Transcript PowerPoint: Physics Word Problem Review

Fall Final Review
WKS: WORD PROBLEMS
d
Average Speed v 
t
1. A rock is dropped from the top of a tall cliff 9
meters above the ground. The ball falls freely and
hits the ground 1.5 seconds later. What is the
average speed of the rock?
d
Average Speed v 
t
1. A rock is dropped from the top of a tall cliff 9
meters above the ground. The ball falls freely and
hits the ground 1.5 seconds later. What is the
average speed of the rock?
d
v
t
9m
v
1 .5 s
v  6m / s
2. A car travels at a constant speed of 20 m/s for 5
seconds. How far did it travel?
2. A car travels at a constant speed of 20 m/s for 5
seconds. How far did it travel?
d
v
t
Rearrange to get d  vt
d  20m / s 5s 
d  100 m
3. A car travels 45 miles in 1 hour. What was the
car’s average speed?
3. A car travels 45 miles in 1 hour. What was the
car’s average speed?
d
v
t
45 miles
v
1hr
v  45miles / hr
Uniformly Accelerated Motion
v  v0  gt
4. You drop a rock off the top of a tall building and it
hits the ground 4 seconds later. What is the
velocity of the rock when it hits the ground?
Uniformly Accelerated Motion
v  v0  gt
4. You drop a rock off the top of a tall building and it
hits the ground 4 seconds later. What is the
velocity of the rock when it hits the ground?
v  v0  gt
time
final velocity
intial velocity
g =10
Uniformly Accelerated Motion
v  v0  gt
4. You drop a rock off the top of a tall building and it
hits the ground 4 seconds later. What is the
velocity of the rock when it hits the ground?
V0=0
v  v0  gt
t=4s
v  0  104s 
V=?
v  40m / s
Newton’s Second Law
F  ma
5. A 2.5 N object is sitting at rest on the table. What
is the force of the table acting on the object? What
is the net force?
Newton’s Second Law
F  ma
5. A 2.5 N object is sitting at rest on the table. What
is the force of the table acting on the object? What
is the net force?
Fg = -2.5N
2.5N
Support
force
-2.5N Force or
gravity
(weight)
2.5N + (-2.5N) = 0N
6. You are ice skating with a friend and he pushes
you across the frictionless ice with a force of 5
Newtons. If your mass is 50 kg, what will you
acceleration be?
6. You are ice skating with a friend and he pushes
you across the frictionless ice with a force of 5
Newtons. If your mass is 50 kg, what will you
acceleration be?
F
a
m
5N
a
50kg
a  0.1 m
s
2
7. If we disregard friction, what force would be
required to accelerate an 10 kg object at 4.0 m/s2 ?
7. If we disregard friction, what force would be
required to accelerate an 10 kg object at 4.0 m/s2 ?
F  ma

F  10kg  4.0m / s
F  40 N
2

Force due to gravity
Fg  mg
8. What is the weight of a 10 kg mass on Earth?
Force due to gravity
Fg  mg
8. What is the weight of a 10 kg mass on Earth?
Fg = weight
Fg  mg
Fg  10kg 10
Fg  100N
9. A rock has a weight of 98 N. What is the mass?
9. A rock has a weight of 98 N. What is the mass?
Fg = weight
Fg  mg
98N
m
10
Rearrange and get
m
m  9.8kg
Fg
g
10. Which has a greater mass. A 20 N object on the
moon, or a 20N object on the Earth? (why)
10. Which has a greater mass. A 20 N object on the
moon, or a 20N object on the Earth? (why)
20N object = Fg of the object
mmoon 
Fg
mearth 
Fg
g
g
mmoon
20 N

