Calculate the net gravitational force on the
shaded ball. Be sure to include the
magnitude and direction. Each ball has a
mass of 20,000 kg.
Chapter Six
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.
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
Your lifting
force and
the box’s
direction
Frictional
Force
Fighting
against the
force of
gravity
Work and Direction
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
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 is the only part
that does work
This
part is
lost
1
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 tray of meals
across the room.
4. A rocket accelerates through space.
5. A rocket travels at a constant speed
through space
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?
How much work is
this man doing
carrying the cat.
Assume the cat
masses 8 kg.
How much work is done lifting a 5.0 kg
gnome to a shelf 2.0 meters above the
floor at a constant speed?
A waiter lifts a 6.00 kg platter 35.0 cm
from the kitchen.
a. Calculate the work done lifting the platter.
(20.6 J)
b. Calculate the work done carrying the
platter to the table.
c. Suppose they had put the platter on a
cart and pushed it to the table. They
used a force of 25.0 N for 10.0 m.
Calculate the work. (250 J)
d. What were they most likely working
against?
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.
a. Calculate the work was done
(96.5 J)
b. If the box masses 45.0 kg,
calculate the normal force
(418 N)
Fx = (30N)(cos50o) = 19 N
Fy = (30N)(sin50o) = 23 N
W = Fxd = (19N)(5m) = 96J
W = Fdcosq
Direction of motion
Fy
q=
50o
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.
a. Calculate the work done. (338 J)
b. Calculate the normal force (the sled
masses 20.0 kg) (105 N)
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
fk
n
mg
Let’s deal with the vertical forces first
Now the horizontal forces
W G = mgxcos90o = 0
W N = FNxcos90o = 0
W fr = Ffrd
W fr = (50 N)(40m) = -2000 J
No work is done in the vertical direction (the
box is not lifted)
W p = Fpdcos 37o
Wp = (100N)(40 m)(cos 37o) = 3200 J
W net = W p + W fr
W net = 3200 J -2000 J = 1200 J
3
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?
No work
in the horizontal
direction.
How much work did gravity do?
W G = -1470 J
W NET = W hiker + W G
W NET = 1470 J -1470 J = 0
The only forces are vertical
0 =Fv-mg
Fv = mg = (15.0 kg)(9.80 m/s2)
Fv = 147 N
W hiker = Fd = (147 N)(10.0 m)
W hiker = 1470 J
Does the Earth Do Work on the
Moon?
W = Fdcosq
W = Fd(cos 90o)
W = Fd(0)
W=0
English Unit of Work
Variable Force
• Foot-pound – English unit of work.
• Pound – unit of Force
• W = Fd = (foot*pound)
Work is really an area:
v
FR
W =∫Fdx
(an integral tells you the
area)
WORK
4
Given the following graph, determine the
work done by the following force in the first
ten meters
Calculate the work:
a) In the first 10 meters
b) In the second 10 meters
c) Overall
Calculate the work:
a) In the first 20 meters
b) From 20 to 40 meters
c) From 40 to 80 meters
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
d) Overall
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 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
fk
n
mg
5
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. (5378 J)
A 70.0 kg student is pulled by Mr. Fredericks with
a force of 500.0 N and an angle of 40.0o with the
ground for 75.0 m to the office. The floor has a
mk of 0.250.
a. Calculate the normal force on the student. (365 N)
b. Calculate the work done by friction. (- 6844 J)
c. Calculate the work done by the pull. (28,725 J)
d. Calculate the net work. (21,900 J)
Energy
Kinetic Energy
Energy – The capacity to do work
– 1 Joule = 1 Newton meter =1 kgm2/s2
– Two types
• Potential Energy
• Kinetic Energy
A cat 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.
• 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?
Deriving Kinetic Energy
W = Fd
W = mad
W = m(vf2 – vi2)d
2d
W = ½ mvf2 – ½ mvi2
W NET = Kf – Ki
W NET = DK
#only if no change in elevation
6
General Form
+
How much work is required to accelerate a
1000-kg car from 20-m/s to 30-m/s?
