Download PowerPoint: Physics Word Problem Review Part 2

Fall Final Review
WKS: WORD PROBLEMS
Part II
1. A car travels at a constant speed of 15 m/s for 10
seconds. How far did it go?
1. A car travels at a constant speed of 15 m/s for 10
seconds. How far did it go?
d
v
t
d
v
t
rearrange
d  vt
d  15m / s 10s 
d  150m
2. How much time will it take to walk 100 meters at a
constant speed of 2 m/s?
2. How much time will it take to walk 100 meters at a
constant speed of 2 m/s?
d
v
t
d
v
t
100m
t
2m / s
rearrange
d
t
v
t  50s
3. You drop an object off the top of a 75 meter tall
building. It falls freely and hits the ground 7 seconds
later. What is the average speed of the object as it
fell?
3. You drop an object off the top of a 75 meter tall
building. It falls freely and hits the ground 7 seconds
later. What is the average speed of the object as it
fell?
d
v
t
75m
v
7s
v  10.7m / s
4. You drop a rock off the top of a tall cliff. How fast
is it traveling 3 seconds later?
4. You drop a rock off the top of a tall cliff. How fast
is it traveling 3 seconds later?
You are asked to calculate the final velocity (v)
Initial velocity: V0 = 0 since you “dropped” the rock
v  v0  gt
v  0  103s 
v  30m / s
5. You drop a 4.0 kg rock from a height of 20.0
meters above the ground. How long will it take to hit
the ground?
5. You drop a 4.0 kg rock from a height of 20.0
meters above the ground. How long will it take to hit
the ground?
In a free fall problem the mass does not change the
acceleration of an object when there is no air
resistance
t
2d
g
220m 
t
10
t  2s
t 4
6. According to the figure below, what is the
acceleration of the block?
6. According to the figure below, what is the
acceleration of the block?
First you need to calculate the net force (20 -10 =10)
F
a
m
10 N
a
20kg
a  0.5m / s
2
7. A net force of 10 Newtons is used to push a car
about 25 yds. The car’s acceleration is 1.0 m/s2.
What is the mass of the car?
7. A net force of 10 Newtons is used to push a car
about 25 yds. The car’s acceleration is 1.0 m/s2.
What is the mass of the car?
F  ma
rearrange
10 N
m
2
1.0m / s
F
m
a
m  10kg
8. You kick a 0.5 kg stationary ball with a force of 20
Newtons. What is the force on your foot?
8. You kick a 0.5 kg stationary ball with a force of 20
Newtons. What is the force on your foot?
Newton’s Third Law states that the forces are equal
and opposite in direction
Force would be 20N
9. If you have a weight of -850 Newtons, and are
standing on the ground. What is the force that the
Earth is pushing back up on you?
9. If you have a weight of -850 Newtons, and are
standing on the ground. What is the force that the
Earth is pushing back up on you?
850 N
-850 N
Remember that weight (force of gravity) is pointed
downward. We can represent down with a (-) sign
10. What happens to the gravitational attraction (the
force) between two objects when the distance
between them increases? decreases?
10. What happens to the gravitational attraction (the
force) between two objects when the distance
between them increases? decreases?
Newton’s Law of Universal Gravitation tells us the
relationship of distance and mass on the
gravitational force
m1m2
F G 2
d
Distance is on the bottom so it is inversely
proportional-distance gets greater, force gets
smaller
When distance decreases, force increases
11. When an object is moving in uniform circular
motion, which way is it accelerating?
11. When an object is moving in uniform circular
motion, which way is it accelerating?
Both acceleration and centripetal force are
pointed to the center of the circle
12. Each of the 3 objects below moves with a velocity
of 1m/s. What is the total kinetic energy of the
system?
12. Each of the 3 objects below moves with a velocity
of 1m/s. What is the total kinetic energy of the
system?
To calculate total kinetic energy you need to add up
the masses of the 3 objects.
1 2
E  mv
2
1
2
E  5kg 1m / s 
2
E  2.5J
13. What is the kinetic energy of an object with a
mass of 5 kg traveling at a speed of 10
meters/second? Assume no other forces act upon
the object.
13. What is the kinetic energy of an object with a
mass of 5 kg traveling at a speed of 10
meters/second? Assume no other forces act upon
the object.
1 2
E  mv
2
1
2
E  5kg 10m / s 
2
E  250 J
14. You lift a heavy rock with a mass of 15 kg to a
height of 5.0 meters above the ground. What is the
potential energy given to the rock?
14. You lift a heavy rock with a mass of 15 kg to a
height of 5.0 meters above the ground. What is the
potential energy given to the rock?
E  mgh
E  15kg105.0m
E  750 J
15. A 1.5 kg rock falls to the ground off the top of a
25 meter tall cliff. What is the Kinetic energy of the
rock right before it hits the ground?
15. A 1.5 kg rock falls to the ground off the top of a
25 meter tall cliff. What is the Kinetic energy of the
rock right before it hits the ground?
We do not have velocity to use in equation for KE.
Calculate the PE knowing it will all be transformed
into KE
E  mgh
E  1.5kg1025m
E  375J
16. A 3.5 kg object is moving with a velocity of 2.0
m/s. What is its momentum?
16. A 3.5 kg object is moving with a velocity of 2.0
m/s. What is its momentum?
p  mv
p  3.5kg2.0m / s 
p  7.0kg  m / s
17. The diagram below represents an inelastic
collision between two blocks on a frictionless air
track. After the collision the two objects “stick”
together. What is the mass of the second box?
17. The diagram below represents an inelastic
collision between two blocks on a frictionless air
track. After the collision the two objects “stick”
together. What is the mass of the second box?
Calculate the total momentum of both boxes
before the collision. It will have the same total
momentum after as well
pblock1  20kg20m / s   400
pblock1  mv
pblock 2  mv
pblock2  ?0  0
ptotal  400  0  400
17. The diagram below represents an inelastic
collision between two blocks on a frictionless air
track. After the collision the two objects “stick”
together. What is the mass of the second box?
Since the total momentum before = 400
Total momentum after = 400
pafter  mv
400  m15
400
mtotal 
 26.7kg
15
18. When the two objects of different mass collide
and couple together, what will be their resultant
velocity?
18. When the two objects of different mass collide
and couple together, what will be their resultant
velocity?
Calculate the total momentum of both boxes
before the collision. It will have the same total
momentum after as well
pblock1  205  100
pblock1  mv
pblock 2  mv
pblock2  50  0
ptotal  100  0  100
18. When the two objects of different mass collide
and couple together, what will be their resultant
velocity?
Since the total momentum before = 100
Total momentum after = 100
pafter  mv
100  15v
100
v
 6.7m / s
15