Work and Energy Chapter 11 Class 9 Science Assignments

Please refer to Work and Energy Chapter 11 Class 9 Science Assignments below. We have provided important questions and answers for Work and Energy which is an important chapter in Class 9 Science. Students should go through the notes and also learn the solved assignment with solved questions provided below. All examination and class tests questions are as per the latest syllabus and books issued by CBSE, NCERT, and KVS. We have also provided Class 9 Science Assignments for all chapters on our website.

Chapter 11 Work and Energy Class 9 Science Assignments

Question. The potential energy of a body is 39600 J. How high is the body if its mass is 20 kg?

Question. How much work is done by a force of 10 N in moving an object through a distance of 4 m in the direction of the force?

Question. What is power? How do you differentiate kilowatt from kilowatt hour?

Question. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?

Question. Can energy be destroyed? Can energy be created?

Question. A cell converts one form of energy into another. Name the two forms.

Question. Calculate the work done in lifting 200 kg of water through a vertical height of 6 m.

Question. What will happen to the kinetic energy of a body if its mass is doubled?

Question. What will happen to the kinetic energy of a body if its velocity is halved?

Question. By how much will the speed of a body, of fixed mass, increase if its kinetic energy becomes four times its initial kinetic energy?

Question. Give one example each of potential energy (i) due to position (ii) due to shape.

Question. What kind of energy transformation takes place when a body is dropped from a certain height?

Question. Can kinetic energy of a body be negative?

Question. When is 1 joule of work said to be done?

Question. What is the energy of a body due to its motion called?

Question. What is the SI unit of kinetic energy?

Question. How does the kinetic energy of a body change if the mass of the body is halved?

Question. A car and a truck are moving with the same velocity of 60 km/hr^–1, which one has more kinetic energy?

Question. What is the relationship between kilowatt and watt?

Question. What is meant by the term horsepower (hp)?

Question. Does work done depend upon the velocity of the body.

Question. State the law of conservation of energy.

Question. In a tug-of-war one team gives way to the other. What work is being done and by whom?

Question. (a) How much work is done when a force of 1 N moves a body through a distance of 1 m in its direction?
(b) Is it possible that a force is acting on a body but still the work done is zero? Explain giving one example.

Question. What will cause greater change in kinetic energy of a body? Changing its mass or changing its velocity?

Question. A spring which is kept compressed by tying its ends together is allowed to be dissolved in an acid. What happens to the potential energy of the spring?

Question. At what rate is electrical energy consumed by a bulb of 60 watt?

Question. Name the energy present in an object due to its position or configuration.

Question. Give one example of potential energy due to position.

Question. Give an example of potential energy due to change in shape.

Question. What type of energy is possessed by a flying bird and a flying aeroplane?

Question. Does the potential energy of a spring increase or decrease when it is compressed?

Question. A spring is compressed, what change is expected in the potential energy of the spring

Question. What is the amount of work done by a force when a

Question. Name the common forms of the mechanical energy.

Question. What is the relationship between megawatt and watt?

Question. What is the relationship between megawatt and joules per second?

Question. A spring which has been kept compressed by tying its ends together is allowed to be dissolved in an acid.What happens to the potential energy of the spring?

Question. Is it possible that a body be in accelerated motion under a force acting on the body, yet no work is being done by the force? Explain your answer giving a suitable example.

Question. What is the work done on a body moving in a circular path?

Question. Does every change in energy of the body involve work?

Question. A body moves along a circular path. How much work is done in doing so? Explain.

Question. When a constant force is applied to a body moving with constant acceleration, is the power of the force constant? If not, how would force have to vary with speed for the power to be constant?

Question. Name the form of energy associated in each case :
(i) A flying bird.
(ii) A man climbing the stairs.
(iii) A compressed watch spring.
(iv) A fast moving object.
Answer.
(i) Mechanical energy
(ii) Mechanical energy
(iii) Potential energy
(iv) Kinetic energy

Question. Define : (a) power (b) work done (c) kinetic energy.Give SI unit of each.
Answer.
(a) The rate of doing work is called power. Its SI unit is watt.
(b) Work is the product of force and displacement. Its SI unit is joule.
(c) It is the energy possessed by a body by virtue of its motion. Its SI unit is joule.

Question. Define power. Write commercial unit and SI unit of electrical energy. An electrical geyser of 1.5 kW works for 2 hours. Find the electrical energy units consumed in a day.
Answer. Power is defined as the rate of doing work. SI unit is joule and kWh is the commercial unit of electrical energy.
Given, P = 1.5 kW,
t = 2 hours,
E = P × t = 1.5 × 2 = 3 kWh

Question. The masses of scooter and bike are in the ratio of 2 : 3 but moving with same speed of 108 km h^–1. Compute the ratio of their kinetic energy.
Answer. The energy possessed by a body by virtue of its motion.
Given, m₁/m₂,2/3 the ratio of KE is equal to the ratio of their masses if they have the same velocity, therefore, ratio of their KE is also 2 : 3.