(Fd)
W = DK + DU
- (fkd) ALWAYS MAKE FRICTION NEGATIVE
(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 140.0 g baseball is caught by a fielder.
The glove moves 25.0 cm. The fielder
experienced an 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.
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 2.50 kg box is given an initial speed of
2.00 m/s. If the mk between the box and
the ground is 0.300, calculate how far the
box will slide.
7
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
U = 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 = DK + DU
W = DU
W = mgyf – mgyi
How much work does a 50.0 kg woman do by
climbing to the top of a 300 m hill?
#only if no change in speed
8
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
Spring Example 1
k = spring constant (measure of the stiffness
of a spring)
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
U = ½ kx2
x = distance stretched from normal length
U = ½ kx2
U = ½ (300 N/m)(0.100 m)2
U = 1.5 N m = 1.5 J
9
When 10 J is used to stretch a spring, the
spring stretchs 20.0 cm. Calculate the
spring constant.
50 N of force is required to stretch a spring
by 0.50 m. Calculate the spring constant.
Calculate how much potential energy is
stored in the spring.
Conservative and
Nonconservative Forces
Conservative 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
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?
Law of Conservation of Mechanical
Energy
W = DK + DU
(assume no friction or
outside force)
0 = Kf - Ki + Uf - Ui
B
Ki + Ui = Kf + Uf
½ mv2i + mgyi = ½ mv2f + mgyf
A
10
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.
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
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)(2m)
v22 = 19.6 m2/s2
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 = 4.4 m/s
v2 = 6.3 m/s
11
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
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)
v22 = (9.8 m/s2)(3 m)(2)
40 m
v2 = 7.7 m/s
a) What is its speed at the bottom of the
hill?
b) At what height will it have half of that
speed?
½ mv21 + mgy1 = ½ mv22 + mgy2
0 + mgy1 = ½ mv22 + 0
gy1 = ½ v22
v2 = \/ 2gy1
= \/ (2)(9.8 m/s2)(40 m)
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)
v2 = 28 m/s
A narwhal on a water slide is at a height of
25.0 m, and assume it starts out with almost
no speed at the top.
a) Calculate the speed at
the bottom. (22.1 m/s)
b) Calculate the height at
which it will have 75% of
that speed. (10.9 m)
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)?
12
½ mv2gr + mgygr = ½ mv2top + mgytop
½ mv2gr + 0 = 0 + mgytop
½ mv2gr = mgytop
½ v2gr = gytop
vgr = \/2gytop
= (2 X 9.8m/s2 X 381 m)1/2
vgr = 86.4 m/s (193 mph)
• What if he fell?
• vmax = 120-130 mph.
• Equivalent to
jumping from about
150 meters (about
450 feet).
Springs: Example 1
Springs and Conservation of Energy
Springs store potential energy
W = DK + DU + DUsp
(work is usually zero)
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 0.100 kg toy dart compresses a spring 6.0
cm. The spring’s constant is 250 N/m.
What speed will the dart leave the gun?
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)
13
Springs and Vertical Movement
• Car springs/struts
• Bungee Cord
• Make the zero the unstretched spring
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.
½ mv2i + mgyi + ½ ky2i = ½ mv2f + mgyf + ½ ky2f
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
55.0 cm
-15.0 cm
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.
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)
(2.28 cm)
14
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
A rock rolls down a road inclined 12o to the
horizontal. It rolls 2400 m along the road.
How fast is the rock going when it reaches
the bottom, striking a lazy physics student?
(322 N/m)
A 78-kg skydiver has a speed of 62 m/s at
an altitude of 870 m above the ground.
a. Determine the kinetic energy possessed
by the skydiver. [150,000 J]
b. Determine the potential energy
possessed by the skydiver. [665,000 J]
c. Determine what height she jumped from.
