Download SPH4U: Lecture 12 Notes

SPH4U
Conservation of Energy
1-D Variable Force Example: Spring

For a spring we recall that Fx = -kx.
F(x)
x1
x2
x
relaxed position
-kx
the mass
F = - k x1
F = - k x2
Review: Springs

Hooke’s Law: The force exerted by a spring is
proportional to the distance the spring is stretched or
compressed from its relaxed position.

FX = -kx
Where x is the displacement from the equilibrium and
k is the constant of proportionality.
relaxed
position
FX = 0
x
More Spring Review

F(x)
The work done by the spring Ws during a
displacement from x1 to x2 is the area under the
F(x) vs x plot between x1 and x2.
Ws 
x1
x2
x2
 F ( x)dx
x1
x2
x
Ws
-kx
In this example it is a negative
number. The spring does
negative work on the mass
  ( kx) dx
x1
1
  kx 2
2
x2
x1
1
Ws   k  x22  x12 
2
Problem: Spring pulls on mass.

A spring (constant k) is stretched a distance d, and a mass m
is hooked to its end. The mass is released (from rest). What
is the speed of the mass when it returns to the relaxed
position if it slides without friction?
m
relaxed position
m stretched position (at rest)
d
m
after release
v
m
vr
back at relaxed position
Solution: Spring pulls on mass.

First find the net work done on the mass during the
motion from x = d to x = 0 (only due to the spring):
1
1
1 2
2
2
2
2
Ws   k  x2  x1    k  0  d   kd
2
2
2
x2
x1
m stretched position (at rest)
d
m
vr
relaxed position
Problem: Spring pulls on mass.

Now find the change in kinetic energy of the mass:
1
1
mv 22  mv12
2
2
1
1
 mv 2r  m02
2
2
1
 mv 2r
2
ΔK 
x2
x1
m stretched position (at rest)
d
m
vr
relaxed position
Solution: Spring pulls on mass.

Now use work kinetic-energy theorem: Wnet = WS = K.
1 2 1
kd  mv r 2
2
2
k
vr  d
m
x2
x1
m stretched position (at rest)
d
m
vr
relaxed position
Springs : Understanding

A spring with spring constant 40 N/m has a relaxed length of
1 m. When the spring is stretched so that it is 1.5 m long, what
force is exerted on a block attached to the end of the spring?
x=0
k
x=1
x=0
x = 1.5
k
M
M
FX = -kx
(a) -20 N
(b) 60 N
(c) -60 N
FX = - (40N/m) ( .5m)
FX = - 20 N
Understanding
Forces and Motion

A block of mass M = 5.1 kg is supported on a frictionless
ramp by a spring having constant k = 125 N/m. When the
ramp is horizontal the equilibrium position of the mass is at
x = 0. When the angle of the ramp is changed to 30o what
is the new equilibrium position of the block x1?
(a) x1 = 20cm
k
(b) x1 = 25cm
(c) x1 = 30cm
x=0
M
q = 30o
Solution



Choose the x-axis to be along downward direction of ramp.
FBD: The total force on the block is zero since it’s at rest.
Consider x-direction: Force of gravity on block is Fx,g = Mg sin(q
Force of spring on block is Fx,s = -kx1
N
y
x
Fx,g = Mg sinq
q
q
Mg
Solution

Since the total force in the x-direction must be 0:
Mg sinq  kx1  0
x1 
5.1kg  9.81m s 2  0.5
x1 
 0.2 m
125 N m
y
x
q
Μg sin θ
k
Work by variable force in 3-D: Nice
to know explanation

Work dWF of a force F acting
through an infinitesimal
displacement r is:
F
r
.
dW = F r

The work of a big displacement through a variable force will
be the integral of a set of infinitesimal displacements:

.
WTOT = F r
Work/Kinetic Energy Theorem for a
Variable Force in 3D
r2
W   F  dr  12 mv22  12 mv12   kinetic energy
r1
Sum up F.dr along path
That’s the work integral
That equals change in KE
For conservative forces, the work
is path independent and depends
only on starting point and end
point
Work by variable force in 3-D:
Newton’s Gravitational Force

