AP State Board new syllabus AP Board Solutions Class 9 Physics 10th Lesson Work and Energy Questions and Answers.
AP 9th Class Physical Science 10th Lesson Questions and Answers Work and Energy
9th Class Physics 10th Lesson Work and Energy Questions and Answers (Exercise)
Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
a) Suma is swimming in a pond.
b) A donkey is carrying a load on its back.
c) A windmill is lifting water from a well.
d) A green plant is carrying out photosynthesis.
e) An engine is pulling a train.
f) Food grains are getting dried in the sun.
g) A sailboat is moving due to wind energy.
Answer:
Suma is swimming in a pond :
a) Suma pushes water backwards (Force) : She goes in forward direction. Displacement is opposite to force so work done is negative.
b) A donkey is carrying a load on its back : Displacement is perpendicular to force, so work done by the donkey is zero.
c) A wind – mill is lifting water from a well : Displacement is indirection of force, so work done is positive.
d) A green plant is carrying out photosynthesis : No work is done as it is a chemical process.
e) An engine is pulling a train : Force and displacement are in the same direction. So, work done is positive.
f) Food grains are getting dried in the sun : Drying of grains in sun doesn’t involve Force & displacement no work done.
g) A sailboat is moving due to wind energy: Wind displaces sailboat, so work done is positive.
Question 2.
An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object ?
Answer:
Here the displacement of the object and the gravitational force acting on the object are perpendicular to each other.
So, work done by the force of gravity on the object is zero.
Question 3.
A battery lights a bulb. Describe the energy changes involved In the process.
Answer:
The energy changes involved In the process are as follows:
Question 4.
Certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms-1. Calculate the work done by the force.
Answer:
Mass m = 20 kg; Initial velocity u = 5 ms-1
Question 5.
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Here the displacement of the object and the gravitational are perpendicular to each other, so work done is zero.
Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy ? Why ?
Answer:
- No, -it doesn’t violate law conservation energy.
- Some of Potential Energy (P.E) is converted into Kinetic Energy (K.E).
- Some of its energy is used to overcome air resistive force.
Question 7.
What are the various energy transformations that occur when you are riding a bicycle ?
Answer:
Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it ? Where is the energy you spend going ?
Answer:
When we push a huge rock.
- the energy is used to overcome the inertia of rock.
- some energy is used to overcome friction.
Question 9.
A certain household has consumed 250 units of energy during a month. How much energy is this in joules ?
Answer:
Energy consumed in a month = 250 units
= 250 kWh (1 unit = 1 kWh)
= 250 x 3.6 x 106 J = 900 x 106 J
= 9 x 108 J [1 kWh = 3.6 x 106 J]
Question 10.
An object of mass 40 kg is raised to a height of 5m above the ground. What is its potential energy ? If the object is allowed to fall, find its kinetic energy when it is half-way down.
Answer:
Object: Mass m = 40 kg; Height h = 5 m
Acceleration due to gravity g = 10 ms-2
Gravitational potential energy P. E = mgh
i) E = 40 x 10 x 5 = 2000J
ii) At half way K.E =
K.E = 1000J
Question 11.
What is the work done by the force of gravity on a satellite moviner round the earth? Justify your answer.
Answer:
The force of gravity acting on a satellite moves the satellite in a circular path. Since, the force of gravity acts at right angle to the displacement of the satellite (Figure), so work done,
W = Fs cos 90° = 0.
Question 12.
Can there be displacement of an object in the absence of any force acting on it ? Think. Discuss this question with your friends and teacher.
Answer:
- Yes, displacement is possible without the involvement of force.
- Celestial bodies move in the space without any external force acting on them.
Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not ? Justify your answer.
Answer:
Work = Force x Displacement.
No, work is done by the person as the force and displacement are perpendicular to each other.
Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours ?
Answer:
Energy consumed = workdone ; t = 10 hrs ; P = 1500 W = 1.5 kWh
Workdone = Power x time
⇒ W = Pt = 1.5 x 10 = 15 kWh
Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest ? What happens to its energy eventually ? Is it a violation of the law of conservation of energy ?
Answer:
When pendulumiraised from A to C due to change in position it acquires P.E.
- When we release the bob its P.E change to K.E and reaches A
- Due to the K.E it moves to ‘B’ and again acquire P.E.
- This motion gets repeated.
- During the transition some energy is utilised to overcome air resistive & gravitational force.
- Due to this it slows down and stops eventually.
- So, the energy is conserved.
