# Maharashtra Board Class 9 Science Chapter 2 Work and Energy Solution

## Maharashtra Board Class 9 Science Solution Chapter 2 – Work and Energy

Balbharati Maharashtra Board Class 9 Science Solution Chapter 2: Work and Energy. Marathi or English Medium Students of Class 9 get here Work and Energy full Exercise Solution.

 Std Maharashtra Class 9 Subject Science Solution Chapter Work and Energy

a) Explain the difference between kinetic energy and potential energy.

Ans:

Kinetic energy:

• Kinetic energy of the body is due to its state of motion.
• The kinetic energy of body depends on the speed of the body who is in motion.
• The kinetic energy of the body can be transferred from one body to other like in collision of two bodies.
• The kinetic energy of body of mass m and moving with speed v is given by,
• Kinetic energy= 1/2 mv2
• The kinetic energy of the body is scalar quantity.

Potential energy:

• The potential energy of the body is due to the rest position or size and shape of the body.
• The potential energy of the body doesn’t depends on its speed.
• The potential energy of body cannot be transferred from one body to other.
• The potential energy of the body of mass m, acceleration due to gravity g and at a height h is given by,
• Potential energy= mgh
• Potential energy is nothing but the work stored in a body.
• Potential energy is the scalar quantity.

b) Derive the formula for the kinetic energy of an object of mass m moving with velocity v.

Ans:

Let us consider the object which is initially at rest and when we applied a force F on it then the acceleration a will be produced in that object. And hence after time t, the object moves with velocity v.

In this time t, the object covers the displacement as s.

Here, the object is at rest initially hence initial velocity u= 0.

• As we applied force and object get displaced hence work has to be done.

This work done is given by,

W= F*s

According to the Newton’s second law of motion, force is given by

F=m*a

• And we know that, second equation of motion which is given by,

s= u*t + 1/2*at2

But, here initial velocity u=0

Thus, s= 1/2*a*t2

Hence, work done by the object is given by,

W= F*s

W= ma*1/2at2

Thus, W= 1/2m*(at)2

• Also, we know that first equation of motion is given by,

v= u + at

Here, u= 0

Hence, v= at

Thus, W= 1/2*m(at)2

W= 1/2*mv2

Thus, work done by the object is nothing but the amount of kinetic energy gained by that object while moving.

Hence, Kinetic energy= 1/2mv2

This is the expression for kinetic energy of the object.

c) Prove that the kinetic energy of freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.

Ans:

• If we have released any object of any mass from any height it directly falls on the ground due to the gravitational force of attraction, such falling of object under the action of acceleration due to gravity is called as free fall.
• Let we observe the kinetic energy and potential energy of freely falling body at different positions.
• Let us consider the body of mass m is released from a height h as we see in figure.
• From figure we can say that, point P is at a height h from the ground, point Q is at a distance x from the point P and Point R is on the ground which is at a distance of (h – x) from the point Q.

Now we discuss in detail the motion of body at points P, Q and R respectively.

At point P, the body is at rest hence its initial velocity u= 0.

As we earlier discussed,

Kinetic energy= 1/2m*u2

Since, u= 0

Thus, kinetic energy= 0

But, potential energy= mgh

Now, at point Q the velocity of the body is vQ and it covers distance x from the point P.

Hence, at point Q

Initial velocity u= 0, v= vQ,  s= x , a=g

We already know that the equation of motion is,

v2 = u2 + 2as

vQ2 = 0+ 2gx

Hence, kinetic energy at point Q is given by,

Kinetic energy= 1/2mvQ2 = 1/2m(2gx)= mgx

And at point Q, the body is at height (h-x) from the ground.

Hence, potential energy is given by,

Potential energy= mg(h-x)

Now, finally at the body reach to point R on the ground.

