phys 152 fall 2005 thursday, sep 22 chapter 6 - work, kinetic energy & power
TRANSCRIPT
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.
Pg 4
Reading Quiz: Sections 6.2 & 6.3
During a short time interval a particle moves along a straight line a distance
sr=2i$+ 2 j$
During that time a constant force acted on the particle:
Fur=4 j$
The work done on the particle was
a. 0
b. 4
c. 8
d. 16
e. .8 2
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
Pg 6
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 = ΔK = Fd
ΔK1 = Fd = m1gd
ΔK2 = Fd = 2m1gd
Pg 7
W = FF rr
• If = 90o no work is done.
Review: Constant Force...
vvNN
TT
vv
No work done by NN
No work done by TT
Pg 8
Work-Kinetic Energy Theorem
€
W = ΔK =12
mv22 −
12
mv12
Work done by the net external (constant) force equals the change in kinetic energy
{NetNet WorkWork done on object} = {changechange in kinetic energy kinetic energy of object}
Pg 9
• First calculate the work done by gravity:
Wg = mgg rr = -mg rr
• Now find the work done bythe hand:
WHAND = FFHAND rr = FHAND rr
Work done by Lifting Example: Lifting a book from the floor to a shelf
mgg
rr FFHAND
vv = const
aa = 0
floor
shelf
Pg 10
Work done by Lifting
Work/Kinetic Energy Theorem: W = K
€
K = K f − Ki = WNET
When lifting a book from the floor to a shelf, the object is stationary before and after the lift:
€
Ki = K f = 0, ΔK = 0, WNET = 0
WNET= 0
WNET = WHAND + Wg
= FHAND rr - mg rr
= (FHAND - mg) rr
WHAND = - Wg
mgg
rr FFHAND
vv = const
aa = 0
floor
shelf
Pg 11
Lifting vs. Lowering
WHAND = - Wg
mgg
rr FFHAND
vv = const
aa = 0
floor
shelf
mgg
rr FFHAND
vv = const
aa = 0
floor
shelfLifting Lowering
Wg = -mg rr
WHAND = FHAND rr
WHAND = - Wg
Wg = mg rr
WHAND = -FHAND rr
€
WNET = 0
Pg 12
rr11rr22
rr33
rrnn
= FF rr 1+ FF rr2 + . . . +FF rrn
= FF (rr11 + rr 2+ . . .+ rrnn)
Work done by gravity...
m
mgg
h j j
W NET = W1 + W2 + . . .+ Wn
Wg = mg h
r
= F r
= F y
Pg 13
Work done by Variable Force: (1D)* When the force was constant, we wrote W = F x
area under F vs. x plot:F
x
Wg
x
€
W = F(x)dxx1
x2
∫
F(x)
x1 x2 dx
* For variable force, we find the area by integrating:
dW = F(x) dx.
Pg 14
€
W =x1
x2
∫ dx
Work/Kinetic Energy Theorem for a Variable Force
dtmma ==FF
dx
€
=mdt
x1
x2
∫ dv
€
=mv1
v2
∫v dv
€
=m12
( − ) =12
m −12
m = ΔKEv22 v1
2 v22 v1
2
dv
dxdxdv dv dv
dxv (chain rule)
€
=mv1
v2
∫ dxdv
dxv
dt=
dt=
Pg 15
Power is the rate at which work is done by a forcePower is the rate at which work is done by a force
PPAVGAVG = W/ = W/t Average Powert Average Power
P = dW/dt Instantaneous PowerP = dW/dt Instantaneous Power
The unit of power is a Joule/second (J/s) which we define as a The unit of power is a Joule/second (J/s) which we define as a Watt (W)Watt (W)
1 W = 1 J/s1 W = 1 J/s
P =
dWdt
=ddt
rF ⋅d
rx( ) =
ddt
(Fur
gvrdt) =
rF ⋅
rv
Power
Pg 16
The force on a particle of mass m is given by
Fur=−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 17
Work done by a Spring
*For a person to hold a spring stretched out or compressed by x from its unstretched length, it requires a force
where k =spring constant measures the stiffness of the spring.
Spring unstretchedx=0
Fs Fp Person pulling
FsFp
Person pushing
kxFp =
Pg 18
Work done by a Spring
The spring exerts a force (restoring force) in the opposite direction:
where k =spring constant measures the stiffness of the spring.
Spring unstretchedx=0
Fs Fp Person pulling
FsFp
Person pushing
kxFs −= Hooke’s law
Pg 19
1-D Variable Force Example: Spring
*For a spring Fx = -kx. ( Hooke’s Law)
k =spring constant
F(x) x2
x
x1
-kxrelaxed position
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.
F(x) x2
x
x1
-kxrelaxed position
F = - k x1
F = - k x2
Ws
Pg 21
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) x2
x
x1
-kx
Ws
€
Ws = F(x)dxx1
x2
∫
= (−kx)dxx1
x2
∫
= −12
kx2
x1
x2
Ws = −12
k x22 − x1
2( )
Pg 22
Work - EnergyA 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.
xv1
m1
m1
Pg 23
Work - Energy
x1v1
m1
m1
Use the fact that WNET = K.
so kx2 = mv2k
mvx 1
11=
In this caseWNET = WSPRING = -1/2 kx2 and K = -1/2 mv2
In the case of x1
Pg 24
Work - Energy
x2v2
m2
m2
If the initial speed of the box were doubled and its mass were halved, how far x2 would the spring compress ?
(a)(a)12
xx =12
x2x =(b)(b) 12x2x =(c)(c)
Pg 25
Work - Energy
x2v2
m2
m2
If the initial speed of the box were doubled and its mass were halved, how far x2 would the spring compress ?
k
mvx = So if v2 = 2v1 and m2 = m1/2
k
2mv
k
2mv2x 1
11
12==
12x2x =
Pg 26
ExampleA 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:
The work is
The work to compress the spring is the same since W is proportional to x 2.
€
k =Fx
=75N
0.03m= 2.5 ×103 N / m
€
W =12
kxmax2 =
12
2.5 ×103 N / m( ) 0.03m( )2 = 1.1J
Pg 27
Example: Compressed SpringA 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?
mg x=0
Fs=kx
N
x=-11
Pg 28
Example: Compressed SpringThe work done to stretch or compress the spring is:
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:
( )( ) J18.2m11.0m/N3602
1kx
2
1W 22 ===
€
Wnet = K f − Ki =12
mv2 − 0 ΔK =12
mv2
€
v =2(ΔK )
m=
2 2.18J( )1.85Kg
= 1.54m / s