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Transcript 
Exam Review Part 2
SPH4C
Question 1
Starting from rest, Jacob accelerates to a
velocity of 4.6 m/s [E] in 2.0 s.
(a) What was Jacob's acceleration?
Question 1
Starting from rest, Jacob accelerates to a
velocity of 4.6 m/s [E] in 2.0 s.
(a) What was Jacob's acceleration?
v1  0 ms
v2  4.6 ms
 t  2 .0 s
a?
Question 1
Starting from rest, Jacob accelerates to a
velocity of 4.6 m/s [E] in 2.0 s.
(a) What was Jacob's acceleration?
v1  0
m
s
v2  4.6 ms
 t  2 .0 s
a?
v2  v1
a
t
4.6 ms  0 ms
a
2.0 s
a  2.3 sm2
Question 1
Starting from rest, Jacob accelerates to a
velocity of 4.6 m/s [E] in 2.0 s.
(a) What was Jacob's acceleration?
v1  0
m
s
v2  4.6 ms
 t  2 .0 s
a?
v2  v1
a
t
4.6 ms  0 ms
a
2.0 s
a  2.3 sm2
Question 1
Starting from rest, Jacob accelerates to a
velocity of 4.6 m/s [E] in 2.0 s.
(a) What was Jacob's acceleration?
v1  0
m
s
v2  4.6 ms
 t  2 .0 s
a?
v2  v1
a
t
4.6 ms  0 ms
a
2.0 s
a  2.3 sm2
Question 1
Starting from rest, Jacob accelerates to a
velocity of 4.6 m/s [E] in 2.0 s.
(a) What was Jacob's acceleration?
v1  0
m
s
v2  4.6 ms
 t  2 .0 s
a?
v2  v1
a
t
4.6 ms  0 ms
a
2.0 s
a  2.3 sm2 [ East ]
Question 1
(b) What was the distance Jacob travelled while
accelerating?
v1  0 ms
v2  4.6 ms
 t  2 .0 s
d  ?
Question 1
(b) What was the distance Jacob travelled while
accelerating?
v1  v2 d

2
t
0 ms  4.6 ms
v1  0 ms
d

2
2.0 s
v2  4.6 ms
 t  2 .0 s
d  ?
d
2.3 
2.0 s
m
s
4.6 m  d
Question 1
(b) What was the distance Jacob travelled while
accelerating?
v1  v2 d

