1. Rank the work done by the force from greatest to least.

a) F=B >A > D > C >E c) E > C > A=D > B >F

b) F > B > A=D >C > E d) F > B > C > A=D > EB Tips: greatest is F Least is D That eliminates answer c)

2. If all of the blocks started at rest, rank the final speed from greatest to least

a) F=B >A > D > C >E c) E > C > A=D > B >F

b) F > B > A=D >C > E d) F > B > C > A=D > EB DK = Wnet, so greatest final speed is the one which has greatest WORK

3. Rank the inertia of each block from greatest to least

a) C=E >A > B > D > F c) C > E > A > B > D >F

b) F > B > A=D >C > E d) F > C=E > A > B > DA Tips: greatest is C and E Least is D Only one applies

4. Rank the normal force applied by the surface to each block from greatest to least

a) F=B > D > A > C > E c) D=A > F=B > C > E

b) E > C > A > F=B > D d) D > F=B > A > C > ED Tips: greatest is D Least is E eliminates a) and b)

5. A coin lies on the top of a turntable at a distance R from the center. The coefficient of static friction between the coin and turntable is ms. What is the maximum speed that the turntable can go before the coin will slide off?

a) c)

b) d)

A (Tips: can eliminate b and d based on wrong units)

gRsm

mgRsm

gR

)( ggRs m

gRvR

mvF

Ff

s

Ns

Cs

m

m

2

Draw a diagram and FBD to solve

6. At the moment of the car’s location its acceleration is?

a) c)

b) d)B

7. At the moment of the car’s location, its velocity is?

a) c)

b) d)C

9. Determine the acceleration of the box as it slides down the ramp.

a) b) c) d)

B Tips: can eliminate a) since it is accel down incline due only to gravity can eliminate d) since it is a force not an acceleration

2/67.3sin smg

agg

maFmg

mafF

maF

k

Nk

kgx

xx

m

m

cossin

sin

Draw a diagram and FBD to solve

2/01.2)(sin smg k m

2/13.2)cos(sin smg k m

2/153)cos(sin smmg k m

a) 5.60 mb) 8.10 m

c) 3.35 m

d) 15.2 m

A Tip: can eliminate b) and d) since without friction max height is 8.1m (EB=ET and all K at bottom transforms to Ug at top). Friction removes energy so that sled will not go as high and h < 8.1

mghmvhmg

mghmvdF

mghmvdf

EEW

Bk

BNk

Bk

TBkf

2

21

2

21

2

21

)16sin/)(16cos(

180cos

m

m

dh /16sin

a) 11 Nb) 25 N

c) 36 N

d) 42 N

Tip: can eliminate a) and b) since FT must be greater than mg if pendulum mass is to move in a circle

2

2

21

2 B

B

BT

vgh

mvmgh

EE

mLLh 585.040cos Isolated system- E is conserved

Object is moving in circle, so there is FC and that is provided by the tension and gravity

LghmmgL

mvmgF

FFF

BT

CgT

22

h

Fg

FT

C

D

a) UO = UA = UB c) UO - UA = 2(UA – UB)

b) UA - UB = 2UO d) UA - UB = UO - UA

This is about the potential differences between the points O, A, and B

Da) 3.4 s

b) 17.1 s

c) 10.6 s

d) 24.5 st

tmgh

tW

tEP chair

)10)(8.9)(100(400

D

If the chair is pulling up the hill at constant velocity, it is balancing gravity

A, the larger the vertical component of the initial velocity, the longer the time in air and therefore the higher.

Rank the balls from smallest to largest maximum height

a) C < B < A

b) A < B < C

c) A = B = C

d) A = C < B

Three soccer balls are kicked to the left at the same speed as shown. Ball A is kicked to the left at an angle of 60o above the horizontal, ball B is kicked 45o above the horizontal, and ball C is kicked 30o above the horizontal.

A, the larger the vertical component of the initial velocity, the longer the time in air and therefore the higher.

Rank the ball’s time in air from smallest to largest

a) C < B < A

b) A < B < C

c) A = B = C

d) A = C < B

Three soccer balls are kicked to the left at the same speed as shown. Ball A is kicked to the left at an angle of 60o above the horizontal, ball B is kicked 45o above the horizontal, and ball C is kicked 30o above the horizontal.

D, the range depends both on the time in air and the horizontal velocity. As the angle is increased, the time in air increases, but horizontal component of velocity decreases. Up until 45o, this tradeoff works to increase range; beyond 45o, range decreases. Complimentary launch angles have the same range. (This is only for projectiles launched over level ground)

Rank the ball’s range from smallest to largest

a) C < B < A

b) A < B < C

c) A = B = C

d) A = C < B

Three soccer balls are kicked to the left at the same speed as shown. Ball A is kicked to the left at an angle of 60o above the horizontal, ball B is kicked 45o above the horizontal, and ball C is kicked 30o above the horizontal.