Question 16.
An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest ?
Answer:
Object: Mass = m ; Velocity = v
To bring the object to rest the work done by the force would be equal and opposite.
W = \(-\frac{1}{2}\)mv2
Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h.
Answer:
Car: Mass m = 1500 kg
Initial velocity u = 60 kmh-1
[ ∵ kmh-1 to ms-1, Multiply with \(\frac{5}{18}\)
= 16.66 ms-1
As it comes to rest.
Final velocity v = 0
To stop the car – ve work must be done by the Force.
Question 18.
In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
Answer:
Work done = 0
Force ⊥ displacement
Work done = Positive
Force and displacement are in the same direction.
Work done = negative
Force and displacement are in opposite direction.
Question 19.
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her ? Why ?
Answer:
Yes, if the several force acting on them constitute a balanced force, then the net force is zero.
Then there will be no acceleration.
Question 20.
Find the energy in joules consumed in 10 hours by four devices of power 500W each.
Answer:
Devices: Power =500 W; No. of devices n = 4; Time of usage t=10 hrs
Total power
[∵ 1 kWh=1000 W]
Question 21.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy ?
Answer:
When an object falling freely
- some of its K.E is utilized to overcome air resistive force.
- some of the energy is converted to sound & heat energy.
- the rest is absorbed by the earth.
9th Class PS 10th Lesson Questions and Answers (InText)
Question 1.
A force of 7N acts on an object. The displacement is say 8m, in the direction of the force (figure). Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer:
Force F = 7N ; Displacement s = 8m
Work done = Force x displacement ⇒ W=Fxs = 7×8 = 56J
Question 2.
When do we say that work is done ?
Answer:
When a force acts on an object and the object moves, we say that the force has done the work on the object.
Question 3.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:
Work done = Force x displacement ⇒ W = Fs
Question 4.
Define 1 J of work.
Answer:
When a force of 1 Newton acts on an object causing a displacement of 1 metre.
Question 5.
A pair of bullocks exerts a force of 140 N on a plough. The held being ploughed is 15 m long. How much work is done in ploughing the length of the field ?
Answer:
Bullocks (Pair): Force exerted by them F = 140 N
Distance ploughed by them s = 15m
W = Fs = 140 x 15 = 2100 J
2100J is the work done by pair of bullocks for ploughing a distance of 15m.
Question 6.
What is the kinetic energy of an object ?
Answer:
Kinetic Energy : The energy of an object by virtue of its motion is called Kinetic Energy.
Question 7.
Write an expression for the kinetic energy of an object.
Answer:
Expression for kinetic energy :
Let us consider a block of mass ‘m’.
- As it is at Rest.
- It’s Initial velocity u = 0.
- Let a force ‘F’ acting on the block.
- Moved with a velocity V from A to B, travelled a distance AB.
- According to Newton’s 2nd law,
Question 8.
The kinetic energy of an object of mass, m moving with a velocity of 5 ms-1 is 25 J. What will be its kinetic energy when its velocity is doubled ? What will be its kinetic energy when its velocity is increased three times ?
Answer:
Question 9.
What is power?
Answer:
Rate of doing work (or) rate of transfer on energy is called Power.
Question 10.
Define 1 watt of power.
Answer:
If work is done at the rate of 1 Joule per second, then the power delivered is 1 watt.
1 watt = \(\frac{1 \text { joule }}{\text { second }} ; P=\frac{W}{t}\)
Question 11.
A lamp consumes 1000J of electrical energy in 10 s. What is its power?
Answer:
Electrical energy consumed by lamp =1000J
Time t = 10 sec
Power P = \(\frac{W}{t}\)=\(\frac{1000}{10}\) = 100 W
Question 12.
Define average power.
Answer:
Total energy consumed by the total time taken is considered as average power.
\(P_{\text {Avg }}=\frac{\text { Total energy }}{\text { Total time }}\)
Examples
Question 1.
A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force (Figure). If the force acts on the object all throughout the displacement. Find the work done by the force.
Answer:
Given data, Force (F) = 5N ; Displacement (s) = 2m
Workdone (W) = Force (F) x Displacement (s)
W = F x s = 5N x 2m = 10 J = W = 10 J
Hence, the work done by the force is 10 J.
Question 2.
A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.
Answer:
Mass of luggage, m = 15 kg and displacement, s = 1.5 m
Work done, W = F x s = mg x s
= 15 kg x 10 ms-2 x 1.5 m = 225 kg ms-2 m = 225 N m = 225 J
Work done is 225 J.
