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High School

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Physics Ex-37

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Energy is the ability to do work. There are three types ofwork that can be done:

Accelerating an object The work done goes to the object in the form of EK (kinetic energy)

thereby increasing the velocity of the object.

Raising an 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.

PART-2

Physics Ex-37

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An ideal system is one which has no friction. Frictionless systemsdo not exist in nature. However, since they simplify the solving ofphysics problems, it is often convenient to assume a system isfrictionless. Once the problem is solved without friction, the effectscaused by friction are added to the system. Remember that in anideal system, there is no loss of energy due to friction.

A real system is one that has friction. All systems in nature arereal systems. That is, all systems in nature lose some energy dueto friction (usually in the form of heat and sound).

The Law of Conservation of Energy tells us that energy cannot becreated nor destroyed. This means that when work is done, all ofthe energy must be accounted for. If the system is ideal, none ofthe work done is wasted. However, if the system is real, some ofthe work done is “lost” (as heat and sound energy).

PART-2

Physics Ex-37

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When an object is accelerated, work is being done.

When an object is decelerated, the kinetic energy it has mustbe dissipated (reduced to zero).

When stopping a car, work is done by the brakes (friction) todissipate the energy. The EK of the car is transformed intoheat and sound.

The fact that the kinetic energy of an object is proportional to thesquare 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 timesgreater (102). This explains the highway slogan that “speed kills”.

PART-2

Physics Ex-37Question-1

Explain the difference between an ideal system and a real system.

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An ideal system is one which has no friction and thus has noenergy loss in the form of heat and sound.

A real system is one which has friction and thus has energyloss in the form of heat and sound.


List the three types of work that can be done and tell whatbecomes of this work.

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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.


A 2 kg cart rests motionless on a horizontal plane.

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Is the cart at rest? a) Yes

What is the resultant force?b) 0

What is the acceleration?c) 0

What is the kinetic energy?d) 0

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A 2 kg cart travels at a constant velocity of 10 m/s from Point-A toPoint-B, a distance of 5 m. Assuming the system is frictionless (f = 0), answer the following questions:

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Was the cart at rest? a) Yes

What was the applied force?b) 0

What was the frictional force?c) 0

What was the resultant force?d) 0

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s = 5 m

“given”

vi = 10 m/s vf = 10 m/s

What was the acceleration?e) 0

2 kg

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f = 0

“constant velocity”

A B

(While traveling from point-A to point-B)

(FR = FA – f = 0 – 0 = 0)



A 2 kg cart travels at a constant velocity of 10 m/s from Point-A toPoint-B, a distance of 5 m. Assuming the system is frictionless (f = 0), answer the following questions:

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What was the initial EK? f) 100 J

What was the final EK?g) 100 J

How much energy did the cart gain?h) 0

How much work was done on the cart?i) 0

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s = 5 m

vi = 10 m/s vf = 10 m/s

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Summarize the amounts of work done:j)

2 kg

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f = 0

A B

1) To accelerate the cart 02) To raise the cart 03) To overcome friction 04) Total work done 0 ClickClickClickClickClick


Starting from rest, a horizontal force of 20 N is applied to a 2 kg cartresulting in a final velocity of 10 m/s. Assuming the system is ideal(frictionless), answer the following questions.

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Was the cart at rest? a) No

What was the applied force?b) 20 N right

What was the horizontal componentof the applied force?

c) 20 N right

What was the frictional force?d) 0

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s = 5 m

“frictionless”

vi = 0 vf = 10 m/s

What was the resultant force?e) 20 N right

2 kg

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f = 0

A B


(FR = FA – f = 20 N – 0 = 20 N)

FA = 20 N



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What was the acceleration of the cart?f) 10 m/s2

What was the initial EK of the cart?g) 0

What was the final EK of the cart?h) 100 J

How much work was done on the cart?i) 100 J

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s = 5 m

vi = 0 vf = 10 m/s

What becomes of the work done on the cart?j)

It is transferred to the cart in the form of EK (faster speed).

2 kg

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f = 0

A B

FA = 20 N

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How much work was done to overcome friction?k) 0

What was the total work done?l) 100 J

Where did the work done come from?m) The applied force

Summarize the amounts of work done:n)

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s = 5 m

vi = 0 vf = 10 m/s

2 kgf = 0

A B

FA = 20 N

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1) To accelerate the cart 100 J2) To raise the cart 03) To overcome friction 04) Total work done 100 J ClickClickClickClick

(Cart was not raised)

(Frictionless system)

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A horizontal force of 20 N is applied to a 2 kg cart whose initial velocity is 8 m/sresulting in a final velocity of 12 m/s. Assuming there is no friction, answer thefollowing questions concerning the cart in going from Point-A to Point-B (4 m).