 12 kg
1.66
mearth
20 N

 2kg
10
Remember that value of g varies on the moon and
the Earth. Moon is only 1/6th that on Earth
11. You are pulling a 50 kg crate with a force of 5 N.
If there is 2 N of friction, what is the acceleration of
the crate?
11. You are pulling a 50 kg crate with a force of 5 N.
If there is 2 N of friction, what is the acceleration of
the crate?
First calculate the net force-
F
a
m
3N
a
50kg
5N – 2N = 3N (to the right)
a  0.06m / s
2
Work
W  Fd
Power
W
P
t
12. How much work is done pushing a 10N crate a
distance of 10 meters?
Work
W  Fd
Power
W
P
t
12. How much work is done pushing a 10N crate a
distance of 10 meters?
W  Fd
W  100 J
W  10N 10m
Remember that the
units for both work and
energy are Joules
Kinetic Energy
1
2
E  mv
2
Gravitational Potential Energy
E  m gh
14. What is the kinetic energy of a 5.0 kg rock
moving at 2 m/s?
Kinetic Energy
1
2
E  mv
2
14. What is the kinetic energy of a 5.0 kg rock
moving at 2 m/s?
1
2
E  5.0kg 2m / s 
2
1
2
E  mv
2
1
E  5.0 4 
2
Remember to start by
squaring velocity
E  10 J
15. A 5 kg rock falls off the top of a 5 meter tall roof.
What is the approximate kinetic energy of the rock
just before it hits the ground?
You are not given the velocity, so you
will need to use the equation for PE
understanding that it will all be
h=5m transformed into KE as it falls
E  m gh
E  250 J
E  5kg 105m
It starts with 250J of PE that
is converted to 250J of KE
16. What is the kinetic energy of a 20 Newton
object that is moving at 5 m/s?
16. What is the kinetic energy of a 20 Newton
object that is moving at 5 m/s?
You will first need to find out what the mass of the
object is using equation for Fg=mg
Fg  mg
rearrange
1
2
E  mv
2
m
Fg
g
20 N
m
 2kg
10
1
2
E  2kg 5m / s 
2
E  25 J
17. What is the gravitational potential energy
given to a 3.0 kg object that is lifted 2.0 meters off of
the ground?
17. What is the gravitational potential energy
given to a 3.0 kg object that is lifted 2.0 meters off of
the ground?
E  m gh
E  3kg 102.0m
E  60 J
Momentum
p  mv
18. What is the momentum of a 5.0 kg ball moving
at 6 m/s?
18. What is the momentum of a 5.0 kg ball moving
at 6 m/s?
p  mv
p  5.0kg 6m / s 
p  30kg  m / s
19. A 75 kg skier is moving down a hill with a
momentum of 100 kg·m/s. What is the velocity of
the skier?
19. A 75 kg skier is moving down a hill with a
momentum of 100 kg·m/s. What is the velocity of
the skier?
p  mv
Rearrange and solve for v
p
v
m
p
v
m

100kg  m / s 
v
75kg 
v  1.33m / s
20. What is the mass of an object that has a
momentum of 60 kg·m/s and a velocity of 10 m/s?
20. What is the mass of an object that has a
momentum of 60 kg·m/s and a velocity of 10 m/s?
p  mv
p
m
v
Rearrange and solve for m
p
m
v

60kg  m / s 
m
10m / s 
m  6kg
Collision in One Dimension
21. What is the final velocity of the coupled railroad
cars after the inelastic collision?
Collision in One Dimension
21. What is the final velocity of the coupled railroad
cars after the inelastic collision?
To solve these problems you
need to remember two
things:
•pbefore = pafter
•Use equation p=mv
Collision in One Dimension
21. What is the final velocity of the coupled railroad
cars after the inelastic collision?
Momentum before =
pcar1  mv pcar1  57  35
pcar 2  mv pcar 2  20  0
ptotal  35 0  35
Remember total pbefore = pafter
pafter  mv
35  5  2v
pafter =35
35
vafter   5m / s
7
22. What is the mass of the second (smaller sized)
box?
22. What is the mass of the second (smaller sized)
box?
To solve these problems you
need to remember two
things:
•pbefore = pafter
•Use equation p=mv
22. What is the mass of the second (smaller sized)
box?
Momentum before =
pblock1  mv
pblock1  26  12
pblock 2  mv
pblock 2  20  0
ptotal  12  0  12
Remember total pbefore = pafter
pafter  mv
12  m3
pafter =12
12
mtotal   4kg
3
Mass of second block: 4kg – 2kg = 2kg
23. What is the velocity of the gray ball after the
elastic collision below? Assume each ball has the
same mass.
23. What is the velocity of the gray ball after the
elastic collision below? Assume each ball has the
same mass.
4 m/s
In order for momentum to be conserved, the total
momentum that exists before will equal the total
after the collision.
If each ball has the same mass, the velocity of the
ball after collision must equal the velocity of the
ball before