Transcript ppt Review 2015 ( Dorine Starace)

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 > E
B
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 > E
B
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 > D
A
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 > E
D
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) m s gR
c)
b) ms mgR
d) ( m s gR  g )
gR
Draw a diagram and FBD to solve
f s  FC
mv 2
m s FN 
R
v  m s gR
A (Tips: can eliminate b and d
based on wrong units)
6. At the moment of the car’s location its acceleration is?
a)
b)
c)
d)
B
7. At the moment of the car’s location, its velocity is?
a)
b)
c)
d)
C
9. Determine the acceleration of
the box as it slides down the
ramp.
a) g sin   3.67m / s 2
2
g
(sin


m
cos

)

2
.
13
m
/
s
b)
k
2
g
(sin


m
)

2
.
01
m
/
s
c)
k
2
mg
(sin


m
cos

)

153
m
/
s
k
Draw a diagram and FBD to solve d)
F
x
 max
Fgx  f k  ma
mg sin   m k FN  ma
g sin   m k g cos   a
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
a) 5.60 m
b) 8.10 m
c)
d)
3.35 m
15.2 m
sin 16  h / d
W f k  EB  ET
2
f k d cos 180  12 mvB  mgh
 m k FN d  mvB  mgh
1
2
2
 m k (mg cos 16)( h / sin 16)  mvB  mgh
1
2
A
2
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
a) 11 N
b) 25 N
c)
d)
FT
h
36 N
42 N
Fg
Isolated system- E is conserved
h  L  L cos 40  0.585m
ET  EB
mgh  12 mvB
2 gh  vB
2
2
Object is moving in circle, so
there is FC and that is provided
by the tension and gravity
FT  Fg  FC
FT  mg  mvB
2
L
 mg 
m2 gh
L
Tip: can eliminate a) and b) since FT must be greater than mg if pendulum
mass is to move in a circle
C
This is about the potential differences between the points O, A, and B
D
a) UO = UA = UB
c) UO - UA = 2(UA – UB)
b) UA - UB = 2UO
d) UA - UB = UO - UA
a) 3.4 s
b) 17.1 s
c) 10.6 s
d) 24.5 s
D
If the chair is pulling up the
hill at constant velocity, it is
balancing gravity
DE Wchair mgh


t
t
t
(100)(9.8)(10)
400 
t
P
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.
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
A, the larger the
vertical component of
the initial velocity, the
longer the time in air
and therefore the
higher.
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.
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
A, the larger the
vertical component of
the initial velocity, the
longer the time in air
and therefore the
higher.
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.
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
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)
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.
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
C
(only for projectiles launched over
level ground)
a) 1000 N
b) 600 N
c) 500 N
(Use g = 10 m/s2)
d) 400 N
f s  m s FN  (0.6)(1000)
f s  600 N
f k  mk FN  (0.4)(1000)
f k  400 N
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.
C
a) g/5
b) g/4
c) g/3
d) g
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.
F
system
 msystema
mg  (4m  m)a
a  g /5
Fg
A
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?
a) (1/5)mg
b) (4/5)mg
c) (1/3)mg
d) mg
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:
 Fx  4ma
FT  4mg / 5
Fg
B
What is the work done on the object when it is moved from 0 to 9.0m?
a) 80 J
b) 185 J
c) 275 J
d) 365 J
B, work is area under F-x curve
If the object started from rest at the origin, what is its speed at 9.0 m
a) 8.6 m/s2
b) 6.1 m/s2
c) 275 m/s2
d) 365 m/s2
A
Wnet  DK
185  Kf  Ki
2
 12 mv f  0
A ball is thrown straight up. At the top of its
path its instantaneous speed is
a) about 20 m/s
b) about 10 m/s
c) about 5 m/s
d) 0 m/s
D
A ball is thrown straight up. At the top of its
path its acceleration is
a) about 20 m/s
b) about 10 m/s
c) about 5 m/s
d) 0 m/s
B
4
a) Less than 0
b) Between 0 and Mg
c) Equal to Mg
d) Greater than Mg
D, net force must be up
C
a) The car
b) The truck
c) Both will stop same distance
d) Cant tell without masses
F
 max
 f k  ma
m k FN  ma
m k mg  ma
x
The acceleration is
independent of the mass
D
a) F/m
b) F/(2m)
c) F/(3m)
d) F/(4m)
What is the acceleration of
the less massive box?
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.
F
system
 msystema
F  4ma
a  F /( 4m)
D
What is the tension force in
the string between the
boxes?
a) F
b) F/2
c) F/3
d) F/4
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:
 F  ma
 F 
FT  ma  m

 4m 
FT  F / 4
a) 0.87g
b) 0.71g
c) 0.5g
d) g
INCLINE
F
 max
Fgx  ma
mg sin   ma
g sin   a
x
B
Tip: can eliminate d) since it is not in free fall on the ramp
a) h=0; v=450m/s
b) h=450m; v=0m/s
c) h=450m; v=10m/s
B
KT  0
v0
U gT  450,000 J
mgh  450,000 J
h  45m
ET  450,000  Eloop
Eloop  450,000  K L  U gL
450,000  K L  200,000 J
K L  250,000 J
KT  250,000 J
2
1
mv
L  250000
2
vL  500m / s
U gT  200,000 J
mgh  200,000 J
h  20m
a)
b)
c)
d)
K = 200,000J; h=20m; v=63.2m/s
K = 0; h=20m; v=0m/s
K = 250,000J; h=20m; v=70.7m/s
K = 200,000J; h=25m; v=70.7m/s
C
v2
aC 
 714.1m / s 2
r
FC  mac  714,000 N
 Fg  FN
714000  mg  FN
704,000 J  FN
Fg
FN
a) 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
B