2
t
0 ms  4.6 ms
v1  0 ms
d

2
2.0 s
v2  4.6 ms
 t  2 .0 s
d  ?
d
2.3 
2.0 s
m
s
4.6 m  d
Question 2
Dalton pushes a crate of weight 320 N across a
horizontal floor with a force of 110 N. The coefficient
of kinetic friction between the crate and the floor is
0.30.
(a) What is the magnitude of the frictional force acting
on the crate?
Fg  320 N
FA  110 N
  0.30
Ff  ?
Question 2
Dalton pushes a crate of weight 320 N across a
horizontal floor with a force of 110 N. The coefficient
of kinetic friction between the crate and the floor is
0.30.
(a) What is the magnitude of the frictional force acting
on the crate?
Fg  320 N
Ff  FN  Fg
FA  110 N
Ff  (0.30)(320 N )  96 N
  0.30
Ff  ?
Question 2
Dalton pushes a crate of weight 320 N across a
horizontal floor with a force of 110 N. The coefficient
of kinetic friction between the crate and the floor is
0.30.
(a) What is the magnitude of the frictional force acting
on the crate?
Fg  320 N
Ff  FN  Fg
FA  110 N
Ff  (0.30)(320 N )  96 N
  0.30
Ff  ?
Question 2
Dalton pushes a crate of weight 320 N across a
horizontal floor with a force of 110 N. The coefficient
of kinetic friction between the crate and the floor is
0.30.
(a) What is the magnitude of the frictional force acting
on the crate?
Fg  320 N
Ff  FN  Fg
FA  110 N
Ff  (0.30)(320 N )  96 N
  0.30
Ff  ?
Question 2
(b) What is the net force on the crate?
Question 2
(b) What is the net force on the crate?
Fnet  ma ?
Question 2
(b) What is the net force on the crate?
Fnet  ma ?
Question 2
(b) What is the net force on the crate?
Fnet  FA  F f
Fnet  110 N  96 N
Fnet  14 N [ forwards]
Question 2
(b) What is the net force on the crate?
Fnet  FA  F f
Fnet  110 N  96 N
Fnet  14 N [ forwards]
Question 3
Eric uses a force of 180 N to push a crate of
weight 320 N at constant speed up a ramp
with a length of 3.6 m and a rise of 1.2 m.
(a) What is the ideal mechanical advantage of
the ramp?
FE  180 N
FL  320 N
d E  3.6 m
d L  1.2 m
Question 3
Eric uses a force of 180 N to push a crate of
weight 320 N at constant speed up a ramp
with a length of 3.6 m and a rise of 1.2 m.
(a) What is the ideal mechanical advantage of
the ramp?
dE
FE  180 N
IMA 
dL
FL  320 N
3 .6 m
d E  3.6 m
IMA 
3
1 .2 m
d L  1.2 m
Question 3
(b) What is the actual mechanical advantage of
the ramp?
FE  180 N
FL  320 N
d E  3.6 m
d L  1.2 m
FL
AMA 
FE
320N
AMA 
 1 .8
180N
Question 4
Victoria is trying to lift a crate using a lever as shown
below. The crate has a weight of 1650 N and is 1.8 m
from the fulcrum.
(a) What is the load torque on the lever?
Question 4
Victoria is trying to lift a crate using a lever as shown
below. The crate has a weight of 1650 N and is 1.8 m
from the fulcrum.
(a) What is the load torque on the lever?
FL  1650 N
d L  1.8 m
TL  ?
Question 4
Victoria is trying to lift a crate using a lever as shown
below. The crate has a weight of 1650 N and is 1.8 m
from the fulcrum.
(a) What is the load torque on the lever?
FL  1650 N
TL  FL d L
d L  1.8 m
TL  (1650 N )(1.8 m)  2970 N  m
TL  ?
Question 4
(b) If the effort force is applied 3.1 m from the
fulcrum, what is its magnitude?
FL  1650 N
d L  1.8 m
d E  3.1 m
FE  ?
Question 4
(b) If the effort force is applied 3.1 m from the
fulcrum, what is its magnitude?
FL  1650 N
d L  1.8 m
d E  3.1 m
FE  ?
FL d L
FE d E  FL d L  FE 
dE
FE

1650N 1.8 m 

 958N
3.1 m
or 960N
Question 4
(b) If the effort force is applied 3.1 m from the
fulcrum, what is its magnitude?
FL  1650 N
d L  1.8 m
d E  3.1 m
FE  ?
FL d L
FE d E  FL d L  FE 
dE
FE

1650N 1.8 m 

 958N
3.1 m
or 960N
Question 5
Connor's car brakes exert a force of 12 000 N
[backwards] over a distance of 24 m along a
level piece of highway.
(a) Determine the work done by the brakes.
F  12000 N
d  24 m
W ?
Question 5
Connor's car brakes exert a force of 12 000 N
[backwards] over a distance of 24 m along a
level piece of highway.
(a) Determine the work done by the brakes.
F  12000 N
d  24 m
W ?
W  Fd
W   12000 N 24 m   288 000 J
Question 5
(b) Is the work in (a) positive or negative? What
does this mean?
The work done is negative.
Question 5
(b) Is the work in (a) positive or negative? What
does this mean?
The work done is negative.
Friction is reducing the energy of the car.
Question 6
Brandan drops a 1.5 kg water balloon from a 14 m
high roof.
(a) What is the gravitational potential energy of the
balloon on the roof?
m  1.5 kg
h  14 m
g  9.8 sm2
Eg  ?
Question 6
Brandan drops a 1.5 kg water balloon from a 14 m
high roof.
(a) What is the gravitational potential energy of the
balloon on the roof?
m  1.5 kg
h  14 m
g  9.8 sm2
Eg  ?
Question 6
Brandan drops a 1.5 kg water balloon from a 14 m
high roof.
(a) What is the gravitational potential energy of the
balloon on the roof?
m  1.5 kg
h  14 m
g  9.8 sm2
Eg  ?
E g  mgh


