Download Joules (N m)

Work and Energy
Work = Force X distance
W = Fd
• Unit - Joules (N m), (kg m2/s2)
• Force must be in the direction of the
motion.
Work and Direction
Lifting a box is work
• Fighting force of
gravity.
• Lifting force in same
direction box
moves.
Your lifting
force and
the boxes
direction
Fighting
against the
force of
gravity
Work and Direction
Pushing a box is work
• Applied force is in the
direction of movement
• Working against force of
friction
• Pushing on ice would be
less work
Frictional
Force
Work and Direction
Carrying a box is not work
Your lifting
force
But this
is the
direction
of
motion
Fighting
against the
force of
gravity
Work and Direction
Mowing the lawn is some
work
• only do work in horizontal
direction.
• The downward push,
because of the angle of
the handle, doesn’t do any
work.
Your
push
This
part is
lost
This is the only part
that does work
Vertical vs. Horizontal Work
Lifting
Gives an object
Potential Energy of
position.
Pushing
Gives an object
Kinetic Energy of
Motion.
Work or Not
1. A teacher pushes against a wall until he
is exhausted.
2. A book falls off the table and falls freely
to the ground.
3. A waiter carried a full try of meals across
the room.
4. A rocket accelerates through space.
5. A rocket travels at a constant speed
through space
Work: Example 1
How much work is done lifting a 5.0 kg
backpack to a shelf 2.0 meters above
the floor?
Superman does 36,750 J of work lifting this
car 2.5m from the ground? What is the
weight of the car? What is the mass of the
car?
Work: Example 2
Mr. Fredericks pulls a
box with 30 N of Force
a distance of 5.0 m, at
an angle of 50o with the
ground. How much
work was done?
Direction of motion
q = 50o
Fx = (30N)(cos50o) = 19 N
Fy = (30N)(sin50o) = 23 N
W = Fxd = (19N)(5m) = 96J
W = Fdcosq
Fy
q = 50o
Fx
A sled is pulled with a force of 100.0 N at an
angle of 65o to the ground for 8.00 m.
Calculate the work done.
(338 J)
Work: Example 3
A 50-kg crate is pulled 40 m with a force of
100 N at an angle of 37o. The floor is
rough and exerts a frictional force of 50 N.
Determine the work done on the crate by
each force and the net work done on the
crate.
Fp
q
Ffr
FN
mg
Let’s deal with the vertical forces first
WG = mgxcos90o = 0
WN = FNxcos90o = 0
No work is done in the vertical direction (the
box is not lifted)
Now the horizontal forces
Wfr = Ffrd
Wfr = (50 N)(40m) = -2000 J
Wp = Fpdcos 37o
Wp = (100N)(40 m)(cos 37o) = 3200 J
Wnet = WG + Wfr
Wnet = 0 + 0 + 3200 J -2000 J = 1200 J
Example 4
A hiker walks up a hill at a constant speed.
He is carrying a 15.0 kg backpack and the
hill is 10.0 m high. How much work did he
do?
Surprisingly, he
does no work
in the horizontal
direction.
The only forces are vertical
SF = FH-mg
ma = FH-mg (no acceleration)
0 =FH-mg
FH = mg = (15.0 kg)(9.80 m/s2)
FH = 147 N
Whiker = Fd = (147 N)(10.0 m)
Whiker = 1470 J
How much work did gravity do?
WG = -1470 J
WNET = Whiker + WG
WNET = 1470 J -1470 J = 0
Does the Earth Do Work on the
Moon?
W = Fdcosq
W = Fd(cos 90o)
W = Fd(0)
W=0
v
FR
English Unit of Work
• Foot-pound – English unit of work.
• Pound – unit of Force
• W = Fd = (foot*pound)
Variable Force
Work is really an area:
W =∫Fdx
(an integral tells you the
area)
WORK
Given the following graph, determine the
work done by the following force in the first
ten meters
Distance (m)
Calculate the work:
a) In the first 10 meters
b) In the second 10 meters
c) Overall
Distance (m)
Calculate the work:
a) In the first 20 meters
b) From 20 to 40 meters
c) From 40 to 80 meters
d) Overall
Calculate the work:
a) In the first 10 meters
b) In the second 10 meters
c) Overall
Calculate the work:
a) In the first 10 meters
b) In the second 10 meters
c) Overall
Distance (m)
A 35.0 kg suitcase is pulled 20.0 m with a force of
100.0 N at an angle of 45.0o. The floor is has a mk
of 0.200.
