Transcript Physics 106P: Lecture 1 Notes

PHYS 152
Fall 2005
Thursday, Sep 22
Chapter 6 - Work, Kinetic Energy & Power
Exam #1
• Thursday, September 29, 7:00 – 8:00 PM.
• Material: Chapters 2-6.3
• Use black lead #2 pencil and calculator.
• Formula sheet provided. You may bring 1
extra sheet of handwritten notes (both sides)
but no other materials
• Sample exams on Web (link on homepage)
Room assignments - to be announced.
Supplemental Instruction
Exam Review Session
Tuesday Sept. 27th
7-9pm
RHPH 172
Pg 3
Reading Quiz: Sections 6.2 & 6.3
During a short time interval a particle moves along a
r
straight line a distance
s  2i$ 2 $j
During that time a constant force acted on the
ur
particle:
F  4 $j
The work done on the particle was
a. 0
b. 4
c. 8
d. 16
e. 8 2 .
Pg 4
Two marbles, one twice as heavy as the the other,
are dropped to the ground from the roof a building.
Just before hitting the ground, the heavier marble
has
(a) as much kinetic energy as the lighter one
(b) twice as much kinetic energy as the lighter one
(c) half as much kinetic energy as the lighter one
(d) four times as much kinetic energy as the lighter
one
(e) impossible to determine
Pg 5
Two marbles, one twice as heavy as the the other, are
dropped to the ground from the roof a building. Just before
hitting the ground, the heavier marble has
(a) as much kinetic energy as the lighter one
(b) twice as much kinetic energy as the lighter one
(c) half as much kinetic energy as the lighter one
(d) four times as much kinetic energy as the lighter one
(e) impossible to determine
W  K  Fd
K1  Fd  m1gd
K 2  Fd  2m1gd
Pg 6
Review: Constant Force...
W = F r
• If  = 90o no work is done.
T
No work done by T
No work done by N
v
v
N
Pg 7
Work-Kinetic Energy Theorem
Work done by the net external (constant) force
equals the change in kinetic energy
1
2 1
2
W  K  mv2  mv1
2
2
{Net Work done on object} = {change in kinetic energy of object}
Pg 8
Work done by Lifting
Example: Lifting a book
from the floor to a shelf
• First calculate the work done by gravity:
Wg = mg r = -mg r
shelf
r
FHAND
v = const
a=0
• Now find the work done by
the hand:
mg
WHAND = FHAND r = FHAND r
floor
Pg 9
Work done by Lifting
Work/Kinetic Energy Theorem: W = K
shelf
r
K  K f  Ki  WNET
FHAND
v = const
a=0
mg
floor
When lifting a book from the floor to a shelf, the object is
stationary before and after the lift:
Ki  K f  0, K  0, WNET  0
WNET= 0
WNET = WHAND + Wg
= FHAND r - mg r
= (FHAND - mg) r
WHAND = - Wg
Pg 10
Lifting vs. Lowering
Lifting
Lowering
shelf
shelf
r
FHAND
r
FHAND
v = const
a=0
v = const
a=0
mg
mg
floor
floor
Wg = -mg r
Wg = mg r
WHAND = FHAND r
WHAND = -FHAND r
WHAND = - Wg
WNET  0
WHAND = - Wg
Pg 11
Work done by gravity...
W NET = W1 + W2 + . . .+ Wn
= F r 1+ F r2 + . . . +F rn
= F (r1 + r 2+ . . .+ rn)
= F r
= F y
Wg = mg h
m
h
r1
r2
r3 r
mg
j
rn
Pg 12
Work done by Variable Force: (1D)
* When the force was constant, we wrote W = F x
area under F vs. x plot:
F
Wg
x
x
* For variable force, we find the area by integrating:
dW = F(x) dx.
W
x2
 F(x)dx
x1
F(x)
x1
dx
x2
Pg 13
Work/Kinetic Energy Theorem for a
Variable Force
W 
x2
 F dx
x1
x2
m
x1
v2
dv
dt
F  ma  m dv
dt
dx
dv dx dv
dv
=
= v dx (chain rule)
dt
dt dx
 m  v dv dx
v1
v2
dx
 m  v dv
v1
1
1
1
2
2
2
 m ( v2  v1 )  m v2  mv12  KE
2
2
2
Pg 14
Power
Power is the rate at which work is done by a force
PAVG = W/t
P = dW/dt
Average Power
Instantaneous Power
The unit of power is a Joule/second (J/s) which we define as a
Watt (W)
1 W = 1 J/s
r r
dW
d r r
d ur r
P