Question. Name the various forms of energy.
Answer.
The various forms of energy are :
(i) Potential energy
(ii) Kinetic energy
(iii) Mechanical energy
(iv) Heat energy
(v) Chemical energy
(vi) Electrical energy
(vii) Light energy.

Question. (a) What is meant by potential energy? Is potential energy vector or scalar quantity?
(b) Give one example of a body having potential energy.
Answer.
(a) The energy possessed by a body by virtue of its position or configuration. It is a scalar quantity.
(b) Stretched string of a bow.

Question. When is the work done by a force said to be negative?Give one situation in which one of the forces acting on the object is doing positive work and the other is doing negative work.
Answer.
We know that work done W = Fs cos q , where q is the angle between F and S. Clearly, W will be –ve, if q is between 90° and 180° because then cos q will be –ve. Consider the case of a body falling under gravity. The body experiences an upward frictional force and downward force due to gravity. Since, the body is moving downwards, the work done by force to gravity will be +ve but that is against the upward thrust will be –ve.

Question. Is it possible that a body is in accelerated motion tinder a force acting on the body, yet no work is being done by the force? Explain your answer giving a suitable example.
Answer.Yes, it is possible, when the force is perpendicular to the direction of motion. The Moon revolving round the Earth under the centripetal force of attraction of the Earth but Earth does not do any work on the motion of the Moon.

Question. Define work. How is work measured? When is work done by a force negative?
Answer.Work is said to be done if force acting on an object displaces it through a certain distance.
It is measured as the product of force and displacement.Work done is negative if force and displacement are in the opposite direction.

Question. An object of mass when raised to height h possess a potential energy of 1200 J. Find the new potential energy :
(a) if the same object is raised to height h/4 .
(b) if the same object is raised to height 4h.
Answer.
PE = 1200 joules
(a) New PE = 1/4 old PE = 1/4 × 1200 = 300 joules
(b) New PE = 4 × old PE = 4 × 1200 = 4800 joules

Question. Explain that the flying bird has; potential and kinetic energy and give their expressions.
Answer.
The potential energy of the bird while flying in the sky is with respect to the Earth. The KE is due to its velocity with which it is flying.
PE = mgh and KE = 1/2 mv²

Question. (a) An arrow moves forward when released from a stretched bow. Explain the transformation of energy in the process.
(b) A boy of mass 50 kg climbs up a vertical height of 100 m. Calculate the amount of potential energy he gains.
Answer.
(a) When the bow is stretched it stores potential energy. When the arrow is released the potential energy stored in the bow gets transformed into the kinetic energy of the arrow.
(b) Given m = 50 kg, h = 100 m,
g = 10 ms^–1, PE = ?
PE = mgh = 50 × 10 × 100 = 5000 J

 Question. What are the factors on which the work done depends?
Answer.
The work done by a force depends upon :
(i) The magnitude of the force.
(ii) The magnitude of the displacement.
(iii) The angle between force and displacement.

Numerical Questions

Question. The heart does 1.5 J of work in each heartbeat. How many times per minute does it beat if its power is 2 watt?
Answer.
Total work = P × t = 120 J,
Number times heartbeat in 1 min.
= Total work done/Work done in each beat
= 120/1.5 = 80 times

Question. Calculate the time taken by 60 W bulb to consume 3000 J of energy.
Answer.
Power = 60 W and Energy consumed = 3000 J
We know that
Power = Energy/Time Taken
Time Taken = Energy Consumed/Power
= 3000/60= 50 sec

Question. A horse exert a force of 200N to pull the cart. If the horse cart system moves with velocity 36 kmh^–1 on the level road, then find the power of horse in term of horse power (1 HP = 746 W).
Answer.
Velocity = 36 kmh^–1 = 10 m/s
W =F × s = 200 × 10 = 2000J
P=W/ t
= 2000J/1sec = 2000 W
746 W = 1 HP
So, 2000 W = 2000/746= 2.68 HP

Question. An electric kettle of 500W is used to heat water everyday for 2 hours. Calculate the number of unit of electrical energy consumed by it in 10 days.
Answer.
E = Pt = 500 W × 10 × 2h
= 10000 Wh
= 10 kWh = 10 unit