Question 3.
An object of mass 15 kg is moving with a uniform velocity of 4 ms-1. What is the kinetic energy possessed by the object ?
Answer:
Mass of the object, m = 15 kg, velocity of the object, v = 4 ms-1
From Eq. Ek = \(\frac{1}{2}\) mv2
= \(\frac{1}{2}\) x 15 kg x 4 ms-1 x 4 ms-1 = 120 J
The kinetic energy of the object is 120 J.
Question 4.
What is the work to be done to increase the velocity of a car from 30 km h-1 to 60 km h-1 if the mass of the car is 1500 kg ?
Answer:
Mass of the car, m = 1500 kg
initial velocity of car, u=30 km h-1 = \(\frac{30 \times 1000 \mathrm{~m}}{60 \times 60 \mathrm{~s}}\) =25/3 ms-1
Similarly, the final velocity of the car, v =60 kmh-1 = 50/3ms-1
Therefore, the initial kinetic energy of the car,
Ekd = \(\frac{1}{2}\) mu2 = \(\frac{1}{2}\) × 1500 kg × (25/3 ms-1)2
=156250/3 J
The final kinetic energy of the car, Ef = \(\frac{1}{2}\) × 1500 kg×(50 / 3ms-1)2
= 625000 / 3 J.
Thus, the work done = Change in kinetic energy
= EKf -EKi = 156250 J
Question 5.
Find the energy possessed by an object of mass 10 kg when it is at a height of 6 m above the ground. Given, g =9.8 ms-2
Answer:
Mass of the object, m =10kg,
Displacement (height), h=6m and acceleration due to gravity, g = 9.8ms-2
From Eq. Potential energy = mgh
Potential energy = mgh=10 kg × 9.8 ms-2 × 6m =588J.
The potential energy is 588 J.
Question 6.
An object of mass 12 kg is at a certain height above the ground. If the potential energy of the object is 480 J, find the height at which the object is with respect to the ground. Given, g=10 ms-2.
Answer:
Mass of the object, m=12 kg, potential energy, Ep=480J
Ep=mgh
480 J = 12 kg× 10 ms-2 × h ⇒ h = \(\mathrm{h}=\frac{480 \mathrm{~J}}{120 \mathrm{~kg} \mathrm{~ms}^{-2}}\) = 4m
The object is at the height of 4m
Question 7.
Two girls, each of weight 400 N climb up a rope through a height of 8 m. We name one of the girls A and the other B. Girl A takes 20 s while B takes 50 s to accomplish this task. What is the power expended by each girl?
Answer:
i) Power expended by girl A:
Weight of the girl, mg = 400N; Displacement (height), h=8 m; Time taken, t=20s
From Eq. power
ii) Power expended by girl B :
Weight of the girl, mg=400N; Displacement (height), h=8m; Time taken, t=50s
Power expended by girl A is 160 W
Power expended by girl B is 64 W
Question 8.
A boy of mass 50 kg runs up a staircase of 45 steps in 9s. If the height of each step is 15 cm, find his power. Take g =10 ms-2
Answer:
Weight of the boy, mg=50 kg × 10 m s-2 =500 N
Height of the staircase, h =45 × 15 / 100 m =6.75 m;
Time taken to climb, t = 9s
From Eq. Power,
Power is 375 W
9th Class Physical Science Chapter 10 Questions and Answers (Lab Activities)
(Page No. 54) Activity 10.1
Question 1.
Kamali is preparing for examinations. She spends lot of time in studies. She reads books, draws diagrams, organises her thoughts etc.,
You are working hard to push a huge rock. Let us say the rock does not move despite all the effort.
You climb up the steps of a staircase and reach the second floor of a building just to see the landscape from there.
We have discussed in the above paragraphs a number of activities which we normally consider to be work in day-to-day life. For each of these activities, ask the following questions and answer them :
i) What is the work being done on ?
Answer: Force.
ii) What is happening to the object ?
Answer: Object is undergoing some change.
iii) Who (what) is doing the work?
Answer: Force
(Page No. 56) Activity 10.2
Question 2.
- Think of some situations from your daily life involving work.
- List them.
- Discuss with your friends whether work is being done in each situation.
- Try to reason out your response.
- If work is done, which is the force acting on the object?
- What is the object on which the work is done ?
- What happens to the object on which work is done ?
Answer:
(Page No. 56) Activity 10.3
Question 3.