a. Calculate the normal force on the suitcase (HINT:
take the Fpy into account). (272 N)
b. Calculate the work done by friction. (-1088 J)
c. Calculate the work done by the pull. (1414 J)
d. Calculate the net work on the suitcase. (326 J)
Fp
q
Ffr
FN
mg
A 50.0 kg child and sled is pulled by a rope
with a tension of 200.0 N and an angle of
35.0o with the ground for 50.0 m. The sled
and snow has a mk of 0.150.
a. Calculate the normal force on the sled. (375 N)
b. Calculate the work done by friction. (- 2813 J)
c. Calculate the work done by the pull. (8191 J)
d. Calculate the net work on the sled. (5379 J)
Energy
Energy – The capacity to do work
– 1 Joule = 1 Newton-meter =1 kg-m2/s2
– Two types
• Potential Energy
• Kinetic Energy
Kinetic Energy
• Definition - Energy of motion
• K = ½ mv2
– m = mass (kg)
– v = speed (m/s)
– Unit = Joules
Questions:
a. If the mass of an object is doubled, what
happens to KE?
b. If the speed is doubled, what happens to
KE?
A kitten is traveling with a kinetic energy of
10.0 J at 2.00 m/s
a. Calculate his mass
b. If Mr. Fredericks (56.0 kg) has the same
kinetic energy, calculate his speed.
Kinetic Energy
If Mr. Fredericks (56 kg), had the same
kinetic energy as the Thing in the previous
example, how fast would he move?
Kinetic Energy and Work
WNET = KEf – KEI
WNET = DKE
How much work is required to accelerate a
1000-kg car from 20-m/s to 30-m/s?
(W = 250,000 J)
A 1000 kg car stops over 50.0 m. The
coefficient of friction is 0.25.
a) Calculate the force of friction. (-2450 N)
b) Calculate the work done by friction.
(-1.225 X 105 J)
c) Calculate the initial speed of the car. (15.7
m/s)
A 400.0 g football travelling at 25.0 m/s is
caught by a player. The players arms go
back 75.0 cm while catching the ball.
a) Calculate the initial kinetic energy of the
ball. (125 J)
b) Calculate the force the ball exerted on the
player’s hand. (167 N)
A 140.0 g baseball is caught by a fielder.
The glove moves 25.0 cm. The fielder
experienced and average force of 204
N.
a. Calculate the kinetic energy of the ball
before being caught. (51 J)
b. Calculate the initial speed of the ball.
(27.0 m/s)
c. Draw a free body diagram of the ball in
the air.
d. Draw a free body diagram of the ball
while being caught by the glove.
Potential Energy
• Definition - Stored energy
• Two Types
– Energy of position
• Gravitational PE - stored by placing something at a
height above the ground
• Mechanical PE - compressing a spring or
mechanical device
– Energy stored in chemical bonds and atomic
nuclei
Potential Energy of Position
PE = mgy
m = mass in kg
g = acceleration of gravity (9.8 m/s2)
y = height from the ground in meters.
Potential Energy of Position
.What would the potential energy of a ball be
at the other steps?
45 J
15 J
Potential Energy and Work
W = DPE
W = mgy
How much work does a 50.0 kg woman do by
climbing to the top of a 300 m hill?
Assume the roller coaster car has a mass
of 1000 kg. Calculate the PE:
a) At pts. A, B, and C
b) Gained from A B
c) Lost B  A
d) Lost B  C
Hooke’s Law
Fx = kx
F = Force exerted on the spring
k = spring constant
x = distance compressed or stretched.
A force of 600 Newtons will compress a
spring 0.5 meters.
a) Calculate the spring constant of the
spring. (1200 N/m)
b) Calculate the force is necessary to
stretch the spring by 2 meters. (2400 N)
c) A force of 40 Newtons will stretch a
different spring 0.1 meter. How far will a
force of 80 Newtons stretch it? (0.2 m)
Springs
PE = ½ kx2
k = spring constant (measure of the stiffness
of a spring)
x = distance stretched from normal length
Spring Example 1
How much potential energy is stored in a
spring if it is compressed 10.0 cm from its
normal length. Assume it has a spring
constant of 300 N/m
PE = ½ kx2
PE = ½ (300 N/m)(0.100 m)2
PE = 1.5 N-m = 1.5 J
When 10 J is used to stretch a spring, the
spring stretchs 20.0 cm. Calculate the
spring constant.
Conservative and
Nonconservative Forces
Conservative Forces
– Work is independent of the path taken
– Gravity, elastic spring
Nonconservative Forces
– Work depends on the path taken
– Friction, air resistance, tension in a cord
– Also called dissipative forces
Conservative Forces
6 floors
4 floors
Imagine carrying a box up
6 floors, then down 2.
Overall, you only
increased the PE by 4
floors (gravity is
conservative).