At point R,

Let vR be the velocity of body, s= h, a= g

And we know that the equation of motion is given by,

v2 = u2 + 2as

vR2 = 0 + 2gh

Thus, kinetic energy at point R is given by,

Kinetic energy= 1/2mvR2

= 1/2m(2gh)= mgh

Now, potential energy is given by,

At point R the height of body from ground is zero.

Hence, h= 0

Potential energy= 0

• Thus, from the above observations we conclude that, the potential energy of the body at point P is totally transformed to it’s kinetic energy at point R.
• Hence, we can say that the kinetic energy of freely falling object on reaching the ground the ground is nothing but the transformation of its initial potential energy.

d) Determine the amount of work done when an object is displaced at an angle of 30° with respect to direction the direction of the applied force.

Ans:

The amount of work done when an object is displaced by the displacement s at an angle of a, when force F is applied on it is given by,

W= F*s*Cosa

Here, the angle a= 30°

Hence, W= F*s*Cos(30°)

W= F*s*√3/2

This is the amount of work done for a given condition of motion.

e) If an object have zero momentum, does it have kinetic energy. Explain your answer.

Ans:

We know that, momentum of the object is the product of mass and it’s velocity.

If m is the mass of object and v is the velocity of the object then it’s momentum P is given by,

P=m*v

Given that, momentum o the object is zero.

P= 0

Hence, m*v= 0

But, mass of object never be zero in any condition.

Hence, v= 0

And, we know that the kinetic energy of the object is given by,

Kinetic energy= 1/2mv2

But, v= 0

Hence, kinetic energy= 0

Thus, we can say that if the momentum of the object is zero then it’s velocity is zero and hence kinetic energy is zero.

f) Why is the work done on an object moving with uniform circular motion zero.

Ans:

• We know that, in a uniform circular motion the body is moving with constant speed along a circular path as shown in figure.
• And we already know that, work is nothing but the product of force, displacement and cos of angle between them. Where force and displacement both are the vector quantities.
• And during the circular motion of body, the direction of velocity changes continuously with total displacement covered by the body in one cycle will be zero.
• Hence, we can say that work= force× displacement ×cosθ
• But when a particle perform circular motion then displacement is perpendicular to direction of force.

And we know that,

Cos(90)= 0

Thus work done vanishes.

• So we can say that, work done on an object moving with uniform circular motion is zero.

2.) Choose one or more correct alternatives.

1) For work to be performed, energy must be

a) transferred from one place to another

b) concentrated

c) transformed from one type to another

d) destroyed

Ans:

a) transferred from one place to another

2) Joule is the unit of

a) Force

b) Work

c) Power

d) Energy

Ans:

b) Work

d) energy

3) Which of the forces involved in dragging heavy object on a smooth, horizontal surface, have the same magnitude.

a) the horizontal applied force

b) gravitational force

c) reaction force in vertical direction

d) force of friction

Ans:

b) gravitational force

c) reaction force in vertical direction

d) force of friction

4) Power is the measure of

a) the rapidity with which work is done.

b) amount of energy required to perform the work

c) the slowness with which work is performed

d) length of time

Ans:

a) the rapidity with which the work is done.

5) While dragging or lifting an object, negative work is done by

a) the applied force

b) gravitational force

c) frictional force

d) reaction force

Ans:

b) gravitational force

c) frictional force

3.) Rewrite the following sentences using proper alternative.

1) The potential energy of your body is least when you are

a) sitting on the chair

b) sitting on the ground

c) sleeping on the ground

d) standing on the ground

Ans:

The potential energy of your body is least when you are sleeping on the ground.

2) The total energy of the object falling freely towards the ground

a) decreases

b) remains unchanged

c) increases

d) increases in the beginning and then decreases

Ans:

The total energy of the object falling freely towards the ground is remains unchanged.

3) If we increase the velocity of a car moving on a flat surface to four times it’s original speed, its potential energy

a) will be twice it’s original energy

b) will not change

c) will be four times it’s original energy

d) will be 16 times it’s original energy

Ans:

If we increase the velocity of a car moving on a flat surface to four times it’s original speed, it’s potential energy will not change.

