Aristo Work and Energy - Module 23 - AIEEE Qs

8/12/2019 Aristo Work and Energy - Module 23 - AIEEE Qs

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Work, Power and Energy

Scene 1

A pencil writes & draws the following (free hand):

Physics

Work, Power and Energy

Show graphics of the following reference image:

A ball is rolling inside a hemispherical bowl

Scene 2

On screen:

AIEEE Questions

Voiceover:

In this session we will cover a few questions related to work and energy that have been asked in the

AIEEE in the recent years.


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Page 2

Scene 3

On screen:

AIEEE 2007

Voiceover:

Here is a question from from AIEEE 2007.

A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a un-compress spring, and

compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is

10000 N/m. The spring is compressed by:

(1) 5.5 cm (2) 2.5 cm (3) 11.0 cm (4) 8.5 cmIt is the kinetic energy of the block which compresses the spring. So, we can say, kinetic energy ofthe block stores as elastic potential energy in the spring.

Scene 4

On screen:

A block slides on a horizontal floor with a speed of . It strikes a un-compress

spring, and compresses it till the block is motionless. The kinetic friction force is and spring

constant is . The spring is compressed by:

Solving we get-

So, option (1) is correct.


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Voiceover:

Let the spring compresses by an amountx . Therefore, we can write-

2 21 1mv kx2 2

2

2m N2kg 4 10000 xs m

Solving we get- x 5.6 cm


Scene 5

On screen:

AIEEE 2006

Voiceover:

Here is another question on work-energy theorem from AIEEE 2006. Look at this question.

A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the

force of gravity during the time the particle goes up is:

(1) 0.5 J (2) -0.5 J (3) -1.25 J (4) 1.25 JHere, particle is projected vertically upwards. Its velocity decreases and finally becomes zero at its

maximum height. So, work done by force of gravity is equal to change in kinetic energy of the

particle.

A particle of mass is thrown vertically upwards with a speed of .

The work done by the force of gravity during the time the particle goes up is:


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On screen:

Voiceover:

Here, particle is projected vertically upwards. Its velocity decreases and finally becomes zero at its

maximum height. So, work done by force of gravity is equal to change in kinetic energy of the

particle.

2 2

g f i f i1 1W K K K mv mv2 2

Here,i

v 5m / s ,f

v 0 and m 100gm 0.1kg

On solving, we get-

gW 1.25 J


Here, , and

On solving, we get-



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Page 5

Scene 7

On screen:

AIEEE 2006

Voiceover:

Yet another question from AIEEE 2006 based on the concept of principle of conservation of

mechanical energy.

The potential energy of a 1 kg particle free move along the X-axis is given by:

()

The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is:

(1) 2 (2) (3) (4)

Here, particle is moving freely along x-axis. Its potential energy function is given as a

function of position x. During the free motion, kinetic energy of the particle increases while

potential energy decreases. And, we have to find maximum speed of the particle.

The potential energy of a particle free move along the axis is given by:

The total mechanical energy of the particle is . Then, the maximum speed (in ) is:


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Scene 8

On screen:

Now, to find minimum potential energy, let us put

That is:

We find that at and at .

At , the potential energy is zero i.e.

And at , the potential energy is

Therefore, potential energy is minimum at .

Now, we can apply principle of conservation of mechanical energy to find maximum speed of

the particle.

Solving we get-



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Voiceover:

Let us first find the points where potential energy is a minimum and kinetic energy is a maximum.

We know that potential energy will be minimum, where the particle is in equilibrium i.e. the position

of the particle where force becomes zero.

Now, the question is:

What kind of force is acting on the particle?

Of course it is a conservative force; because particle is moving in one-dimension and is a function of

position. And, we know that the conservative force is negative of change in potential energy.

That is:

dV x

F xdx

4 2

3d x xF x x xdx 4 2

Now, to find minimum potential energy, let us put F x 0

That is: 3F x 0 x x x x 1 x 1

We find that F x 0 at X 0 and at X 1 .

At X 0 , the potential energy is zero i.e. V x 0

And at X 1 , the potential energy is 1

V x 14

Therefore, potential energy is minimum at X 1 .

Now, we can apply principle of conservation of mechanical energy to find maximum speed of the

particle.

max min

T.E K.E P.E


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Page 8

2

max

1 12 J mv

2 4

Solving we get-

max

3v

2


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Aristo Work and Energy - Module 23 - AIEEE Qs