Fall Final Review WKS: WORD PROBLEMS. Average Speed 1. A rock is dropped from the top of a tall...

Fall Final ReviewWKS: WORD PROBLEMS

Average Speed

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?

t

dv

Average Speed

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?

t

dv

t

dv

s

mv

5.1

9 smv /6

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?

t

dv Rearrange to get vtd

ssmd 5/20 md 100

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?

t

dv

hr

milesv

1

45 hrmilesv /45

Uniformly Accelerated Motion

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?

gtvv 0



gtvv 0

gtvv 0final velocity

intial velocity

time

g =10



gtvv 0

gtvv 0

sv 4100

V0=0

V=?

t=4s

smv /40

Newton’s Second Law

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?

maF

Newton’s Second Law

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?

maF

Fg = -2.5N 2.5N

-2.5N Force or gravity (weight)

2.5N + (-2.5N) = 0N

Support force

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?

m

Fa

kg

Na

50

5 21.0

sma

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 ?

maF 2/0.410 smkgF

NF 40

Force due to gravity

8. What is the weight of a 10 kg mass on Earth?

mgFg

Force due to gravity

8. What is the weight of a 10 kg mass on Earth?

mgFg

Fg = weight mgFg

1010kgFg

NFg 100

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?

mgFg

Fg = weight

g

Fm gRearrange and get

10

98Nm kgm 8.9

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)

g

Fm gmoon

g

Fm gearth

kgN

mmoon 1266.1

20

kgN

mearth 210

20

Remember that value of g varies on the moon and the Earth. Moon is only 1/6th that on Earth

20N object = Fg of the object

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- 5N – 2N = 3N (to the right)

m

Fa

kg

Na

50

3 2/06.0 sma

Work

Power

12. How much work is done pushing a 10N crate a distance of 10 meters?

FdW

t

WP

Work

Power

12. How much work is done pushing a 10N crate a distance of 10 meters?

FdW

t

WP

FdW mNW 1010

JW 100Remember that the units for both work and energy are Joules

Kinetic Energy

Gravitational Potential Energy

14. What is the kinetic energy of a 5.0 kg rock moving at 2 m/s?

2

2

1mvE

mghE

Kinetic Energy

14. What is the kinetic energy of a 5.0 kg rock moving at 2 m/s?

2

2

1mvE

2

2

1mvE 2/20.5

2

1smkgE

40.52

1E Remember to start by

squaring velocity

JE 10

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 transformed into KE as it falls h=5m

mghE mkgE 5105

JE 250 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

mgFg g

Fm grearrange kg

Nm 2

10

20

2

2

1mvE 2/52

2

1smkgE

JE 25

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?

mghE mkgE 0.2103

JE 60

Momentum

18. What is the momentum of a 5.0 kg ball moving at 6 m/s?

mvp

18. What is the momentum of a 5.0 kg ball moving at 6 m/s?

mvp smkgp /60.5

smkgp /30

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?

mvp Rearrange and solve for v

m

pv

m

pv

kg

smkgv

75

/100

smv /33.1

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?

mvp Rearrange and solve for m

v

pm

v

pm

sm

smkgm

/10

/60

kgm 6

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


Momentum before =

mvpcar 1 35751 carp

mvpcar 2 0022 carp

35035 totalp

Remember total pbefore = pafter pafter =35

mvpafter v2535 smvafter /57

35

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


Momentum before =mvpblock 1 12621 blockp

mvpblock 2 0022 blockp

12012 totalpRemember total pbefore = pafter pafter =12

mvpafter 312 m kgmtotal 43

12

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.

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

4 m/s

Fall Final Review WKS: WORD PROBLEMS. Average Speed 1. A rock is dropped from the top of a tall...

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