Download Potential Energy and Conservation of Energy

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.7 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)
- The gravitational force Fg= mg does work on the object.
=> Ki + ΣW = Kf. (Work.Kinetic Energy Theorem )
y=yi-yf
=> 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
yi
a distance of ‘y’,
yf
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
1
Etot = Ki + Ui = Kf + Uf
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
h
K=
~*
Ug =1000J K=0J
Total Energy = U g + K = 1000J
Ug =500J
Total Energy = U g + K = 1000J
K=500J
Ug =250J K=750J
Total Energy = U g + K = 1000J
Ug =0J
Total Energy = U g + K = 1000J
W=Fdcosφ
K=1000J
f = µFN
K = mv²
Ug = mgh
Ki +ΣW=Kf
Ki + Ui = Kf + Uf
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?
2
W=Fdcosφ
f = µFN
K = mv²
Ug = mgh
Ki +ΣW=Kf
Ki + Ui = Kf + Uf
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
FN
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.
φ
h=2m
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 =
Finally, solve for vf
(remember that h=dsin30º)
ans) 6.26m/s
3
W=Fdcosφ
f = µFN
K = mv²
Ki +ΣW=Kf
Ug = mgh
Ki + Ui = Kf + Uf
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 =
,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=60°
θ=30°
case-2
case-1
Solution) Ki + ΣW = Kf where Ki=0
Only force doing work on block is gravity Fg
=> (mgcosφ)d=
=> mgh=
case-3
=> Wg = Kf
since φ=90°‒ θ, cosφ=sinθ and sinθ=h/d
. So vf =
=
= 9.9m/s for all cases
4
W=Fdcosφ
f = µFN
K = mv²
Ki +ΣW=Kf
Ug = mgh
Ki + Ui = Kf + Uf
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
5
W=Fdcosφ
f = µFN
K = mv²
Ug = mgh
Ki +ΣW=Kf
Ki + Ui = Kf + Uf
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.
a) 2.8m/s
b) 3.6m/s
c) 4.5m/s
d) 6.1m/s
36.9°
A
h
B
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)
a) 1.14m
vi=5m/s
h=?
b) 4.61m
vf =13m/s
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
6
d) 35.4m
W=Fcosφ×d
f =µFN
K= mv²
Ug=mgh
Ki+ΣW=Kf
Ws = kxi2– kxf 2 Us = kx2
∆K+∆U =0
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 = kxi2 – kxf 2
Since the elastic potential energy is expressed as Us = kx2,
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
ii) Solution using Conservation of Mech. E method
i) Solution using W-E method
Ki + ΣW = Kf
Only force doing work is the spring force Fs
Ki + Usi = Kf + Usf , here Ki= Usf =0 => Usi = Kf
=>
= Kf
=> Ki + Ws = Kf
If we choose the initial state when the moment the block is
released, then Ki =0, Ws=
=>
and xi=0.02m, xf=0
Ans)0.2J, 0.5m/s
= Kf then find Kf and solve for vf
7
W=Fcosφ×d
f =µFN
K= mv²
Ug=mgh
Ki+ΣW=Kf
Ws = kxi2– kxf 2 Us = kx2
∆K+∆U =0
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.
xf =0.15m
If we choose the initial state when the moment block comes in
contact with spring, then xi=0 and xf=0.15m.
vf =?
xi=0
vf =6m/s
Ans) v=3.67m/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?
3m
Ans) 0.116m
8
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 (see p.4)
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 and elastic force.
- 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
=> 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 = ∆E
- We can see that the total mechanical energy is not conserved because the Wapp changes ∆E.
9
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
W=Fcosφ×d
f =µFN K= mv² Ug=mgh
Ki+ΣW=Kf
Ws = kxi2– kxf 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 => ∆K + ∆U = Wapp
If there are both friction and applied force => ∆K + ∆U = Wapp - fkd
FN
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)
10
fk
φ
h=2m
30º
Fg
Solution) ∆K + ∆Ug = - fkd.
=>Kf - Ki + Ugf - Ugi = - fkd , here Ki=Ugf =0 => Kf - Ugi = - fkd
So rearrange according to Kf
=>
=mghi - fkd , then solve for vf
W=Fcosφ×d
f =µFN K= mv²
ans) 4.82m/s
Ug=mgh
∆K + ∆U =0: for conservative forces
Ki+ΣW=Kf
Ws = kxi2– kxf 2
Us = kx2
∆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
11
W=Fcosφ×d
f =µFN K= mv²
Ug=mgh
∆K + ∆U =0: for conservative forces
Ki+ΣW=Kf
Ws= kxi2– kxf 2
Us = kx2
∆K + ∆U = Wapp - fkd: for non-conservative forces
m1=3kg
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
m2=1kg
1m
Solution is on the next page for your reference
12
W=Fcosφ×d
f =µFN K= mv²
Ug=mgh
∆K + ∆U =0: for conservative forces
Ki+ΣW=Kf
Ws= kxi2– kxf 2
Us = kx2
∆K + ∆U = Wapp - fkd: for non-conservative forces
FN
Method I. Using Newton's Law
m1 : ∑Fx= T = m1a
∑Fy= FN - m1g = 0
m2 : ∑Fy= m2g – T= m2a
m1=3kg
T
m1g
T
m2=1kg
From the above two equations, eliminating tension
force 'T' and solving for acceleration 'a' gives
a=
1m
m2g
= 2.45m/s2
Using 2ad = vf 2 – vi 2 to find vf, where vi=0 and replacing 'd' for 'h' gives
vf =
2
= 2.21m/s
2
Since K= mv , Kf = ×1×2.21 =2.44J
m1=3kg
T
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.
T
Isolated System for m1!
m2=1kg
Isolated system for m2!
m1: Ki+WT =Kf
=>0+Thcos0°=½m1v1f 2
=>Th=½mv1f 2 -----------------(1)
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= mv2, Kf = ×1×2.212=2.44J
13
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.
m1=3kg
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'
m2=1kg
single
system!!
Conservation of mechanical energy states
1m
K1i+Ug1i + K2i+Ug2i = K1f +Ug1f + K2f +Ug2f
Since Ug1i = Ug1f , and K1i =K2i =0
Ug2i = K1f + K2f
m2ghi =
+
The speed of m1 and m2 are the same since there connected with the string, so v1=v2=v,
m2ghi =
+
Solving for vf gives
vf =2.21m/s
2
2
Since K= mv , Kf = ×1×2.21 =2.44J
Power
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
=
14
Instantaneous power can be expressed as
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 2 hours.
(Remember that the energy is power used multiplied by the time during which it was used)
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
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
15
d) 830N
1hp = 746W
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
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
16
d) 734W
3m/s