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Part 2 /3
Physics Ex-37
PART-2
Energy is the ability to do work. There are three types of
work that can be done:
 Accelerating an object
 Raising an object
The work done goes to the object
in the form of EK (kinetic energy)
thereby increasing the velocity
of the object.
The work done goes to the object in the
form of EP (potential energy) thereby
increasing the height of the object
(from a reference line).
 To overcome friction
The work done to overcome friction
is “lost” as heat and sound energy.
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Physics Ex-37
PART-2
An ideal system is one which has no friction. Frictionless systems
do not exist in nature. However, since they simplify the solving of
physics problems, it is often convenient to assume a system is
frictionless. Once the problem is solved without friction, the effects
caused by friction are added to the system. Remember that in an
ideal system, there is no loss of energy due to friction.
A real system is one that has friction. All systems in nature are
real systems. That is, all systems in nature lose some energy due
to friction (usually in the form of heat and sound).
The Law of Conservation of Energy tells us that energy cannot be
created nor destroyed. This means that when work is done, all of
the energy must be accounted for. If the system is ideal, none of
the work done is wasted. However, if the system is real, some of
the work done is “lost” (as heat and sound energy).
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Physics Ex-37
PART-2
When an object is accelerated, work is being done.
When an object is decelerated, the kinetic energy it has must
be dissipated (reduced to zero).
When stopping a car, work is done by the brakes (friction) to
dissipate the energy. The EK of the car is transformed into
heat and sound.
The fact that the kinetic energy of an object is proportional to the
square of its speed has important consequences. Doubling the
speed of a car, for example, quadruples its kinetic energy (22).
If the velocity increases ten times, then the EK becomes 100 times
greater (102). This explains the highway slogan that “speed kills”.
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Question-1
Physics Ex-37
Explain the difference between an ideal system and a real system.
An ideal system is one which has no friction and thus has no
energy loss in the form of heat and sound.
A real system is one which has friction and thus has energy
loss in the form of heat and sound.
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Physics Ex-37
Question-2
List the three types of work that can be done and tell what
becomes of this work.
Accelerating an object The work goes to the object in the form of EK.
Raising an object The work goes to the object in the form of EP.
To overcome friction The work is lost in the form of heat and sound
energy.
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Physics Ex-37
Question-3
A 2 kg cart rests motionless on a horizontal plane.
a) Is the cart at rest?
Yes
b) What is the resultant force?
0
c) What is the acceleration?
0
d) What is the kinetic energy?
0
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Physics Ex-37
Question-4
A 2 kg cart travels at a constant velocity of 10 m/s from Point-A to
Point-B, a distance of 5 m. Assuming the system is frictionless (f = 0),
answer the following questions:
vi = 10 m/s
vf = 10 m/s
2 kg
f=0
A
B
s=5m
a) Was the cart at rest?
Yes “constant velocity”
(While traveling from point-A to point-B)
b) What was the applied force?
0
c) What was the frictional force?
0
“given”
d) What was the resultant force?
0
(FR = FA – f = 0 – 0 = 0)
e) What was the acceleration?
0
“constant velocity”
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Physics Ex-37
Question-4
A 2 kg cart travels at a constant velocity of 10 m/s from Point-A to
Point-B, a distance of 5 m. Assuming the system is frictionless (f = 0),
answer the following questions:
vi = 10 m/s
vf = 10 m/s
2 kg
f=0
A
B
s=5m
f) What was the initial EK?
100 J
g) What was the final EK?
100 J
h) How much energy did the cart gain?
0
i) How much work was done on the cart?
0
j) Summarize the amounts of work done:
1) To accelerate the cart
2) To raise the cart
3) To overcome friction
4) Total work done
0
0
0
0
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Physics Ex-37
Question-5
Starting from rest, a horizontal force of 20 N is applied to a 2 kg cart
resulting in a final velocity of 10 m/s. Assuming the system is ideal
(frictionless), answer the following questions.
vi = 0
vf = 10 m/s
FA = 20 N
f=0
A
2 kg
B
s=5m
No
a) Was the cart at rest?