- Think of situations when the object is not displaced in spite of a force acting on it.
- Also think of situations when an object gets displaced in the absence of a force acting on it.
- List all the situations that you can think of for each.
- Discuss with your friends whether work is done in these situations.
Answer:
1) Situations where an object is not displaced inspite of a force acting on it :
- when a book is placed on a table.
- a heavy box placed on a rough surface may not move because the force of friction is greater than the applied force.
2) Situations, where an object gets displaced in the absence of a force acting on it :
- Some objects, such as a rolling ball or a falling apple, may move without the application of an external force. This is due to the object’s inherent properties, such as its shape or weight, that cause it to move.
- A ball rolling on a smooth surface will continue to roll for a short distance even after the external force that set it in motion is removed.
(Page No. 58) Activity 10.4
Question 4.
– Lift an object up. Work is done by the force exerted by you on the object. The object moves upwards. The force you exerted is in the direction of displacement. However, there is the force of gravity acting on the object.
1) Which one of these forces is doing positive work ?
2) Which one is doing negative work ?
– Give reasons.
Answer:
1) Muscular Force has done the positive work.
2) Gravitational Force has done negative work.
(Page No. 60) Activity 10.5
Question 5.
- A few sources of energy are listed above. There are many other sources of energy. List them.
- Discuss in small groups how certain sources of energy are due to the Sun.
- Are there sources of energy which are not due to the Sun?
Answer:
1) There are many other sources of energy besides those listed, including: Geothermal energy – energy harnessed from heat within the earth’s crust. Hydroelectric power – energy generated from the movement of water. Nuclear power – energy generated through nuclear reactions. Biomass energy – energy from organic matter such as wood, crops and waste. Wind power – energy generated from the wind’s movement.
2) Most of the sources of energy on Earth are ultimately derived from the Sun. For example, solar energy is directly harnessed from the Sun’s radiation, while biomass energy is the result of photosynthesis, which converts solar energy into chemical energy stored in organic matter. Wind power is also indirectly linked to the Sun, as differences in atmospheric pressure caused by uneven heating from the Sun create wind.
3) However, there are some sources of energy that are not directly linked to the Sun. For example, geothermal energy, tidal power etc.
(Page No. 62) Activity 10.6
Question 6.
Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from height of about 25 cm. The ball creates a depression.
- Repeat this activity from heights of 50 cm, 1m and 1.5 m.
- Ensure that all the depressions are distinctly visible.
- Mark the depressions to indicate the height from which the ball was dropped.
- Compare their depths.
- Which one of them is deepest ?
- Which one is shallowest? Why ?
- What has caused the ball to make a deeper dent ?
- Discuss and analyse.
Answer:
1) Ball dropped from 1.5 m height create deepest depression on sand.
2) Ball dropped from 25 cm is the shallowest, it is dropped from a very small height.
3) Height [greater the height, greater the depression].
(Page No. 62) Activity 10.7
Question 7.
- Set up the apparatus as shown in the adjacent figure.
- Place a wooden block of known mass in front of the trolley at a convenient fixed distance.
- Place a known mass on the pan so that the trolley starts moving.
- The trolley moves forward and hits the wooden block.
- Fix a stop on the table in such a manner that the trolley stops after hitting the block. The block gets displaced.
- Note down the displacement of the block. This means work is done on the block by the trolley as the block has gained energy.
– From where does this energy come?
Answer: The trolley got energy due to its motion.
– Repeat this activity by increasing the mass on the pan. In which case is the displacement more?
Answer: We can observe that when mass increases the displacement also increases.
– In which case is the work done more?
Answer: Work – done will be more when we increase the mass on the pan.
– In this activity, the moving trolley does work and hence it possesses energy.
(Page No. 66) Activity 10.8
Question 8.
1) Take a rubber band.
2) Hold it at one end and pull from the other. The band stretches.
3) Release the band at one of the ends.
4) What happens?
Answer: The band comes to its original shape.
5) The band will tend to regain its original length. Obviously the band had acquired energy in its stretched position.
6) How did it acquire energy when stretched ?
Answer: Rubber band got energy due to change in its shape.
(Page No. 66) Activity 10.9
Question 9.
1) Take a slinky as shown in the adjacent figure.
2) Ask a friend to hold one of its ends.
You hold the other end and move away from your friend. Now you release the slinky.
3) What happened?
Answer: The slinky comes back and retains its original shape. Due to change in its shape.