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c) 20 N right

What was the frictional force?d) 0

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s = 4 m

“frictionless”

vi = 8 m/s vf = 12 m/s


2 kg

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f = 0

A B


(FR = FA – f = 20 N – 0 = 20 N)

FA = 20 N



s = 4 m

vi = 8 m/s vf = 12 m/s

2 kgf = 0

A B

FA = 20 N


What was the initial EK of the cart?g) 64 J




It is transferred to the cart in the form of EK (faster speed).

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s = 4 m

vi = 8 m/s vf = 12 m/s

2 kgf = 0

A B

FA = 20 N

How much work was done to overcome friction?k) 0




1) To accelerate the cart 80 J2) To raise the cart 03) To overcome friction 04) Total work done 80 J


(Frictionless system)



A 2 kg cart travels at a constant velocity of 10 m/s for a distanceof 5 m. If the frictional force is 2 N, answer the following questionsconcerning the cart in going from Point-A to Point-B.

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c) 2 N right

What was the frictional force?d) 2 N left

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s = 5 m

vi = 10 m/s vf = 10 m/s

What was the resultant force?e) 0

2 kg

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f = 2 N

A B


(FR = FA – f = 2 N – 2 N = 0)

FA = ?

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What was the acceleration of the cart?f) 0




What becomes of the work done?j)

It is used to overcome friction (lost energy).


s = 5 m

vi = 10 m/s vf = 10 m/s

2 kgf = 2 N

A B

FA = ?




How much work was done to overcome friction?k) 10 J




1) To accelerate the cart 02) To raise the cart 03) To overcome friction 10 J4) Total work done 10 J





s = 5 m

vi = 10 m/s vf = 10 m/s

2 kgf = 2 N

A B

FA = ?

(Cart was not accelerated)


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, answerthe following questions concerning the cart in going from A to B.

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c) 20 N right


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s = 40.5 m

vi = 0 vf = 27 m/s


2 kg

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f = 2 N

A B


(FR = FA – f = 20 N – 2 N = 18 N)

FA = 20 N

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“velocity not constant”

Physics Ex-37


What was the initial EK of the cart?g) 0




It is used to accelerate the cart (increase its velocity from 0 to 27 m/s).

Question-8


s = 40.5 m

vi = 0 vf = 27 m/s

2 kgf = 2 N

A B

FA = 20 N







1) To accelerate the cart 729 J2) To raise the cart 03) To overcome friction 81 J4) Total work done 810 J




s = 40.5 m

vi = 0 vf = 27 m/s

2 kgf = 2 N

A B

FA = 20 N



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c) 20 N right


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s = 16 m

vi = 6 m/s vf = 18 m/s


2 kg

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f = 2 N

A B


(FR = FA – f = 20 N – 2 N = 18 N)

FA = 20 N

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“velocity not constant”

Physics Ex-37






It is used to accelerate the cart (increase its velocity from 6 to 18 m/s).


Question-9


s = 16 m

vi = 6 m/s vf = 18 m/s

2 kgf = 2 N

A B

FA = 20 N

Physics Ex-37





1) To accelerate the cart 288 J2) To raise the cart 03) To overcome friction 32 J4) Total work done 320 J



Question-9


s = 16 m

vi = 6 m/s vf = 18 m/s

2 kgf = 2 N

A B

FA = 20 N


A force is applied, at 60o from the horizontal, to a 20 kg cart causing it to travelat a constant velocity of 10 m/s. If the force of friction is 2 N, answer thefollowing questions concerning the cart in going from A to B (5 m).

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What was the applied force?b) 4 N [E 60o N]


c) 2 N right


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s = 5 m

vi = 10 m/s vf = 10 m/s

What was the resultant force?e) 0

20 kg

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f = 2 N

A B


(FR = FA – f = 2 N – 2 N = 0)

FA = ?

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“velocity is constant”

60o

FH = 2 N

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Physics Ex-37

What was the acceleration of the cart?f) 0

What was the initial EK of the cart?g) 1 000 J

What was the final EK of the cart?h) 1 000 J



It is used to overcome friction.


Question-10


s = 5 m

vi = 10 m/s vf = 10 m/s

20 kgf = 2 N

A B

FA = ?

60o

FH = 2 N

“velocity constant”

Physics Ex-37





1) To accelerate the cart 02) To raise the cart 03) To overcome friction 10 J4) Total work done 10 J



Question-10


s = 5 m

vi = 10 m/s vf = 10 m/s

20 kgf = 2 N

A B

FA = ?

60o

FH = 2 N

(Cart was not accelerated)

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