Chhattisgarh State Board Class 9 Science Chapter 12 Work and Energy Exercise Multiple Choice, Fill in the Blanks, Questions and Answers here.

**Chhattisgarh State Class 9 Science Chapter 12 Solution**

**1) Pick an appropriate option**

**(i) If an object displaces 2 m in the directing of 10 N applied force, the work done on the object**

**by the force is,**

**(a) 15 J (b) 20 J (c) .20 J (d) 5 J.**

Ans: – option (b) 20j.

**(ii) A body of mass 5 kg. is in a constant motion at 2 m/s. What will be its kinetic energy?**

(a) 10 Joule (b) 15 Joule (c) 5 Joule (d) 20 Joule.

Ans: – option (a) 10j.

**(iii) A body of 12 kg. mass is placed at a certain height from the surface of earth. If its potential energy is 480 J, then what will be its height w.r.t. to earth’s surface?**

(a) 6 meter (b) 9 meter (c) 5 meter (d) 4 meter.

Ans: – option (d) 4 meters.

**(iv) You burn a bulb of 100 w daily for 5 hours. What will be the energy spent in units by the?**

**bulb in one day?**

(a) 0.4 unit (b) 0.5 unit (c) 0.05 unit (d) 0.01unitt.

Ans: – option (b) 0.5 unit.

2) Fill in the blanks-

(i) SI unit of work is …..joule/sec……………

(ii) 1 kwh is equivalent to ……..36×10^5………….. joules.

(iii) The total energy of an object remains …. constant…………..

(iv) If the direction of applied force of an object and its displacement are opposite then its work will be ……….. negative…………..

**3) What do you understand by kinetic energy? Establish the equation for kinetic energy for an objectin motion.**

Ans: – The work done of a body due to its motion is known as the kinetic energy.

An object of mass ‘m’, applying force ‘F’, the object displaces a distance ‘S’. Then, the work done on the object is

W = FS …………. (i)

Let the acceleration created on the body be ‘a’ due so the applied force. Following is a relation between the constant acceleration ‘a’, initial velocity is ‘u’, final velocity is ‘v’ and displacement is ‘s’.

V^2-u^2= 2as ……………. (ii)

s =v^2 – u^2/2a……………. (iii)

We know from the second law of motion,

F = ma ……………. (iv)

Putting the value of S and F from eq. (iii) and (iv) into eq. (i),

We can write the work done as: –

W=ma (v^2 – u^2)/2a

W=1/2mv^2.

As it starts from rest.

If this work done is equal to the kinetic energy, then,

Ek=1/2×mv^2.

**4) The energy consumed at a house is 250 units in one month. How much will it be in Joules?**

Ans: – As we all know 1kwh=3.6×10^6 j.

So, 250 unit = 250 kwh.

Or, 250 kwh= 250×3.6×10^6 = 9×10^8 j.

**5) (a) What is energy conservation law? Explain.**

Ans: – The energy conservation law States that every can never be created nor be destroyed. The energy just inly transforms from one form to another form. Like the for solar panel the solar energy transfer to electrical energy.

**(b) The potential energy of a freely falling body drops gradually. Does it contradict the law ofenergy conservation? Explain with reasons.**

Ans: – The potential energy of freely falling body is equal to the multiplication of its masses, acceleration due to gravity, height. So, the potential energy varies with the changing of height. And by this phenomenon the laws of energy conservation don’t violates.

**6) Explain potential energy and establish its equation.**

Ans: – The stored energy in an object while doing work against the gravitational acceleration is known as the potential energy of that object.

some force is needed to raise an object of mass m upwards; the minimum required force will be equal to the weight mg. Suppose that thework needed against gravitational force to raise the object to a height h is ‘w’.

Then, work done w = Force × displacement

= mgh

Since the work done on the object is mgh, so the potential energy of the object is also mgh. We willcall this potential energy Ep

Ep= mgh.

**7) (a) If the velocity of a particle is doubled, what will be its kinetic energy?**

Ans: – As the kinetic energy is directly proportional to the square of velocity of the particles. So, if this velocity is double then the kinetic energy will be four times of the previous one.

** (b)If the work done on the particle is zero, then what will be its velocity?**

Ans: – As the work done is zero then there will be no displacement of that object. So, if there is no displacement velocity will be zero.

**8) On applying brakes of a car which is moving at a speed v, it stops after covering a distance d.Calculate, what will be the distance covered by the car after applying brakes if it was moving at aspeed of 2 v.**

Ans: – As we all know, v^2 = u^2 + 2as;

Here u =initial velocity= v; v= final velocity=0; s=d;

Then, 0=v^2 + 2ad; or, a= -(v^2 / 2d) ;

In the second case,

V=0; u=3v, a=v^2 / 2d, S=?;

So, 0=9v^2 -2×a×S;

Or, S= 9v^2 / (2×v^2 /2d) = 9d.

The distance covered by the car applying break is 9d.

**9) A man holds a sack of rice on his head for 30 minutes and gets tired. Has he done any work?Explain your answer with reasons.**

Ans: – The work done is equal to the multiplication of force acting on and displacement. So, in this case as the man doesn’t move after holding it on his head there will be no displacement at all. So,there will be no work done.

**10) On applying a force of 8 Newton on a body, it moves in the direction of motion and displaces for 4m. Calculate the work done.**

Ans: – The work done (w) = Force × distance = F.S.

Here F= 8N, S=4m.

So, work done= 8×5 = 40N/sec.

**11) A body of 10 kg. mass is raised against the gravitational force of earth to a height of 10 m. Howmuch work is done in this case?**

Ans: – The gravitational potential = mgh;

Here, m=mass = 10kg, height=h= 10m, g=10 m/s^2;

So potential energy= 10 × 10 × 10 = 10^3j .

**12) Two body of masses 10 kg. and 15 kg. are raised to a height of 5 m. and 2 m. respectively abovethe surface of earth. Calculate the change in their potential energies.**

Ans: – Potential energy= mgh; where m is mass, h is height and g are gravitational potential.

So the PE of first = 10×5×10=500j.

And potential energy of second= 15×2×10=300j.

So, the change in potential is = 500 – 300 = 200;

**13) By using the information given in the figure calculate the velocity of the ball at the position D. (use thelaw of conservation of mechanical energy).**

Ans: –

**14) On applying a force of 15 N for 6 sec. a man displaces a box by 8 m. Calculate his power.**

Ans: – The power= work done / time;

Here work done= force× displacement = 15×8 = 120;

Time= 6sec.

So, power= 120/6 = 20 watt.

**15) To change the energy of a body of 0.5 kg. by 1 joule, we will have to raise it to what height? (g = 10m/s2).**

Ans: – As we all know that potential energy= mass× height× gravitation.

Mass= 0.5kg, height=?; gravitation=g=10m/s^2. PE=1J.

So, 1= 0.5× 10× H,

H= 1/5 = 0.2 meter.