4) The work done on an object does not depend on

a) displacement

b) applied force

c) initial velocity of the object

d) the angle between force and displacement

Ans:

The work done on an object does not depend on the initial velocity of the object.

4.) Study the following activity and answer the questions.

a) Take two aluminium channels of different lengths.

b) Place the lower ends of the channel on floor and hold their upper ends at the same height.

c) Now, take two balls of the same size and weight and release them from the top ends of the channel. They will roll down and cover the same distance.

Questions:

1) At the moment of releasing the balls which energy do the balls have?

2) As the balls role down which energy is converted into which other form of energy.

3) Why do the balls covers the same distance on rolling down?

4) What is the form of the eventual total energy of the balls?

5) Which law related to energy does the above activity demonstrate.

Ans:

1) At the moment of releasing the balls, they are having potential energy.

2) As the balls role down the potential energy of the ball is converted into kinetic energy.

3) As the balls are identical in shape, size and mass and they are released from same height their potential energy will be same. Hence, the velocity with which they come to ground will also be same. Hence, their distance covered after rolling is also same.

4) The total energy of the balls will be eventually kinetic energy.

5) The above activity shows the law of conservation of energy. When balls are released the total energy will be potential energy, when they rolls down posses both kinetic energy and potential energy. And when balls reach the ground the potential energy is totally converted into kinetic energy. This shows the law of conservation of energy.

5.) Solve the following examples.

a) An electric pump has 2KW power. How much water will the pump lift every minute to a height of 10m.

Ans:

Given that,

Power= 2KW

And we know that,

Power= work/time

Hence, work= power*time

In each minute i.e. 60 seconds the work done is,

W= p*t= 2*1000*60= 120000J

And we know, the work done here is stored in the form of potential energy of the water.

Potential energy= mgh= 120000J

Thus, m= 120000/9.8*10= 1224.5 kg

Thus, the pump lifts 1224.5 kg of water to a height of 10m in one minute.

b) If 1200 W electric iron is used daily for 30 minutes, how much total electricity is consumed in the month of April.

Ans:

Given that,

1200 W electric iron is used daily for 30 minutes.

In the month of April there are 30 days.

Hence, electricity consumed in the month of April will be,

E= P*T

E= 1200*30*60*30= 64.8*106 J

And all we know that,

1 unit= 3.6*106 J

Electricity consumed in units is given by,

E= 64.8*106/3.6*106= 18 units

Hence, the total electricity consumed in the month of April is 18units.

c) If the energy of the ball falling from a height of 10meters is reduced by 40%, how high it will rebound?

Ans:

Given that,

Height of the ball = 10 m

At height h the potential energy of the ball of mass m is given by,

Potential energy= mgh

Then, reduced potential energy of the ball is given by,

Reduced potential energy= 40% of mgh

= (100-40)/100 mgh

= 0.6*10= 6 m

Thus, the height of the ball when it rebounds is 6m.

d) The velocity of the car increases from 54km/hr to 72km/hr. How much is the work done if the mass of the car is 1500 kg.

Ans:

Given that,

Initial velocity v1= 54km/hr= 15m/s

final velocity v2= 72km/hr = 20m/s

Mass of car= 1500 kg

The work done by the car is nothing but the change in kinetic energy of the car.

W= 1/2m*v12 – 1/2m*v22

W= 1/2*1500*(400 – 225)

W= 750*175= 131250J

Thus, the work done by the car is 131250J.

e) Ravi applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by the Ravi?

Ans:

Given that,

Force applied= 10N

Displacement= 30cm = 0.3 m

Work done is given by,

W= force*displacement

W= 10*0.3= 3J

The work done by the Ravi will be 3J.

Updated: August 10, 2021 — 1:09 pm