(While traveling from point-A to point-B)
b) What was the applied force?
20 N right
c) What was the horizontal component
of the applied force?
20 N right
d) What was the frictional force?
e) What was the resultant force?
0
“frictionless”
20 N right
(FR = FA – f = 20 N – 0 = 20 N)
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Physics Ex-37
Question-5
Starting from rest, a horizontal force of 20 N is applied to a 2 kg cart
resulting in a final velocity of 10 m/s. Assuming the system is ideal
(frictionless), answer the following questions.
vi = 0
vf = 10 m/s
FA = 20 N
f=0
A
s=5m
2 kg
B
f) What was the acceleration of the cart? 10 m/s2
g) What was the initial EK of the cart?
0
h) What was the final EK of the cart?
100 J
i) How much work was done on the cart?
100 J
j) What becomes of the work done on the cart?
It is transferred to the cart in the form of EK (faster speed).
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Physics Ex-37
Question-5
Starting from rest, a horizontal force of 20 N is applied to a 2 kg cart
resulting in a final velocity of 10 m/s. Assuming the system is ideal
(frictionless), answer the following questions.
vi = 0
vf = 10 m/s
FA = 20 N
2 kg
f=0
A
B
s=5m
k) How much work was done to overcome friction?
l) What was the total work done?
m) Where did the work done come from?
0
100 J
The applied force
n) Summarize the amounts of work done:
1) To accelerate the cart
2) To raise the cart
3) To overcome friction
4) Total work done
100 J
(Cart was not raised)
0
0
(Frictionless system)
100 J
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Physics Ex-37
Question-6
A horizontal force of 20 N is applied to a 2 kg cart whose initial velocity is 8 m/s
resulting in a final velocity of 12 m/s. Assuming there is no friction, answer the
following questions concerning the cart in going from Point-A to Point-B (4 m).
vi = 8 m/s
vf = 12 m/s
FA = 20 N
f=0
A
2 kg
B
s=4m
No
a) Was the cart at rest?
(While traveling from point-A to point-B)
b) What was the applied force?
20 N right
c) What was the horizontal component
of the applied force?
20 N right
d) What was the frictional force?
e) What was the resultant force?
0
“frictionless”
20 N right
(FR = FA – f = 20 N – 0 = 20 N)
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Physics Ex-37
Question-6
A horizontal force of 20 N is applied to a 2 kg cart whose initial velocity is 8 m/s
resulting in a final velocity of 12 m/s. Assuming there is no friction, answer the
following questions concerning the cart in going from Point-A to Point-B (4 m).
vi = 8 m/s
vf = 12 m/s
FA = 20 N
f=0
A
2 kg
s=4m
B
f) What was the acceleration of the cart? 10 m/s2
g) What was the initial EK of the cart?
64 J
h) What was the final EK of the cart?
144 J
i) How much work was done on the cart?
80 J
j) What becomes of the work done on the cart?
It is transferred to the cart in the form of EK (faster speed).
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Physics Ex-37
Question-6
A horizontal force of 20 N is applied to a 2 kg cart whose initial velocity is 8 m/s
resulting in a final velocity of 12 m/s. Assuming there is no friction, answer the
following questions concerning the cart in going from Point-A to Point-B (4 m).
vi = 8 m/s
vf = 12 m/s
FA = 20 N
2 kg
f=0
A
B
s=4m
k) How much work was done to overcome friction?
l) What was the total work done?
m) Where did the work done come from?
0
80 J
The applied force
n) Summarize the amounts of work done:
1) To accelerate the cart
2) To raise the cart
3) To overcome friction
4) Total work done
80 J
(Cart was not raised)
0
0
(Frictionless system)
80 J
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Physics Ex-37
Question-7
A 2 kg cart travels at a constant velocity of 10 m/s for a distance
of 5 m. If the frictional force is 2 N, answer the following questions
concerning the cart in going from Point-A to Point-B.
vi = 10 m/s
FA = ?