Integrate dWg to find the total work done by gravity in a “big”
displacement:

R2

R2
Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1)
R1
R1
Fg(R2)
R2
Fg(R1)
R1
M
m
Work by variable force in 3-D:
Newton’s Gravitational Force

Work done depends only on R1 and R2, not on the path taken.
 1 1 
Wg  GMm   
 R2 R1 
m
R2
R1
M
Potential Energy

For any conservative force F we can define a
potential energy function U in the following way:
W =


 F.dr = -U
The work done by a conservative force is equal and opposite to
the change in the potential energy function.
r2
This can be written as:
F.dr
r2
U = U2 - U1 = -W = -
r1
r1
U1
U2
Gravitational Potential Energy

So we see that the change in U near the Earth’s surface is:
U = -Wg = mg y = mg(y2 -y1).

So U = mg y + U0 where U0 is an arbitrary constant.

Having an arbitrary constant U0 is equivalent to saying that we can
choose the y location where U = 0 to be anywhere we want to.
 Floor level of 400
 Hazel St
 Science Office (potential is zero here, for sure!)
m
y2
y1
Wg = -mg y
Conservative Forces:

We have seen that the work done by gravity does not
depend on the path taken.
m
R2
 1 1 
Wg  GMm   
 R2 R1 
R1
M
m
h
Wg = -mgh
Understanding
Work & Energy
A rock is dropped from a distance RE above the surface of the earth, and
is observed to have kinetic energy K1 when it hits the ground. An
identical rock is dropped from twice the height (2RE) above the earth’s
surface and has kinetic energy K2 when it hits. RE is the radius of the
earth.
What is K2 / K1?
(a)
The easiest way to solve
this problem is to use
the W=K property.
2
2RE
(b)
(c)
3
2
4
3
RE
RE
Be careful!
Solution

Since energy is conserved, K = WG.
 1
1 
WG =GMm 
 
 R 2 R1 
Do not use mgh formula as this
only works when h is very small.
2RE
RE
RE
 1
1 
ΔK=c 


R
R
 2
1 
Where c = GMm is the
same for both rocks
For the first rock:
 1
1 
1 1
K1 =c 

  c 
2 RE
 R E 2R E 
For the second rock:
 1
1 
2 1
K 2 =c 


c



R
3R
3
RE
 E
E 
2
K2 3 4
= 
K1 1 3
2
Conservative Forces:

In general, if the work done does not depend on the path taken (only
depends the initial and final distances between objects), the force
involved is said to be conservative.
 1 1 
Wg  GMm   
 R2 R1 

Gravity is a conservative force:

Gravity near the Earth’s surface:

A spring produces a conservative force:
W g   mg  y
1
Ws   k  x22  x12 
2
Conservative Forces:


A force that offers the opportunity of two-way conversion
between kinetic and potential energies is called a
conservative force.
The work done by a conservative force always has these
properties:
 It can always be expressed as the difference between
the initial and final values of potential energy function.
 It is reversible.
 It is independent of the path of the body and depends
only on the starting and ending points.
 When the starting and ending points are the same,
the total work is zero.
Conservative Forces:


When the only forces that do work are conservative forces, then the
total mechanical energy is E = K + U
Conservative forces have the nice property of being able to be
defined in terms of a potential energy. The usual definition of
potential energy is through the work-energy theorem as for kinetic
energy, i.e. W = Ui - Uf.
Wtotal  U1  U 2
Wtotal  K 2  K1
U1  U 2  K 2  K1
U1  K1  U 2  K 2
NonConservative Forces:


Not all forces are conservative.
Consider the friction force applied to a crate, the total
work done by friction force when sliding the crate up a
ramp and back down is not zero. (when the direction of
the motion reverses so does the friction force, and the
friction does negative work in both directions.)
Conservative Forces:

We have seen that the work done by a conservative force
does not depend on the path taken.
W2
W 1 = W2