4) How did the slinky acquire energy when stretched ?
Answer: When we stretch the slinky we apply a force on it which does work on the slinky.
5) Would the slinky acquire energy when it is compressed ?
Answer: Yes, it acquires energy when it is compressed.
(Page No. 66) Activity 10.10
Question 10.
1) Take a toy car. Wind it using its key.
2) Place the car on the ground.
3) Did it move?
Answer: Yes. It moves.
4) From where did it acquire energy?
Answer: It acquires energy from our body while winding the key.
5) Does the energy acquired depend on the number of windings ?
Answer: Yes, the energy acquired depend on the number of windings.
6) How can you test this?
Answer: When we wind the key lesser no.of turns it goes small distance.
If we wind the key greater (full) it goes greater distance.
This is due greater change in shape of the windings.
(Page No. 66) Activity 10.11
Question 11.
1) Lift an object through a certain height. The object can now do work. It begins to fall when released.
2) This implies that it has acquired some energy. If raised to a greater height it can do more work and hence possesses more energy.
3) From where did it get the energy? Think and discuss.
Answer: It got energy due to change in its position (height).
(Page No. 66) Activity 10.12
Question 12.
- Take a bamboo stick and make a bow as shown in the adjacent figure.
- Place an arrow made of a light stick on it with one end supported by the stretched string.
- Now stretch the string and release the arrow.
- Notice the arrow flying off the bow.
- Notice the change in the shape of the bow.
- The potential energy stored in the bow due to the change of shape is thus used in the form of kinetic energy in throwing off the arrow.
Self – Interpretation
(Page No. 66) Activity 10.12
Question 12.
- Take a bamboo stick and make a bow as shown in the adjacent figure.
- Place an arrow made of a light stick on it with one end supported by the stretched string.
- Now stretch the string and release the arrow.
- Notice the arrow flying off the bow.
- Notice the change in the shape of the bow.
- The potential energy stored in the bow due to the change of shape is thus used in the form of kinetic energy in throwing off the arrow.
Self – Interpretation
(Page No. 70) Activity 10.14
Question 14.
1) Many of the human activities and the gadgets we use involve conversion of energy from one form to another.
2) Make a list of such activities and gadgets.
3) Identify in each activity/gadget the kind of energy conversion that takes place.
Answer:
1) Cell phone – Chemical energy – Electrical energy.
2) Mixer / Grinder – Electrical energy – Mechanical energy.
3) Speakers – Electrical energy – sound energy
(Page No. 72) Activity 10.15
Question 15.
An object of mass 20 kg is dropped from a height of 4m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case. For simplifying the calculations, take the value of g as 10 ms-2
Height at which object is located
m |
Potential energy
(EP = mgh)
J |
Kinetic energy
(Ek = mv/2)
J |
EP + EK
J |
4 |
|
|
|
3 |
|
|
|
2 |
|
|
|
1 |
|
|
|
Just above the ground |
|
|
|
Answer:
Height at which object is located |
Potential energy
(EP = mgh)
J |
Kinetic energy
(Ek = mv/2)
J |
EP + EK
J |
4 |
800 |
0 |
800 + 0 = 800 |
3 |
600 |
200 |
600 + 200 = 800 |
2 |
400 |
400 |
400 + 400 = 800 |
1 |
200 |
600 |
200 + 600 = 800 |
ground |
0 |
800 |
0 + 800 = 800 |
(Page No. 72) Activity 10.16
Question 16.
1) Consider two children, say A and B. Let us say they weigh the same. Both start climbing up a rope separately. Both reach a height of 8 m. Let us say A takes 15s while B takes 20s to accomplish the task.
2) What is the work done by each?
Answer: Work done by both of them is same.
3) The work done is the same. However, A has taken less time than B to do the work.
4) Who has done more work in a given time, say in 1s?
Answer: Work done By A & B is W = mg but A has taken less time, so ‘A’ does more work.
(Page No. 74) Activity 10.17
Question 17.
1) Take a close look at the electric meter installed in your house. Observe its features closely.
2) Take the readings of the meter each day at 6.30 am and 6.30 p.m.
3) Do this activity for about a week.
4) How many ‘units’ are consumed during the day time?
5) Tabulate your observations.
Last Meter Reading |
First Meter Reading |
No.of units Consumed per day |
|
|
|
6) Now, multiply no. of units per day × 30.
7) You can get total units consumed for a month.
8) Compare this with electricity bill you received.
Self Interpretation.