A
vf = 10 m/s
f=2N
2 kg
B
s=5m
Yes “constant velocity”
a) Was the cart at rest?
(While traveling from point-A to point-B)
b) What was the applied force?
2 N right
c) What was the horizontal component
of the applied force?
2 N right
d) What was the frictional force?
2 N left
e) What was the resultant force?
0
(FR = FA – f = 2 N – 2 N = 0)
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Physics Ex-37
Question-7
A 2 kg cart travels at a constant velocity of 10 m/s for a distance
of 5 m. If the frictional force is 2 N, answer the following questions
concerning the cart in going from Point-A to Point-B.
vi = 10 m/s
A
FA = ?
vf = 10 m/s
f=2N
s=5m
2 kg
B
f) What was the acceleration of the cart?
0
“constant velocity”
g) What was the initial EK of the cart?
100 J
h) What was the final EK of the cart?
100 J
i) How much work was done on the cart?
0
j) What becomes of the work done?
It is used to overcome friction (lost energy).
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Physics Ex-37
Question-7
A 2 kg cart travels at a constant velocity of 10 m/s for a distance
of 5 m. If the frictional force is 2 N, answer the following questions
concerning the cart in going from Point-A to Point-B.
vi = 10 m/s
A
vf = 10 m/s
FA = ?
f=2N
2 kg
B
s=5m
k) How much work was done to overcome friction?
10 J
l) What was the total work done?
10 J
The applied force
m) Where did the work done come from?
n) Summarize the amounts of work done:
1) To accelerate the cart
2) To raise the cart
3) To overcome friction
4) Total work done
0
0
10 J
10 J
(Cart was not accelerated)
(Cart was not raised)
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Physics Ex-37
Question-8
A horizontal force of 20 N is applied to a 2 kg cart, initially at rest,
giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer
the following questions concerning the cart in going from A to B.
vi = 0
A
FA = 20 N
vf = 27 m/s
f=2N
2 kg
s = 40.5 m
B
No “velocity not constant”
a) Was the cart at rest?
(While traveling from point-A to point-B)
b) What was the applied force?
20 N right
c) What was the horizontal component
of the applied force?
20 N right
d) What was the frictional force?
e) What was the resultant force?
2 N left
18 N right
(FR = FA – f = 20 N – 2 N = 18 N)
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Physics Ex-37
Question-8
A horizontal force of 20 N is applied to a 2 kg cart, initially at rest,
giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer
the following questions concerning the cart in going from A to B.
vi = 0
A
FA = 20 N
vf = 27 m/s
f=2N
2 kg
s = 40.5 m
B
f) What was the acceleration of the cart?
g) What was the initial EK of the cart?
9 m/s2
0
h) What was the final EK of the cart?
729 J
i) How much work was done on the cart?
729 J
j) What becomes of the work done on the cart?
It is used to accelerate the cart (increase its velocity from 0 to 27 m/s).
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Physics Ex-37
Question-8
A horizontal force of 20 N is applied to a 2 kg cart, initially at rest,
giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer
the following questions concerning the cart in going from A to B.
vi = 0
A
FA = 20 N
vf = 27 m/s
f=2N
s = 40.5 m
2 kg
B
k) How much work was done to overcome friction?
81 J
l) What was the total work done?
810 J
m) Where did the work done come from?
The applied force
n) Summarize the amounts of work done:
1) To accelerate the cart
2) To raise the cart
3) To overcome friction
4) Total work done
729 J
(Cart was not raised)
0
81 J
810 J
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Physics Ex-37
Question-9
A horizontal force of 20 N is applied to a 2 kg cart, initially at rest,
giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer
the following questions concerning the cart in going from A to B.
vi = 6 m/s
FA = 20 N
A
vf = 18 m/s
f=2N
2 kg
B
s = 16 m
No “velocity not constant”
a) Was the cart at rest?