Therefore the work done by a
conservative force in a closed
path is 0.
W1
W2
WNET = W1 - W2
= W1 - W1 = 0
W1
Potential energy change from one point to
another does not depend on path
Understanding
Conservative Forces

The pictures below show force vectors at different points in
space for two forces. Which one is conservative ?
(a) 1
(b) 2
y
(c) both
y
x
(1)
x
(2)
Solution

Consider the work done by force when moving along different
paths in each case:
No work is done when going
perpendicular to force.
WA = WB
(1)
WA > WB
(2)
Solution

In fact, you could make money on type (2) if it ever existed:
 Work done by this force in a “round trip” is > 0!
 Free kinetic energy!!
WNET = 10 J = K
W=0
W = 15 J
W = -5 J
W=0
Potential Energy Recap:

For any conservative force we can define a potential energy function
U such that:
F.dr
S2
U = U2 - U1 = -W = -
S1

The potential energy function U is always defined only up to an
additive constant.

You can choose the location where U = 0 to be anywhere
convenient.
Conservative Forces & Potential
Energies
(stuff you should know):
Force
F
Fg = -mg
Fg = 
GMm
R2
Fs = -kx
Work (done by force)
W
-mg(y2-y1)
 1 1 
GMm   
 R2 R1 
1
 k  x22  x12 
2
Change in P.E
U = U2 - U1
P.E. function
U
mg(y2-y1)
 1 1
GMm   
 R2 R1 
1
k  x22  x12 
2
mgy + C

GMm
C
R
1 2
kx  C
2
(R is the center-to-center distance, x is the spring stretch)
Understanding
Potential Energy

All springs and masses are identical. (Gravity acts down).
 Which of the systems below has the most potential energy
stored in its spring(s), relative to the relaxed position?
(a) 1
(b) 2
(c) same
(1)
(2)
Solution

The displacement of (1) from equilibrium will be half of that of (2) (each
spring exerts half of the force needed to balance mg)
0
d
2d
(1)
The spring P.E. is
twice as big in (2) !
(2)
1
2
The potential energy stored in (1) is: 2  kd 2  kd 2
The potential energy stored in (2) is:
1
2
k  2d   2kd 2
2
Conservation of Mechanical Energy

If only conservative forces are present, the total kinetic plus potential
energy of is conserved, i.e. the total “mechanical energy” is
conserved (def. of ME).
 (note: E=Emechanical throughout this discussion)
E=K+U
E = K + U
 using K = W
= W + U
= W + (-W) = 0  using U = -W
E = K + U is constant!!!


Both K and U can change, but E = K + U remains constant.
But we’ll see that if dissipative forces act, then energy can be “lost” to
other modes (thermal, sound, etc) changing Emechanical and external
forces can change Emechanical
Example: The simple pendulum

Suppose we release a mass m from rest a distance h1
above its lowest possible point.
 What is the maximum speed of the mass and where
does this happen?
 To what height h2 does it rise on the other side?
m
h1
h2
v
Example: The simple pendulum


Kinetic+potential energy is conserved since gravity is a
conservative force (E = K + U is constant)
Choose y = 0 at the bottom of the swing,
and U = 0 at y = 0 (arbitrary choice)
E = 1/2mv2 + mgy
y
y=0
h1
h2
v
Example: The simple pendulum

E = 1/2mv2 + mgy.
 Initially, y = h1 and v = 0, so E = mgh1.
 Since E = mgh1 initially, E = mgh1 always since energy is
conserved.
y
y=0
Example: The simple pendulum
 1/2mv2 will be
 So at y = 0
maximum at the bottom of the swing.
1/ mv2 = mgh
v2 = 2gh1
2
1
v 
2 gh1
y
y = h1
y=0
h1
v
Example: The simple pendulum


Since E = mgh1 = 1/2mv2 + mgy it is clear that the maximum
height on the other side will be at y = h1 = h2 and v = 0.
The ball returns to its original height.
y
y = h1 = h 2
y=0
Example: The simple pendulum

