference

m1=3kg

m2=1kg

1m

13


1

2mv² Ug=mgh Ki+ΣW=Kf Ws=

1

2kxi

2– 1

2kxf

2 Us =

1

2kx2

∆K + ∆U =0: for conservative forces ∆K + ∆U = Wapp - fkd: for non-conservative forces

Method I. Using Newton's Law

m1 : ∑Fx= T = m1a ∑Fy= FN - m1g = 0

m2 : ∑Fy= m2g – T= m2a

From the above two equations, eliminating tension

force 'T' and solving for acceleration 'a' gives

FN m1=3kg

T

m1g T

m2=1kg

1m m2g

a = 𝑚2𝑔

𝑚1+𝑚2 = 2.45m/s2

Using 2ad = vf 2 – vi

2 to find vf, where vi=0 and replacing 'd' for 'h' gives

vf =√2𝑎ℎ = 2.21m/s

Since K= 1

2mv2, Kf =

1

2×1×2.212=2.44J

Method II. Using Work.Kinetic Energy.

Set up the equations for m1 and m2 separately

where the tension force T is an applied force from

outside the system.

m1: Ki+WT =Kf

=>0+Thcos0°=½m1v1f 2

=>Th=½mv1f 2 -----------------(1)

m1=3kg

T

T

Isolated System for m1!

m2=1kg

Isolated system for m2!

1m m2g

m2: Ki+WT +Wg=Kf

=>0+ Thcos180°+Fghcos0°=½m2v2f 2

=> – Th+mgh=½m2v2f 2 -----------------(2)

The speed of m1 and m2 are the same since they are connected with the string, so setting v1=v2=v,

combining (1) and (2)gives

½(m1+m2)vf 2=m2gh

Solving for vf gives

vf =2.21m/s

Since K= 1

2mv2, Kf =

1

2×1×2.212=2.44J

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How you choose your system makes a big difference

The tension force T can be an applied force or an internal force depending on how you choose your

system.

If you choose each mass separately as an isolated system, the tension force must be solved as an applied

force( =similar to method II) and the mechanical energy is not conserved.

If you choose the system that encompasses both masses as below, then the tension force becomes an

internal force and the mechanical energy is conserved.

Method III. Using Energy Conservation (system

as whole)

First, the gravitational potential energy and kinetic

energy for m1 is 'Ug1' and 'K1'. And the

gravitational potential energy and kinetic energy

for m2 is 'Ug2' and 'K2'

Conservation of mechanical energy states

K1i+Ug1i + K2i+Ug2i = K1f +Ug1f + K2f +Ug2f

Since Ug1i = Ug1f , and K1i =K2i =0

m1=3kg

m2=1kg

single

system!! 1m

Ug2i = K1f + K2f

m2ghi = 1

2𝑚1𝑣1𝑓

2 +1

2𝑚2𝑣2𝑓

2

The speed of m1 and m2 are the same since they are connected with the string, so v1=v2=v,

m2ghi = 1

2𝑚1𝑣𝑓

2+1

2𝑚2𝑣𝑓

2

Solving for vf gives

vf =2.21m/s

Since K= 1

2mv2, Kf =

1

2×1×2.212=2.44J

Relationship between Conservative Forces and Potential Energy

When an object is free-falling in the air, the only force acting on the object is Fg (assuming air resistance

is negligible). Since Fg is a conservative force, the work done by Fg will be equal to the negative change

in gravitational potential energy(−∆Ug).

Wg = –∆Ug : work done by gravity = the negative change in Ug

This can also be applied to a spring-block system, where there is no friction and applied force acting in

the system.

Ws = –∆Us : work done by gravity = the negative change in Ug

15

In general, the work done by internal conservative force will be equal to the negative change in potential

energy

Wint = ∫ 𝑭𝒙 𝒅𝒙𝒙𝒇

𝒙𝒊= –∆U

where Fx is a conservative force.

The above relationship between work and force can be rewritten as

Fx = −𝒅𝑼

𝒅𝒙

Q18) To stretch a certain nonlinear spring by an amount x requires a force F given by Fx = 40x - 6x2,

where Fx is in newtons and x is in meters. What is the change in potential energy when the spring is

stretched 2 meters from its equilibrium position?

a) 16 J b) 28 J c) 56 J d) 64 J e) 80 J

Q19) A single conservative force Fx = (6.0x – 12) N (x is in m) acts on a particle moving along the x axis.

The potential energy associated with this force is assigned a value of +20J at x=0. What is the potential

energy at x=3.0m?

a) +11J b) +29J c) +9J d) -9J e) +20J

Q20) A 0.40kg particle moves under the influence of a single conservative force. At point A where the

particle has a speed of 10m/s, the potential energy associated with the conservative force is +40J. As the

particle moves from A to B, the force does +25J of work on the particle. What is the value of the potential

energy at point B?

a) +65J b) +15J c) +35J d) +45J e) -40J

16

Power see p.232

Two cars were climbing a roadway hill and both reached the top, but one car reached the top in lesser

time than the other. Both cars have done equal amount of work, but the rate amount of work done is

different.

From a practical viewpoint, it is interesting to know not only the work done by vehicles but also the rate

at which it is done.

The time rate of doing work is call power.

The average power is defined as

�̅� = 𝑊

𝛥𝑡

Instantaneous power can be expressed as

𝑃 = 𝑑𝑊

𝑑𝑡= 𝐅 ∘

𝑑𝐫

𝑑𝑡= 𝐅 ∘ 𝐯

The standard unit for power is joules per second(J/s), also call watt(W)

1W = 1J/s =1kg·m2/s3

One horsepower(hp) is equal to 746 watts, so

1hp = 746W

Example) A 300W light bulb running for 12 hours would be converted to 0.3kW·12h = 3.6kWh of

electrical energy used. Or 300W·12×3600s=1.3×107J of energy used in 12 hours.

(Remember that the energy is power used multiplied by the time during which it was used)

𝑃 = 𝑑𝑊

𝑑𝑡= 𝐅 ∘

𝑑𝐫

𝑑𝑡= 𝐅 ∘ 𝐯 1hp = 746W

Q1) A 700N Marine in basic training climbs a 10m vertical rope at a constant speed in 8s. What is his

power output?

a) 875W b) 724W c) 650W d) 572W

17

𝑃 = 𝑑𝑊

𝑑𝑡= 𝐅 ∘

𝑑𝐫

𝑑𝑡= 𝐅 ∘ 𝐯 1hp = 746W

Q2) A certain automobile engine delivers 2.24×104W(=30hp) to its wheels when moving at a constant

speed of 27m/s. What is the resistive force acting on the automobile?

a) 540N b) 664N c) 750N d) 830N

Q3) A skier of mass 70kg is pulled up a smooth frictionless slope by a motor-driven cable.

i) How much work is required for him to be pulled a distance of 60m up a 30° slope at a constant speed of

2m/s?

a) 1.30×104J b) 2.06×104J c) 2.70×104J d) 3.35×104J

ii) A motor of what power is required to perform this task?

a) 450W b) 569W c) 686W d) 734W

Q4) Power Delivered by an Elevator Motor

An elevator car has a mass of 1000kg and is carrying passengers

having a combined mass of 800kg. A constant frictional force of

4000N retards its motion upward. What must be the minimum power

delivered by the motor to lift the elevator car at constant speed of

3m/s?

a) 6.48×104W b) 4.23×104W c) 2.78×104W d) 1.06×104W

3m/s

potential energy and conservation of energyq3) a boy throws up a 0.15kg stone from the top of a 20m...

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