[1066 m]
d. Determine the speed when she is 500.0
m above the ground. [105 m/s]
A skier glides down a frictionless hill of 100
meters, then ascends another hill, of height
90 meters. What is the speed of the skier
when it reaches the top of the second hill?
[14 m/s]
A child’s toy shoots 2.70 g ping pong balls.
When a ball is loaded into the tube, it
compresses the spring (k = 18 N/m). by
9.5 cm. If you shoot a ping pong ball
straight up out of this toy, how high will it
go?
[2.97 m]
15
Law of Conservation of Mechanical
Energy
If nonconservative forces act, use:
W fr = DK + DU
(W fr is negative)
Ki + Ui = Kf + Uf + W fr
½ mv2i + mgyi = ½ mv2f + mgyf + Ffrd
Starting energy
Some of the starting
energy lost to friction
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
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 left side of the slide.
3.5 m
6.0 m
To calculate the coefficient of friction:
Ffr = mmgcosq (mgcosq is the Normal Force)
tan q = 3.5/6.0
q = 35.7o
m = Ffr/mgcosq
m = 241 N/[(100 kg X 9.8 m/s2)(cos 35.7o)]
m = 0.303
16
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
[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
17
Power
Power = Work
time
P=W
t
• Metric Unit: Joules/s = Watt.
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?
A donkey performs 15,000 J of
work pulling a wagon for 20 s.
What is the donkey’s power?
An eagle snatches a 5.00 kg
rabbit from the ground and
flies up 10 m with it in 0.75
seconds, what is the eagle’s
power?
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
First we need to find the height they rise:
y
50 m
y = 5.23 m
y
W 60 = mgy = (60.0 kg)(9.8m/s2)(5.23 m)
W 60 = 3075 J
sin 6.0o =
W 100 = mgy = (100.0 kg)(9.8m/s2)(5.23 m)
W 100 = 5125 J
Remember that work is a change in energy
W = DU
W = Uf – Ui
W = mgyf – 0
18
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
Horsepower
• The English Unit of power is horsepower
A 20.0 kg child in a 30.0 kg wagon is pushed
from rest. 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: Example 1
How much horsepower is required to power
a 100 Watt lightbulb?
• Foot-lb = Horsepower (hp)
second
• 1 hp = 746 Watts
• 1 hp = ½ Columbus (who sailed in 1492)
Horsepower
Consider a 100 hp car engine that can go
from 0 to 60 mi/hr in 20 seconds.
Horsepower: Example 2
Suppose you have a 1.34 hp lawnmower.
How many Watts of power does it have?
A 400 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.
19
Mr. McKeown carries a 200 N box of
Halloween costumes 5 meters upstairs in
3 seconds. What is his 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?
You 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 3000 kWh for
a given month. How many Joules of
energy did you use that month? (Hint:
Start from the power formula)
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)
P = Fv
Calculate the power required to travel on a
flat road against a drag force of 700 N, at a
speed of 22 m/s.
20
Calculate the drag force if a car has 200 hp
and travels at 25.0 m/s.
Calculate the force required to climb a 10.0o
incline. Assume an air drag of 700 N and a
speed of 22 m/s. However, include the force
of gravity. The car has a mass of 1400. kg.
Also calculate the power.
(3082 N, 67,804 W)
36.0 g
Formula Wrap-Up
W = Fd
+ (Fd)
W = DK + DU (+ DUsp )
- (fkd)
P = W/t
P = Fv (moving against air drag)
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
21
52.
54.
58.
60.
62.
74.
12 m/s
0.31
25.5 s
0.134 hp
21 kW
8.0 m/s
g(2h – L) = v2/2
2g(2X0.8 – L) = v2
2g(1.6 – L) = v2
22
23
24
The value of the acceleration due to gravity was
determined and compared to the accepted value of 9.8
m/s2. Instantaneous velocities were determined using
a photogate timer, and a velocity-time graph was
generated. The slope of the graph was the
experimental value, 9.4 m/s2. The lab was accurate
with a 4.0 % error.
25