Transcript Review Questions - Review so Far

Physics 12 – Review so Far Day
1. Please form a group of 2-3, no more no less
2. Collect whiteboards, pens, and erasers
3. Answer the following questions with your group
4. Make sure to be an active participant in your group!
Q1: Work
A pulling force is applied to a crate at an angle of 25o above the
horizontal. The crate is dragged across the deck a distance of 2.5 m.
If the amount of work done after it has been moved is 1 210 J, what
was the applied force?
Q1: Work
A pulling force is applied to a crate at an angle of 25o above the
horizontal. The crate is dragged across the deck a distance of 2.5 m.
If the amount of work done after it has been moved is 1 210 J, what
was the applied force?
This is a simple problem, all we have to do is solve it for F!
W = FDd cosq
F=
W
d cosq
=
1 210 Nm
(
o
2.5 m cos25
)
=
530 N
Q2. Work and Kinetic Energy
A net 6 500 N force is applied to a resting 1 500 kg car, moving it
forward. What is its kinetic energy and speed after being displaced
150 m?
Q2. Work and Kinetic Energy
A net 6 500 N force is applied to a resting 1 500 kg car, moving it
forward. What is its kinetic energy and speed after being displaced
150 m?
First we calculate the amount of work the force does, this will give us
the amount of kinetic energy the car ends up with.
(
W = FDd = 6500 N 150 m
1
W = mv 2
2
v=
2W
m
v=
)
= 975 000 J
æ
kgm ö
2 ç 975 000 2 m÷
è
ø
s
1 500 kg
=
36
m
s
Q3. Conservation of Energy I
The first hill on a roller coaster is 94 m tall, the second hill is 68 m tall.
If it starts from rest on the first hill, what theoretical speed will the roller
coaster have on the second hill?
Q3. Conservation of Energy I
The first hill on a roller coaster is 94 m tall, the second hill is 68 m tall.
If it starts from rest on the first hill, what theoretical speed will the roller
coaster have on the second hill?
1 2
1 2
mgyo + mvo = mgy + mv
2
2
1 2
g y0 = g y + v
2
v=
1 2
gy0 - g y = v
2
æ
mö
2 ç 9.8 2 ÷ 94 m - 68 m
è
s ø
(
)
=
v=
23
(
2g y0 - y
m
s
)
Q4. Conservation of Energy II
A sled and rider together
have a mass of 87 kg. They
are atop a hill elevated at
42.5o. They slide down the
slope a distance of 35 m and
reach the bottom. Find the
speed at the bottom of hill.
Assume no friction.
d
y

Q4. Conservation of Energy II
A sled and rider together
have a mass of 87 kg. They
are atop a hill elevated at
42.5o. They slide down the
slope a distance of 35 m and
reach the bottom. Find the
speed at the bottom of hill.
Assume no friction.
v=
22
m
s
d
y

Q5. Power Basics I
A human fly climbs up the outside of tall building to thrill the teeming
hordes of earthlings below who fear he will fall to his doom. So if
the 52 kg human fly takes 18 minutes to climb a 350 m building,
how much power did he develop in the climb?
Q5. Power Basics I
A human fly climbs up the outside of tall building to thrill the teeming
hordes of earthlings below who fear he will fall to his doom. So if
the 52 kg human fly takes 18 minutes to climb a 350 m building,
how much power did he develop in the climb?
W
P=
t
mgy
=
t
æ
æ 1 ö æ 1 min ö
mö
= 52 kg ç 9.8 2 ÷ 350 m ç
è s ø
è 18 min ÷ø çè 60 s ÷ø
=
170 W
Q6. Power Basics II
A 47 kg bicycle rider develops 0.26 hp. She rides the Featherlite
250 which has a mass of 2.3 kg. Anyway, the rider must climb a
235 m hill. How much time will this take?
æ 746 W ö
P = 0.26 hp ç
÷
è 1 hp ø
= 194 W
Q6. Power Basics II
A 47 kg bicycle rider develops 0.26 hp. She rides the Featherlite
250 which has a mass of 2.3 kg. Anyway, the rider must climb a
235 m hill. How much time will this take?
æ 746 W ö
P = 0.26 hp ç
= 194 W
÷
è 1 hp ø
æ
mö
47
kg
+
2.3
kg
9.8
(
) çè s2 ÷ø ( 235 m)
J
t=
= 585.2
J
J
194
s
s
=
590 s
Q7. Power… a little more challenging
A 15.5 kg block is pulled across a flat deck at a constant speed of 3.0
m/s with a rope. The rope is horizontal to the deck. The coefficient of
friction is 0.330. How much power does it take to do this?
Q7. Power… a little more challenging
A 15.5 kg block is pulled across a flat deck at a constant speed of 3.0
m/s with a rope. The rope is horizontal to the deck. The coefficient of
friction is 0.330. How much power does it take to do this?
æ
mö
F f = T = m mg = 0.330 15.5 kg ç 9.8 2 ÷
è
s ø
æ mö
P = Fv = 50.1 N ç 3.0 ÷ = 150 W
sø
è
(
)
n
= 50.1 N
f
T
mg
Q8: Random!
A person pushes a 10.0 kg cart a distance of 20.0 meters by exerting
a 60.0 Newton horizontal force. The frictional resistance force is 50.0
Newtons. How much work is done by the person exerting a force on
the cart? How much kinetic energy does the cart have at the end of
the 20.0 meters if it started from rest:
W = Fapp d = (60.0N )(20.0m) = 1200J
DEk = Fnet d = (Fapp - Ff )d = (60.0N - 50.0N )(20.0m) = 200J
Q9: Random!
Assuming an efficiency of 25% for the muscle system in the process
of converting food energy into mechanical work, how much energy
would be used by a person of mass 75 kg (weight 165lb) in the
process of climbing four flights of stairs for a total height of 15
metes? Find the answer in joules and convert to dietary calories (1
dietary calorie = 4186 Joules)
Wout = DE = DE p = mgh = (75kg)(9.8m / s 2 )(15m) = 11025 J
Eff =
Wout
W
11025kJ
x100% = 25% ; Win = out =
= 44100J
Win
0.25
0.25
44100J x
1 cal
= 10.5 cal
4186J
The Sojourner rover