Transcript Work and Energy

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Define kinetic energy and potential energy,
along with the appropriate units in each
system.
Describe the relationship between work
and kinetic energy, and apply the WORKENERGY THEOREM.
Define and apply the concept of POWER,
along with the appropriate units.
Energy
Energy is anything that can be converted
into work; i.e., anything that can exert a
force through a distance.
Energy is the capability for doing work.
Potential Energy: Ability to do work by
virtue of position or condition.
A suspended weight
A stretched bow
Gravitational Potential Energy
What is the P.E. of a 50-kg
person at a height of 480 m?
U = mgh = (50 kg)(9.8 m/s2)(480 m)
U = 235 kJ
Kinetic Energy: Ability to do work by
virtue of motion. (Mass with velocity)
A speeding car
or a space rocket
What is the kinetic energy of a 5-g bullet
traveling at 200 m/s?
5g
200 m/s
K  mv  (0.005 kg)(200 m/s)
1
2
2
1
2
K = 100 J
What is the kinetic energy of a 1000-kg
car traveling at 14.1 m/s?
K  12 mv 2  12 (1000 kg)(14.1 m/s) 2
K = 99.4 J
2
A resultant force changes the velocity of an
object and does work on that object.
vo
m
vf
x
F
F
m
a
Work  Fx  (ma) x;
Work  mv  mv
1
2
2
f
1
2
v v
2
0
2
f
2
0
2x
Work is equal
to the change
in ½mv2
Work  mv  mv
1
2
2
f
1
2
2
0
If we define kinetic energy as ½mv2 then we
can state a very important physical principle:
The Work-Energy Theorem: The work
done by a resultant force is equal to the
change in kinetic energy that it produces.
Work = ½
0
mvf2 -
6 cm
80 m/s
½ mvo
F x = - ½ mvo2
x
2
F=?
F (0.06 m) cos 1800 = - ½ (0.02 kg)(80 m/s)2
F (0.06 m)(-1) = -64 J
F = 1067 N
Work to stop bullet = change in K.E. for bullet
Work = F(cos q) x
f = mk.n = mk mg
Work = - mk mg x
2 Work
-½ DK
mv=
o = -mk mg x
25 m
f
0 Work = DK
DK = ½ mvf2 - ½ mvo2
vo =
vo = 2(0.7)(9.8 m/s2)(25 m)
2mkgx
vo = 18.5 m/s
x
fn
h
mg
300
Plan: We must calculate both
the resultant work and the
net displacement x. Then the
velocity can be found from
the fact that Work = DK.
Resultant work = (Resultant force down the
plane) x (the displacement
down the plane)
f
x
n
h
mg
300
h
x
300
From trig, we know that the Sin 300 = h/x and:
h
sin 30 
x
0
20 m
x
 40 m
0
sin 30
Draw free-body diagram to find the resultant force:
f
n
mg cos
h
mg
f
x = 40 m
300
300
300
y
n
mg sin 300
x
mg
Wx = (4 kg)(9.8 m/s2)(sin 300) = 19.6 N
Wy = (4 kg)(9.8 m/s2)(cos 300) = 33.9 N
f
33.9 N
300
n
Resultant force down
plane: 19.6 N - f
19.6 N
Recall that fk = mk n
y
mg
x
SFy = 0 or
n = 33.9 N
Resultant Force = 19.6 N – mkn ; and mk = 0.2
Resultant Force = 19.6 N – (0.2)(33.9 N) = 12.8 N
Resultant Force Down Plane = 12.8 N
x
FR
300
(Work)R = FRx
Net Work = (12.8 N)(40 m)
Net Work = 512 J
Finally, we are able to apply the work-energy
theorem to find the final velocity:
0
Work  mv  mv
1
2
2
f
1
2
2
0
x
fn
Resultant Work = 512 J
Work done on block equals
the change in K. E. of block.
h
mg
300
0
½ mvf2 - ½ mvo2 = Work
½(4 kg)vf2 = 512 J
½ mvf2 = 512 J
vf = 16 m/s
Power is defined as the rate at which work is done:
(P = dW/dt )
t
F
Work F r
m
4s
10 kg
h
20 m
mg
Power 
time

t
mgr (10kg)(9.8m/s 2 )(20m)
P

t
4s
P  490 J/s or 490 watts (W)
Power of 1 W is work done at rate of 1 J/s
One watt (W) is work done at the rate of
one joule per second.
1 W = 1 J/s and 1 kW = 1000 W
One ft lb/s is an older (USCS) unit of power.
One horsepower is work done at the rate of
550 ft lb/s. ( 1 hp = 550 ft lb/s )
What power is consumed in lifting
a 70-kg robber 1.6 m in 0.50 s?
Fh mgh
P

t
t
2
(70 kg)(9.8 m/s )(1.6 m)
P
0.50 s
Power Consumed: P = 2220 W
Recognize that work is equal to
the change in kinetic energy:
Work  mv  mv
P
1
2
2
f
mv 2f
(100 kg)(30 m/s) 2

4s
t
1
2
2
0
Work
P
m = 100 kg
t
1
2
1
2
Power Consumed: P = 1.22 kW
Recall that average or constant velocity is
distance covered per unit of time v = x/t.
P=
Fx
t
=F
x
t
P  Fv
If power varies with time, then calculus is
needed to integrate over time. (Optional)
Since P = dW/dt:
Work   P(t ) dt
v = 4 m/s
P = F v = mg v
P = (900 kg)(9.8 m/s2)(4 m/s)
P = 35.3 kW
 550ft  lb/s 
4hp 
  2200ft  lb/s
 1hp

Work
P
; Work  Pt
t
Work  (2200ft  lb/s)(60s)
Work = 132,000 ft lb
Potential Energy: Ability to do work
U

mgh
by virtue of position or condition.
Kinetic Energy: Ability to do work by
K  12 mv 2
virtue of motion. (Mass with velocity)
The Work-Energy Theorem: The work done by
a resultant force is equal to the change in
kinetic energy that it produces.
Work = ½ mvf2 - ½ mvo2
Power is defined as the rate at which P  Work
t
work is done: (P = dW/dt )
Work F r
Power 

time
t
P= F v
Power of 1 W is work done at rate of 1 J/s