Physics P
Review
4
Review 4 Energy and Its Conservation
Energy and Its Conservation
1.
A player pushes a 250–g hockey puck
over frictionless ice with a constant force,
causing it to accelerate at 24 m/s2 over a
distance of 50 cm.
a.
Find the work done by the hockey
player on the puck.
b.
What is the change in the kinetic
energy of the puck?
2.
You exert a horizontal force of 4.6 N on a
textbook as you slide it 0.6 m across a
library table to a friend. Calculate the
work you do on the book.
3.
An electric motor lifts an elevator at a
constant speed of 15 m/s. The engine must
exert a force of 9000 N in order to balance
the weight of the elevator and the friction
in the elevator cable. What power does the
motor produce in kW?
4.
Leah is helping to build a water habitat in
a neighborhood park. The habitat includes
an upper pond connected to a lower pond,
3.2 m below, by a trickling stream with
several small cascades. At a homebuilding store, she finds a 45–W pump
that has a maximum circulation rate of
1900 L of water per hour. Can the pump
develop enough power to raise the water
from the lower to the upper pond? (The
mass density of water, ρ, is 1.00 kg/L.)
5.
A gardener lifts a 25–kg bag of sand to a
height of 1.1 m, carries it across the yard a
distance of 15 m and sets it down against
the wall.
a.
How much work does the gardener
do when he lifts the bag of sand?
b.
How much total work is done after
the gardener sets down the bag of
sand?
6.
A 0.149–kg baseball is thrown at a speed
of 6.5 m/s. The batter hits the ball and it
flies into the outfield at a speed of
19.2 m/s. How much work is done on the
baseball?
7.
A 6–g block initially at rest is pulled to the
right along a frictionless horizontal
surface by a constant horizontal force of
1.2 × 10–2 N for a distance of 3 cm.
a.
What is the work done by the force?
b.
What is the change in the kinetic
energy of the block?
c.
What is the speed of the block after
the force is removed?
8.
Zeke slides down a snow hill on a rubber
mat. Zeke’s mass is 76 kg and the mass of
the mat is 2 kg. Zeke starts from rest at the
crest of the hill. You may ignore friction.
a.
What is the change in the
gravitational potential energy of
Zeke and the mat when they slide to
1.2 m below the crest?
b.
What is the change in the kinetic
energy of Zeke and the mat when
they slide to 1.2 m below the crest?
c.
How fast are Zeke and the mat
moving when they are 1.2 m below
the crest?
9.
Kuan stands on the edge of a building’s
roof, 12 m above the ground, and throws a
149–g baseball straight down. The ball
hits the ground at a speed of 18 m/s. What
was the initial speed of the ball?
10.
Meena releases her 10.5–kg toboggan
from rest on a hill. The toboggan glides
down the frictionless slope of the hill, and
at the bottom of the slope it moves along a
rough horizontal surface, which exerts a
constant frictional force on the toboggan.
a.
When the toboggan is released from
a height of 15 m, it travels 6 m
along the horizontal surface before
coming to rest. How much work
does the frictional force do on the
toboggan?
b.
From what height should the
toboggan be released so that it stops
after traveling 10 m on the
horizontal surface?
Physics P
1a.
Review 4 Energy and Its Conservation
W = Fd
W = mad
W = 0.25%kg 24%m/s2 0.5%m
)(
(
W = 3"J
1b.
2.
(
)(
(
)
7a.
)(
W = mgh
7c.
KE = 12 mv 2
(
) (
W = 270$J
5b.
W = ΔPEG + ΔKE
W =0
) (
)
)
)
ΔKE=0.000&36&J
v=
v=
2KE
m
(
2 0.000$36$J
)
0.006$kg
v = 0.35%m/s
)
8a.
ΔPE G = PEf − PE i
ΔPE G = mghf − mghi
(
)(
ΔPE G = 0 − 76$kg$+$2$kg g 1.2$m
ΔPE G = !920%J
8b.
W = 25#kg g 1.1#m
)(
W = ΔKE
Since the pond only requires 16.6 W the
45 W pump will suffice.
W = ΔPEG
(
7b.
)
W
P=
t
ΔPE G
P=
t
mgh
P=
t
1900$kg g 3.2$m
P=
3600$s
P = 16.6$W
) (
W = Fd
W = 0.012%N 0.03%m
W = 0.00036%J
P = 135$000$W$=$135$kW
5a.
)(
W = 24#J
W
t
Fd
P=
t
P = Fv
P = 9000#N 15#m/s
(
)
2
2
W = 12 0.149'kg ⎡⎢ 19.2'm/s − 6.5'm/s ⎤⎥
⎣
⎦
P=
(
4.
W = 12 mv f2 − 12 mv i2
W = 12 m v f2 − v i2
W = Fd
W = 4.6$N 0.6$m
(
W = ΔKE
W = KEf − KE i
)
W = ΔKE
ΔKE=3%J
W = 2.76%J
3.
)(
6.
ΔPEG + ΔKE = 0
ΔKE = −ΔPEG
ΔKE = 920$J
)
Physics P
8c.
Review 4 Energy and Its Conservation
KE = 12 mv 2
2KE
m
v=
(
2 920$J
v=
)
76$kg$+$2$kg
v = 4.8$m/s
9.
ΔPEG + ΔKE = 0
PE f − PE i + KE f − KE i = 0
0 − mghi + 12 mv f2 − 12 mv i2 = 0
− ghi + 12 v f2 − 12 v i2 = 0
v i = v f2 − 2ghi
(18#m/s) − 2g (12#m)
2
vi =
v i = 9.4$m/s
10a. The toboggan starts with PEG, that gets
converted completely into KE, which then
becomes heat as the toboggan slides to
rest.
W = ΔPEG
Fd = mgh
F=
F=
mgh
d
10.5%kg g 15%m
(
F = 260$N
mgh
d
Fd
h=
mg
) (
6%m
10b. F =
h=
(260$N)(10$m)
(10.5$kg ) g
h = 25#m
)