What if you went up 8 and
down 4?
Nonconservative Forces
Will it take more work to push the box from 1
to 2 on path A or path B? Or are they the
same?
B
A
Law of Conservation of Mechanical
Energy
KE1 + PE1 = KE2 + PE2
½ mv21 + mgy1 = ½ mv22 + mgy2
(assumes no friction)
Mr. Fredericks (56.0 kg) climbs up a 2.50 m
slide. Calculate the speed at which he will
leave the slide at the bottom. Neglect
friction.
2.50 m
Suppose Mr. Fredericks wanted to leave the
slide with a speed of 10.0 m/s. Calculate
the height of the slide without using the
mass. Neglect friction.
Law of Cons. Energy: Ex 1
Mr. Fredericks (100 kg)
jumps from a 3 meter
cliff. Calculate his
velocity when he is:
a) 2 m above the ground
b) 1 m above the ground
c) Just before he hits (0 m)
a) 2 m above the ground
½ mv21 + mgy1 = ½ mv22 + mgy2
0 + mgy1 = ½ mv22 + mgy2
mgy1 = m(½v22 + gy2)
gy1 = (½v22 + gy2)
(9.8 m/s2)(3m) = ½v22 + (9.8 m/s2)(2m)
v22 = 19.6 m2/s2
v2 = 4.4 m/s
b) 1 m above the ground
½ mv21 + mgy1 = ½ mv22 + mgy2
0 + mgy1 = ½ mv22 + mgy2
mgy1 = m(½v22 + gy2)
gy1 = (½v22 + gy2)
(9.8 m/s2)(3m) = ½v22 + (9.8 m/s2)(1m)
v22 = 39.4 m2/s2
v2 = 6.3 m/s
c) 0 m above the ground (just before he hits)
½ mv21 + mgy1 = ½ mv22 + mgy2
0 + mgy1 = ½ mv22 + 0
mgy1 = ½ mv22
(All PE gets turned to
KE)
gy1 = ½v22
v22 = (9.8 m/s2)(3 m)
v2 = 7.7 m/s
Law of Cons. Energy: Ex 2
A rollercoaster starts at a height of 40 m.
a) What is its speed at the bottom of the
hill? (28.0 m/s)
b) At what height will it have half of that
speed? (30.0 m)
40 m
a) What is its speed at the bottom of the
hill?
½ mv21 + mgy1 = ½ mv22 + mgy2
0 + mgy1 = ½ mv22 + 0
gy1 = ½ v22
v2 = \/ 2gy1
= \/ (2)(9.8 m/s2)(40 m)
v2 = 28 m/s
b) At what height will it have half of that
speed?
½ mv21 + mgy1 = ½ mv22 + mgy2
0 + mgy1 = ½ mv22 + mgy2
gy1 = ½ v22 + gy2
(9.8 m/s2)(40 m) = ½ (14 m/s)2 + (9.8 m/s2)y2
y2 = 30 m (above the low point)
Law of Cons. Energy: Ex 3
Superman is supposed to
be able to leap tall
buildings in a single
bound. What initial
velocity would he need
to jump to the top of the
Empire State Building
(381 m)?
½ mv2gr + mgygr = ½ mv2top + mgytop
½ mv2gr + 0 = 0 + mgytop
½ mv2gr = mgytop
½ v2gr = gytop
vgr = \/2gytop
vgr = 86.4 m/s
= (2 X 9.8m/s2 X 381 m)1/2
So how fast is that in mph?
Would an elephant have to jump faster?
We can ask if the
reverse is true: if
King Kong falls off
the Empire State
Building, will he hit
at 193 mph?
• Air drag, vmax = 120-130 mph.
• Equivalent to jumping from about 150
meters (about 450 feet). Over 150 meters
is academic.
Springs and Conservation of
Energy
• Springs store potential energy
½ mv21 + ½ kx21 = ½ mv22 + ½ kx22
• Use only for horizontal problems
Springs: Example 1
A 0.100 kg toy dart is compresses a spring
6.0 cm. The spring’s constant is 250 N/m.
What speed will the dart leave the gun?
½ mv21 + ½ kx21 = ½ mv22 + ½ kx22
0 + ½ kx21 = ½ mv22 + 0
½ kx21 = ½ mv22
kx21 = mv22
v22 = kx21/m =(250 N/m)(0.060m)2/(0.100 kg)
v22 = 9.0 m/s2
v2 = 3.0 m/s
A 1.50 kg block slides along a smooth table
and collides with a spring (k = 1000 N/m). If
block had an initial speed of 1.20 m/s, how
far was the spring compressed?