E g  1.5kg  9.8sm2 14m   205.8J
Question 6
(b) What is the speed of the balloon when it hits
the ground?
E g  Ek  205.8 J
m  1.5kg
v?
Question 6
(b) What is the speed of the balloon when it hits
the ground?
E g  Ek  205.8 J
m  1.5kg
v?
2 Ek
Ek  mv  v 
m
1
2
2
2205.8J 
v
 16.6 ms
1.5kg
Question 7
9-V is supplied to a circuit containing a single light
bulb. The current through the circuit is 3.0 A.
(a) What is the resistance of the bulb?
V  9V
I  3.0A
R?
Question 7
9-V is supplied to a circuit containing a single light
bulb. The current through the circuit is 3.0 A.
(a) What is the resistance of the bulb?
V  9V
I  3.0A
R?
V
V  IR  R 
I
9V
R
 3
3.0 A
Question 7
9-V is supplied to a circuit containing a single light
bulb. The current through the circuit is 3.0 A.
(a) What is the resistance of the bulb?
V  9V
I  3.0A
R?
V
V  IR  R 
I
9V
R
 3
3.0 A
Question 7
(b) What is the power consumed by the bulb?
V  9V
I  3.0A
P?
Question 7
(b) What is the power consumed by the bulb?
V  9V
I  3.0A
P?
P  VI
P  9 V 3.0 A  27 W
Question 8
You have three resistors, each with a resistance
of 12-.
(a) What is their resistance if the resistors are in
series?
Question 8
You have three resistors, each with a resistance
of 12-.
(a) What is their resistance if the resistors are in
series?
Req  12   12   12   36 
Question 8
(b) What is their resistance if the resistors are in
parallel?
1
1
1
1
3
1





Req 12  12  12  12  4 
Req  4 
Question 9
In a manometer filled with water, the difference
between the heights of the two columns is 6 cm
(0.06 m).
(a) What is the gauge pressure on the lower
column?
Question 9
In a manometer filled with water, the difference
between the heights of the two columns is 6 cm
(0.06 m).
(a) What is the gauge pressure on the lower
column?
p  Dhg
p  (1000mkg3 )(0.06m)(9.8 sm2 )  588Pa
or 0.588kPa
Question 9
In a manometer filled with water, the difference
between the heights of the two columns is 6 cm
(0.06 m).
(a) What is the gauge pressure on the lower
column?
D  1000 mkg3
p  Dhg
h  0.06 m
p  (1000mkg3 )(0.06m)(9.8 sm2 )  588Pa
g  9.8 sm2
P?
or 0.588kPa
Question 9
In a manometer filled with water, the difference
between the heights of the two columns is 6 cm
(0.06 m).
(a) What is the gauge pressure on the lower
column?
D  1000 mkg3
p  Dhg
h  0.06 m
p  (1000mkg3 )(0.06m)(9.8 sm2 )  588Pa
g  9.8 sm2
P?
or 0.588kPa
Question 9
In a manometer filled with water, the difference
between the heights of the two columns is 6 cm
(0.06 m).
(a) What is the gauge pressure on the lower
column?
D  1000 mkg3
p  Dhg
h  0.06 m
p  (1000mkg3 )(0.06m)(9.8 sm2 )  588Pa
g  9.8 sm2
P?
or 0.588 kPa
Question 9
(b) What is the absolute pressure on the lower
column?
p abs  p g  patm
pabs  0.588 kPa  101.3 kPa
pabs  101.888 kPa or 101.9 kPa
Question 10
In a hydraulic press, the master cylinder with an
area of 0.04 m2 applies a force of magnitude
110 N.
(a) What is the pressure applied by the master
cylinder?
A  0.04 m
F  110 N
P?
2
F
P
A
110N
P
 2750 Pa
2
0.04 m
Question 10
In a hydraulic press, the master cylinder with an
area of 0.04 m2 applies a force of magnitude
110 N.
(a) What is the pressure applied by the master
cylinder?
A  0.04 m
F  110 N
P?
2
F
P
A
110N
P
 2750 Pa
2
0.04 m
Question 10
(a) What is the magnitude of the force on the
slave cylinder if it has an area of 0.24 m2?
As  0.04 m
2
Fs  110 N
AL  0.24 m
FL  ?
2
Question 10
(a) What is the magnitude of the force on the
slave cylinder if it has an area of 0.24 m2?
As  0.04 m
2
Fs  110 N
AL  0.24 m
FL  ?
2
Fs
FL Fs

 FL   AL
AL As
As
110N
2
FL 
 0.24 m  660N
2
0.04 m
Question 10
(a) What is the magnitude of the force on the
slave cylinder if it has an area of 0.24 m2?
As  0.04 m
2
Fs  110 N
AL  0.24 m
FL  ?
2
Fs
FL Fs

 FL   AL
AL As
As
110N
2
FL 
 0.24 m  660N
2
0.04 m
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