(While traveling from point-A to point-B)
b) What was the applied force?
20 N right
c) What was the horizontal component
of the applied force?
20 N right
d) What was the frictional force?
e) What was the resultant force?
2 N left
18 N right
(FR = FA – f = 20 N – 2 N = 18 N)
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Physics Ex-37
Question-9
A horizontal force of 20 N is applied to a 2 kg cart, initially at rest,
giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer
the following questions concerning the cart in going from A to B.
vi = 6 m/s
A
FA = 20 N
vf = 18 m/s
f=2N
2 kg
s = 16 m
B
f) What was the acceleration of the cart?
9 m/s2
g) What was the initial EK of the cart?
36 J
h) What was the final EK of the cart?
324 J
i) How much work was done on the cart?
288 J
j) What becomes of the work done on the cart?
It is used to accelerate the cart (increase its velocity from 6 to 18 m/s).
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Physics Ex-37
Question-9
A horizontal force of 20 N is applied to a 2 kg cart, initially at rest,
giving it a final velocity of 27 m/s. If the frictional force is 2 N, answer
the following questions concerning the cart in going from A to B.
vi = 6 m/s
A
vf = 18 m/s
FA = 20 N
f=2N
s = 16 m
2 kg
B
k) How much work was done to overcome friction?
32 J
l) What was the total work done?
320 J
m) Where did the work done come from?
The applied force
n) Summarize the amounts of work done:
1) To accelerate the cart
2) To raise the cart
3) To overcome friction
4) Total work done
288 J
(Cart was not raised)
0
32 J
320 J
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Physics Ex-37
Question-10
A force is applied, at 60o from the horizontal, to a 20 kg cart causing it to travel
at a constant velocity of 10 m/s. If the force of friction is 2 N, answer the
following questions concerning the cart in going from A to B (5 m).
FA = ?
vi = 10 m/s
60o
A
FH = 2 N
vf = 10 m/s
f=2N
20 kg
B
s=5m
Yes “velocity is constant”
a) Was the cart at rest?
(While traveling from point-A to point-B)
b) What was the applied force?
4 N [E 60o N]
c) What was the horizontal component
of the applied force?
2 N right
d) What was the frictional force?
2 N left
e) What was the resultant force?
0
(FR = FA – f = 2 N – 2 N = 0)
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Physics Ex-37
Question-10
A force is applied, at 60o from the horizontal, to a 20 kg cart causing it to travel
at a constant velocity of 10 m/s. If the force of friction is 2 N, answer the
following questions concerning the cart in going from A to B (5 m).
FA = ?
vi = 10 m/s
60o
A
FH = 2 N
vf = 10 m/s
f=2N
s=5m
20 kg
B
f) What was the acceleration of the cart?
0
“velocity constant”
g) What was the initial EK of the cart?
1 000 J
h) What was the final EK of the cart?
1 000 J
i) How much work was done on the cart?
0
j) What becomes of the work done on the cart?
It is used to overcome friction.
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Physics Ex-37
Question-10
A force is applied, at 60o from the horizontal, to a 20 kg cart causing it to travel
at a constant velocity of 10 m/s. If the force of friction is 2 N, answer the
following questions concerning the cart in going from A to B (5 m).
FA = ?
vi = 10 m/s
60o
A
FH = 2 N
vf = 10 m/s
f=2N
s=5m
20 kg
B
k) How much work was done to overcome friction?
10 J
l) What was the total work done?
10 J
m) Where did the work done come from?
The applied force
n) Summarize the amounts of work done:
1) To accelerate the cart
2) To raise the cart
3) To overcome friction
4) Total work done
(Cart was not accelerated)
0
(Cart was not raised)
0
10 J
10 J
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