(4.65 cm)
Springs and Vertical Movement
• Car springs/struts
• Bungee Cord
• Make the zero the unstretched spring
½ mv21 + mgy1 + ½ ky21 = ½ mv22 + mgy2 + ½ ky22
A 2.60 kg ball is dropped. It falls 55.0 cm
before hitting a spring. It compresses the
spring 15.0 cm before coming to rest.
Calculate the spring constant.
55.0 cm
-15.0 cm
½ mv21 + mgy1 + ½ ky21 = ½ mv22 + mgy2 + ½ ky22
0 + mgy1 + 0 = 0 + mgy2 + ½ ky22
mgy1 = mgy2 + ½ ky22
(2.60 kg)(9.8m/s2)(0.550m) =
(2.60 kg)(9.8m/s2)(-0.150m)+ ½ k(-0.150m)2
14.0 J = -3.82 J + (0.01125m2)(k)
17.82 J = (0.01125m2)(k)
k = 1580 N/m
A vertical spring has a spring constant of 450
N/m and is mounted on the floor. A 0.30 kg
block is dropped from rest and compresses
the spring by 2.50 cm. Calculate the height
from which the block was dropped. Use the
unstretched spring length as your zero point.
(2.28 cm)
A 300 g block is dropped from rest 40.0 cm
above a spring. It compresses the spring by
9.50 cm. Calculate the spring constant
(322 N/m)
A vertical spring has a spring constant of
895 N/m is compressed by 15.0 cm.
a) Calculate the upward speed can it give to
a 0.360 kg ball when released. (7.28 m/s)
b) Calculate how high above the top of the
uncompressed spring the ball will fly. (2.70 m)
Law of Conservation of Mechanical
Energy
If nonconservative forces act, use:
KE1 + PE1 = KE2 + PE2 + Wfr
½ mv21 + mgy1 = ½ mv22 + mgy2 + Ffrd
Starting energy
Some of the starting
energy lost to friction
Mr. Fredericks (100 kg) slides
down a 3.5 m tall slide. He
leaves the slide at a speed
of 6.3 m/s.
a) Calculate the Force of
friction. (241 N)
b) Calculate the coefficient of
friction for the slide.(0.303)
c) Would there be more or
less friction if you raise the
3.5 m
left side of the slide.
6.0 m
½ mv21 + mgy1 = ½ mv22 + mgy2 + Ffrd
0 + mgy1 = ½ mv22 + 0 + Ffrd
mgy1 = ½ mv22 + Ffrd
Ffrd = mgy1 - ½ mv22
Ffr = (mgy1 - ½ mv22)/d
Ffr = [(100 X 9.8 X 3.5) – (½ X 100 X 6.32)]/6.0
Ffr = 241 N
To calculate the coefficient of friction:
Ffr = mmgcosq (mgcosq is the Normal Force)
tan q = 3.5/6.0
q = 30.3o
m = Ffr/mgcosq
m = 241 N/[(100 kg X 9.8 m/s2)(cos 30.3o)]
m = 0.28
Friction: Example 2
A 1000 kg roller coaster starts from a height of
40 m. On the second hill, it only rises to a
height of 25 m. Calculate the force of
friction. Assume a travel distance of 400 m.
½ mv21 + mgy1 = ½ mv22 + mgy2 + Ffrd
0 + mgy1 = 0 + mgy2 + Ffrd
mgy1 = mgy2 + Ffrd
Ffr = (mgy1 - mgy2)/d
Ffr = (1000 X 9.8 X 40 – 1000 X 9.8 X 25)/400
Ffr =368 N
A delivery boy wishes to slide a 2.00 kg package
up a 3.00 m long ramp. The ramp makes a 20o
angle with the ground, and has a coefficient of
friction of 0.40.
a) Calculate the height of the ramp (1.03 m)
b) Calculate the normal force on the package
(18.4 N)
c) Calculate the minimum speed the package
must have to go up the ramp. Remember to
take Potential Energy and the energy lost to
friction into account. (6.5 m/s)
d) Draw free-body diagrams for the box at the
bottom and in the middle of the ramp.
Mr. Fredericks (56.0 kg) slides down a 2.50 m
tall slide. The slide has a coefficient of 0.250
and makes a 45.0o angle with the ground.
a) Calculate the length of the slide. (3.53 m)
b) Calculate the normal force. (388 N)
c) Calculate the force of friction. (97.0 N)
d) Calculate how fast he will leave the slide.
(6.06 m/s)
2.50 m
Power
• Definition – rate at which work is done
– A powerful engine can do a lot of work
quickly.
– Running and walking up the steps require the
same amount of work.