F  dx  (F gvdt)  F  v
dt
dt
dt


Pg 15
The force on a particle of mass m is given by
ur
F  10xi$
Choose the correct statement:
a. The work done will be the same going from x=1 to
x=2 as it is for going from x=0 to x=1.
b. The work done will be the same going from x=1 to
x=2 as it is for going from x=-1 to x=-2.
c. The average power is the same as the instantaneous
power.
d. None of the above are correct.
Pg 16
Work done by a Spring
Fp
x=0
Spring unstretched
Fs Fp Person
pulling
Person
Fs pushing
*For a person to hold a spring stretched out or compressed by x
from its unstretched length, it requires a force
Fp  kx
where k =spring constant measures the stiffness of the
spring.
Pg 17
Work done by a Spring
Fp
x=0
Spring unstretched
Fs Fp Person
pulling
Person
Fs pushing
The spring exerts a force (restoring force) in the opposite
direction:
Fs  kx
Hooke’s law
where k =spring constant measures the stiffness of the spring.
Pg 18
1-D Variable Force Example: Spring
*For a spring Fx = -kx. ( Hooke’s Law)
k =spring constant
F(x)
x1
x2
x
relaxed position
-kx
F = - k x1
F = - k x2
Pg 19
1-D Variable Force Example: Spring
*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.
F(x)
x1
x2
x
Ws
relaxed position
-kx
F = - k x1
F = - k x2
Pg 20
1-D Variable Force Example: Spring
*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 
x2

F( x)dx
x1
x2

 (kx)dx
x1
F(x)
x1
1 2
  kx
2
x2
x
Ws
-kx

x2
x1
1
Ws   k x22  x12
2

Pg 21
Work - Energy
A box sliding on a horizontal frictionless surface runs into a
fixed spring, compressing it a distance x1 from its relaxed
position while momentarily coming to rest.
v1
x
m1
m1
Pg 22
Work - Energy
Use the fact that WNET = K.
In this case
WNET = WSPRING = -1/2 kx2 and K = -1/2 mv2
so
kx2 =
mv2
v1
In the case of x1
x1  v1
m1
k
x1
m1
m1
Pg 23
Work - Energy
If the initial speed of the box were doubled and its mass
were halved, how far x2 would the spring compress ?
(a)
x2  x1
v2
(b) x2  2 x1
(c) x2  2 x1
x2
m2
m2
Pg 24
Work - Energy
If the initial speed of the box were doubled and its mass
were halved, how far x2 would the spring compress ?
m
x v
k
So if v2 = 2v1 and m2 = m1/2
x 2  2v 1
m1 2
k
 v1
2m1
k
x2  2 x1
v2
x2
m2
m2
Pg 25
Example
A person pulls on a spring. It requires a force of 75N to stretch
it by 3 cm. How much work does the person do? If the
person compresses the spring by 3 cm how much work
does the person do?
Calculate the spring constant:
F 75N
3
k 
 2.5  10 N / m
x 0.03m
The work is


1 2
1
2
3
W  kxmax  2.5  10 N / m 0.03m   1.1J
2
2
The work to compress the spring is the same since W is
proportional to x2.
Pg 26
Example: Compressed Spring
A horizontal spring has k=360N/m. (a) How much
work is required to compress a spring from x=0 to
x=-11 cm? (b) If a 1.85 kg block is placed against
the spring and the spring is released what will be the
speed of the block when it separates from the spring
at x=0?
N
Fs=kx
x=-11
mg
x=0
Pg 27
Example: Compressed Spring
The work done to stretch or compress the spring is:
1 2 1
2
W  kx  360N / m 0.11m   2.18J
2
2
In returning to its uncompressed length the spring will do
work W=2.18J on the block.
According to the work-energy principle the block acquires
kinetic energy:
Wnet
1 2
1 2
 K f  K i  mv  0 K  mv
2
2
2(K )
22.18J 
v

 1.54m / s
m
1.85Kg
Pg 28