Question. Calculate the cost of using a 2 kWh immersion rod for heating water in a house for one hour each day for 60 days if the rate is 1.50 per unit kWh.
Answer.
E = Pt = 2 kW × 60 × 1 h
= 120 kWh = 120 unit
The cost of using a 2 kWh immersion rod for heating
water = 120 × 1.5 = Rs 180

Question. In an experiment to measure his power, a student records the time taken by him in running up a flight of steps on a staircase.
Answer. Use the following data to calculate the power of the student :
Number of steps = 28,
Height of each step = 20 cm,
Time taken = 5.4 s,
Mass of student = 55 kg,
Acceleration due to gravity = 9.8 ms^–2
P=W/ t = mgh/t
[55×9.8× (28×0. 20)]/5.4 
= 559 J

Question. A bullet of mass 15 g has a speed of 400 m/s. What is its kinetic energy? The bullet strikes a thick target and is brought to rest in 2 cm, calculate the average net force acting on the bullet. What happens to kinetic energy originally in the bullet?
Answer.
K.E. = 1/2 mv²
= 0.5 × 0.015 kg × (400 × 400)
= 1200 J.
Work done = Change in K.E.
As final velocity = 0
So, change in KE = K_f – K_i = 1200 J
Therefore, F × d = 1200
(where F is the average force.)
F=1200/2×10^-2
= 6 × 10⁴N.
The kinetic energy is eventually converted to heat energy.

Question. The power of a heart which beats 72 times in a minute is 1.2 kW. Calculate the work done by heart for each beat. (1 kJ)
Answer.
P = 1200 W and t = 60 s
W = P × t = 1200 × 60 = 72000J
In 72 times heartbeats 72000 J energy used
In 1 beat = 72000/72= 1000J
Work done by the heart in every beat is 1 KJ.

Question. When loading a truck, a man lifts boxes of 100 N each through a height of 1.5 m.
(a) How much work does he do in lifting one box?
(b) How much energy is transferred when one box is lifted?
(c) If the man lifts 4 boxes per minute, at what power is he working? (g = 10 m s^–2)
Answer.
(a) Work done in lifting one box = F × d = 100 × 1.5
= 150 J.
(b) W = E = 150 J.
(c) Power = Work done/ Time 60
= (150×4)/4 = 10 W

Question. (a) Define average power.
(b) A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
(c) Give the commercial unit of energy in joules.
Answer.
(a) Average power is defined as the ratio of total energy consumed to the total time taken.
(b) P=E/ t
= 1000/10 = 100 W
(c) 1 kWh = 3.6 × 106 J

Question. Calculate the electricity bill amount for a month of 31 days, if the following devices are used as specified.
(a) 3 bulbs of 40 W for 6 hours.
(b) 4 tubelights of 50 W for 8 hours,
(c) A TV of 120 W for 6 hours.
The rate of electricity is Rs 2.50 per unit.
Answer.
E_l = P × t = 0.04 × 6 × 3
= 0.72 kWh
E₂ = P × t = 0.05 × 8 × 4
= 1.60 kWh
E₃ = 0.12 × 6 = 0.72 kWh
Total E = 0.72 + 1.6 + 0.72 = 3.04 kWh
Cost in 31 days = rate × E
= 3.04 × 2.50 × 31 = Rs 235.60

Question. (a) What is meant by mechanical energy? State its two forms. State the law of conservation of energy.Give an example in which we observe a continuous change of one form of energy into another and vice-versa.
(b) Calculate the amount of work required to stop a car of 1000 kg moving with a speed of 72 km/h.
Answer.
(a) It is the sum of KE and PE of an object. It states that energy can neither be created nor be destroyed. We observe a continuous change in energy in a simple pendulum. At the mean position, the energy is totally kinetic while at the extreme position it is totally potential. As the pendulum oscillates its energy continuously changes between kinetic and potential.
(b) Given m = 1000 kg, u = 72 kmh^–1 = 20 ms^-1,
v = 0
Work done = change in kinetic energy

= – 200000 J = – 2 × 10⁵ J

Question. (a) State the law of conservation of energy.
(b) What is the work done to increase the velocity of a car from 36 km h^–1 to 72 kmh^–1 if the mass of the car is 1500 kg? Does the work done by the force have a negative or a positive magnitude?
(c) Where does an oscillating pendulum have maximum PE and RE?
Answer.
(a) It states that energy can neither be created nor be
destroyed.
(b) Given, m = 1500 kg, u = 36 kmh–1 = 10 ms^–1,v = 72 k mh^–1 = 20 ms^–1, W = ?
Work done = change in kinetic energy
= –2m (v₂ – u₂ )
= –2 × 1500 × (20² – 10²)
= 225000 J
The work done is positive.
(c) It has maximum potential energy at its extreme position and maximum kinetic energy at the mean position.