– Running up steps requires more Power
Power
Power = Work
time
P=W
t
• Metric Unit: Joules/s = Watt.
Power: Example 1
A donkey performs 15,000
J of work pulling a
wagon for 20 s. What is
the donkey’s power?
Power: Example 2
If the Angel
snatches
Juggernaut’s
helmet (10 kg) and
flies up 10 m with
it in 0.75 seconds,
what is the
Angel’s power?
Suppose Mr. Fredericks (56.0 kg) can run up the
Empire State Building in 2.00 seconds.The
Empire State Building is 380 m tall.
a. Calculate the work done (2.09 X 105 J)
b. Calculate the power (1.05 X 105 W)
c. How could the power be increased?
Example 4
A 60-kg runner and a 100-kg runner run 50
m up a 6.0o slope. Calculate the work
done by both runners.
50 m
6.0o
y
First we need to find the height they rise:
sin 6.0o =
y
50 m
y = 5.23 m
Remember that work is a change in energy
W = DPE
W = PEf – PEi
W = mgy – 0
W60 = mgy = (60.0 kg)(9.8m/s2)(5.23 m)
W60 = 3075 J
W100 = mgy = (100.0 kg)(9.8m/s2)(5.23 m)
W100 = 5125 J
If it takes the 60 kg man 6 seconds, and the
100 kg man 10 seconds, who has more
power?
P = W/t
P60 = 3075 J/6 s = 513 W
P100 = 5125 J/10 s = 513 W
A 50.0 kg child/wagon is pushed from rest in
a wagon. The wagon leaves with a speed
of 6.00 m/s, and the person pushing has a
power of 100.0 Watts.
a) Calculate the work done on the child and
wagon. (900. J)
b) Calculate the time of the push. (9 s)
c) If the coefficient of kinetic friction is 0.25,
calculate how far the wagon will travel.
(7.35 m)
Horsepower
• The English Unit of power is horsepower
• Foot-lb = Horsepower (hp)
second
• 1 hp = 746 Watts
• 1 hp = ½ Columbus (who sailed in 1492)
Horsepower: Example 1
How much horsepower is required to power
a 100 Watt lightbulb?
Horsepower
Consider a 40 hp car engine that can go
from 0 to 60 mi/hr in 20 seconds.
A 160 hp car could go from zero to 60 mi/hr
in 5 seconds.
4 times as powerful means it can do the
same work in ¼ the time.
Horsepower: Example 2
Suppose you have a 1.34 hp lawnmower.
How many Watts of power does it have?
Horsepower: Example 3
A crane lifts a 200 N box 5 meters in 3
seconds. What is the crane’s power in
Watts and in horsepower?
You leave a 60 W lightbulb on outside your
house for 12 hours a day for an entire
month (30 days). How many kiloWatt
hours will you be charged for?
Decide to turn your computer (75 W) off
every night. Assume that it is off for 9
hours every night for 30 days. How many
kilowatt-hours did you save?
Suppose you are charged for 1500 kWh for
a given month. How many Joules of
energy did you use that month? (Hint:
Start from the power formula, and just use
one hour)
Suppose you are charged for 3000 kWh for
a given month. How many Joules of
energy did you use that month? (Hint:
Start from the power formula)
You push a 30.0 kg child up a slide that is 4.00 m long (hypotenuse)
and has an angle of 33.0o. The slide has a mk of 0.22.
a. Calculate the height of the slide. (2.18 m)
b. Calculate the initial speed that is needed to push the child to that
height. Include friction and be sure to calculate the normal force.
(7.56 m/s)
c. What percentage of the initial energy was lost to friction? (25.3%)
d. If you pushed for 1.3 s, what is your power in Watts? (659 W)
e. What is his power in hp? (0.884 hp)
2. a) 1.1 X 103 J b) 5.4 X 103 J
4. 25 N
6. 7.8 J
8. 170 J
10.a) 420 N b) -1800 J
c) -4100 J
d) 5900 J e) 0
12. 5000 J
18. Square root 2, 4
20. -4.67 X 105 J
22. 44 m/s
24. 2.25
26. -1.1 N
26.
30.
32.
34.
36.
38.
40.
42.
50.
52.
54.
-1.1 N
71 J
a) 45.3 J b) 12.3 J
c) 45.3 J
a) ½ k(x2-xo2) b) ½ kxo2 (Same)
49.5 m/s
6.5 m/s
vb = 24 m/s, vc = 9.9 m/s, vd = 19 m/s
a) 100 N/m
b) 22 m/s2
530 J
12 m/s
0.31
74. 8.0 m/s
58.
60.
62.
74.
25.5 s
0.134 hp
360 W
8.0 m/s