The ball will oscillate back and forth. The limits on its
height and speed are a consequence of the sharing of
energy between K and U.
E = 1/2mv2 + mgy = K + U = constant.
y
Example: Airtrack & Glider

A glider of mass M is initially at rest on a horizontal frictionless
track. A mass m is attached to it with a massless string hung
over a massless pulley as shown. What is the speed v of M
after m has fallen a distance d ?
v
M
m
d
v
Example: Airtrack & Glider


Kinetic+potential energy is conserved since all forces are
conservative.
Choose initial configuration to have U=0.
K = -U
1
 m  M v 2  mgd
2
v
M
m
d
2mgd
v
mM
Problem: Hotwheel

A toy car slides on the frictionless track shown below. It
starts at rest, drops a distance d, moves horizontally at
speed v1, rises a distance h, and ends up moving
horizontally with speed v2.
 Find v1 and v2.
v2
d
v1
h
Problem: Hotwheel...



K+U energy is conserved, so E = 0
K = - U
Moving down a distance d, U = -mgd, K = 1/2mv12
Solving for the speed:
v1 
2 gd
d
v1
h
Problem: Hotwheel...



At the end, we are a distance d - h below our starting point.
U = -mg(d - h), K = 1/2mv22
Solving for the speed:
v2 
2 g d  h 
d-h
d
v2
h
Hooke’s Law (review)
The magnitude of the force exerted by the spring is
directly proportional to the distance the spring has
moved from its equilibrium.
Fx  kx
Force is opposite to the
direction spring is moved
Fx  kx
This is the Force applied to the
spring
Example
A 0.085 kg mass is hung from a vertical spring that is
allowed to stretch slowly from its unstretched equilibrium
position until it comes to its new equilibrium position
0.20 m below its initial one.
a) Determine the force constant of the spring?
b) If the ball is returned to the spring’s initial unstreched
equilibrium position and then allowed to fall, what is
the Net Force on the mass when it has dropped
0.082 m?
c) Determine the acceleration of the mass at position b)
Solution
F=-kx
F
y
0
mg    kx   0
kx  mg
mg
k
x

M

0.20m
N
 4.165
m

F=mg
a) Determine the force constant of the spring?
Therefore k= 4.2 N/m
 0.085kg   9.8
N
kg 
Solution
F=-kx
F=mg
F
N
 mg   kx 

N 
N
 0.085kg   9.8    4.165   0.082m 
kg  
m

 0.4915 N
b) If the ball is returned to the spring’s initial unstreched
equilibrium position and then allowed to fall, what is the Net
Force on the mass when it has dropped 0.082 m?
Therefore F= 0.49 N
Solution
F=-kx
F
y
 ma y
0.4915 N  ma y
F=mg
0.4915 N
ay 
0.085kg
m
 5.782 2
s
c) Determine the acceleration of the mass at position b)
Therefore a= 5.8 m/s2 down
Elastic Potential Energy
(review)
The energy stored in objects that are stretched,
compressed, bent, or twisted.
1 2
U s  kx
2
Understanding
A 0.10 kg mass is hung from a vertical spring (k=9.6
N/m). The mass is held so that the spring is at its
unstretched equilibrium position. The mass is then
allowed to fall. Neglect the mass of the spring.
a) How much elastic potential energy is stored in the
spring when the mass has fallen 11 cm?
b) What is the speed of the mass when it has fallen
11cm?
Solution
x=0 cm
x=11 cm
M
1 2
U s  kx
2
1
N
2
  9.6   0.11m 
2
m
 5.8 102 J
a) How much elastic potential energy is stored in the
spring when the mass has fallen 11 cm?
Solution
x=0 cm
M
x=11 cm
Ei  E f
1 2 1 2
mv f  kx
2
2
2mgh  kx 2  mv 2f
mgh 
2mgh  kx 2
vf 
m
m
N
2


2  0.10kg   9.8 2   0.11m    9.6   0.11m 
s 
m



 0.10kg 
 1.0
m
s
b) What is the speed of the mass when it has fallen 11cm?
Flash