C (only for projectiles launched over level ground)

Rank the ball’s impact speed from smallest to largest

a) C < B < A

b) A < B < C

c) A = B = C

d) A = C < B

Three soccer balls are kicked to the left at the same speed as shown. Ball A is kicked to the left at an angle of 60o above the horizontal, ball B is kicked 45o above the horizontal, and ball C is kicked 30o above the horizontal.

C

(Use g = 10 m/s2)

a) 1000 N

b) 600 N

c) 500 N

d) 400 N

)1000)(4.0()1000)(6.0(

Nkk

Nss

FfFf

mm

NfNf

k

s

400600

The 500 N force is not strong enough to overcome the max static friction and therefore the block remains stationary with the 500 N force being balanced by fs.

A

a) g/5

b) g/4

c) g/3

d) g

5/)4(

gaammmg

amF systemsystem

Fg

If we treat the two blocks, string and pulley as a system then the tension is an internal force (force acting between the parts of the system) and it does not contribute to the net force.

B

a) (1/5)mg

b) (4/5)mg

c) (1/3)mg

d) mg

5/44

mgFmaFx

T

Fg

To find tension, we cannot look at the whole system since tension is internal to the system. With the acceleration of the system known (and therefore we know acceleration of each block), pick just one block to find the force of tension. Best to pick the easier block to analyze. For the 4m block:

A block of mass 4m can move without friction on a horizontal table. This block is attached to another block of mass m by a string that passes over a frictionless pulley. If the masses of the string and pulley are negligible, what is the magnitude of the tension of the string?

B, work is area under F-x curvea) 80 J

b) 185 J

c) 275 J

d) 365 J

What is the work done on the object when it is moved from 0 to 9.0m?

Aa) 8.6 m/s2

b) 6.1 m/s2

c) 275 m/s2

d) 365 m/s2

If the object started from rest at the origin, what is its speed at 9.0 m

0185

221

D

f

net

mvKiKf

KW

D

a) about 20 m/s

b) about 10 m/s

c) about 5 m/s

d) 0 m/s

A ball is thrown straight up. At the top of its path its instantaneous speed is

B

a) about 20 m/s

b) about 10 m/s

c) about 5 m/s

d) 0 m/s

A ball is thrown straight up. At the top of its path its acceleration is

4

D, net force must be up

a) Less than 0

b) Between 0 and Mg

c) Equal to Mg

d) Greater than Mg

a) The car

b) The truck

c) Both will stop same distance

d) Cant tell without masses

C

mamgmaFmafmaF

k

Nk

k

xx

mm

The acceleration is independent of the mass

D

a) F/m

b) F/(2m)

c) F/(3m)

d) F/(4m))4/(

4mFa

maFamF systemsystem

Since the string is ideal, the blocks have the same acceleration and we can treat the two blocks and string as a system. The tension is an internal force (force acting between the parts of the system) and it does not contribute to the net force of the system.

What is the acceleration of the less massive box?

D

a) F

b) F/2

c) F/3

d) F/44/

4FF

mFmmaF

maF

T

T

What is the tension force in the string between the boxes?

To find tension, we use the acceleration found and look at just one of the blocks. Best to pick the easier block to analyze. For the m block:

B

a) 0.87g

b) 0.71g

c) 0.5g

d) g

agmamg

maFmaF

gx

xx

sinsin

Tip: can eliminate d) since it is not in free fall on the ramp

INCLINE

B

a) h=0; v=450m/s

b) h=450m; v=0m/s

c) h=450m; v=10m/s

00

vKT

mhJmghJU gT

45000,450000,450

C

a) K = 200,000J; h=20m; v=63.2m/sb) K = 0; h=20m; v=0m/sc) K = 250,000J; h=20m; v=70.7m/sd) K = 200,000J; h=25m; v=70.7m/s

smvmv

JK

L

L

T

/500250000000,250

221

mhJmghJU gT

20000,200000,200

JKJK

UKEEE

L

L

gLLloop

loopT

000,250000,200000,450

000,450000,450

Ba) ac = 714m/s2; Fc=71,400N; FN=71,400N

b) ac = 714m/s2; Fc=71,400N; FN=70,400N

c) ac = 714m/s2; Fc=71,400N; FN=72,400N

N

N

Ng

cC

C

FJFmgFF

NmaF

smrva

000,704714000

000,714

/1.714 22

Fg

FN