Class 9 Science Chapter 10 Question Answers - Work and Energy

Q1: When the work is said to be done?
Ans: When a force acts on an object and moves it in the same direction that of force then work is said to be done.

Q2: What will be the expression for the work done when a force acts on an object in the direction of its motion.
Ans: Work done = Force × Displacement
If W is the work done, F is the force applied on object and d is the displacement, then the expression of work done will be
W = F × d

Q3: Explain 1 joule of work done.
Ans: When a force of 1 N (Newton) is applied on an object and that object displaces upto a distance of 1 m (meter) in the same direction of its displacement, then 1 joule (J) of work is done on the object.

Q4: How much work is done in ploughing a 15 m long field when a pair of bullocks applies a force of 140 N on the plough?
Ans: Since Work done (W) = Force (F) × Displacement (d)
Hence, Work done in ploughing (W) = 140 N × 15 m =  2100 J

Q5: The force acting on the object is 7 N, and the displacement of the object occurs in the direction of the force is 8 m. Suppose that force acts on the object through displacement, then how much work was done in this case?
Ans:  As we know, Work done (W) = Force (F) × Displacement (d)
Thus, Work done in the given case (W) = 7 N × 8 m =  56 J

Q6: Define kinetic energy of an object.
Ans: The kinetic energy of an object is a kind of mechanical energy that exists in the object due to its state of motion (movement).

Q7: Write down the kinetic energy expression of an object.
Ans: If m is the mass of an moving object and v is its velocity, then the expression of its kinetic energy (KE) will be
K.E = 1/2mv²

Q8: Define power.
Ans: The rate by which work is done refers to power. It is expressed by P.
Power = Work done/Time
P = W/t

Q9: What is 1 watt of power?
Ans: When an object is doing work at the rate of 1 J/s, then the power of that body or object is 1 watt (where watt is the unit of power).

Q10: An object is thrown at an angle to the ground, moves along a curve and falls back to the ground. The start and end points of the object path are on the same horizontal line. How much work is done by the gravity on that object?  
Ans: There must be a displacement to calculate the work, but since the vertical displacement in this case is zero (because the start and end points are on the same horizontal line), the work done by gravity is zero.

Q11: How does the state of energy get changed when a battery lights up a bulb?
Ans: The chemical energy of the battery is converted into heat and light energy of the bulb in the given case.

Q12: Calculate the work done by the force that changes the velocity of a moving body from 5 ms^-1 to 2 ms^-1. The body has a mass of 20 kg.
Ans: Since work done by force = Change in the kinetic energy of the moving body
Therefore, Work done by force =

Q13: An object having 10 kg weight is moved from point A to point B on the table. If the distance between A and B is horizontal, what work does gravity do to the object?  Give the reason for the answer.
Ans: Since the work done by gravity on the object depends on the change in the vertical height of the object, the vertical height of the object will not change. Because the connection level of A and B is at the same height, the work done is zero.

Q14: The potential energy of an object decreases gradually in a free fall. How does this violate the law of conservation of energy?
Ans: This does not violate the law of conservation of energy, because the potential energy of an object in free fall gradually decreases with gradual changes until the kinetic energy of the object maintains the state of free fall, that is, the total energy of the object remains conserved.

Q15: What energy conversion occurs when riding a bicycle?  
Ans: Our muscle energy is converted into mechanical energy while riding a bicycle.

Q16:  Does energy transfer occur when you push a huge rock with all your strength without moving it? Where did the energy you applied go?  
Ans: As long as you push a big rock with all your strength and do not move it, energy transfer will not occur, because cell energy is only used for muscle contraction and relaxation, and also for releasing heat (sweating).

Q17:A household uses 250 units of energy in a month. How much energy is used  by that house in joules?
Ans: Energy consumption by a house = 250 kWh
Since, 1 kWh = 3.6 × 10⁶ J
hence, 250kWh=  250 × 3.6 × 10⁶ = 9 × 10⁸ J

Q18: The output power of the electric heater is 1500 watts. How much energy does it consume in 10 hours?  
Ans: Power of electric heater (p) = 1500W = 1.5kW
Energy = Power × Time = 1.5kW × 10 hours = 15 kWh

Q19: An object of mass m moves at a constant speed v. How much work does the subject need to do to make it stable?  
Ans: For an object to be stationary, the work done must be equal to the kinetic energy of the moving object.
The kinetic energy of any object is equal to
K.E=1/2mv², where m is the mass of the body and v is its velocity.

Q20: Sony said that even if different forces act on the object, the acceleration of the object can be zero. Do you agree with her, if yes, why?  
Ans: Yes, we agree with Soni, because the displacement of an object becomes zero when many balancing forces act on that object.

Q21: Calculate the energy (in kilowatt hours) consumed by four 500 W devices in 10 hours.
Ans: Since, Energy = Power × Time
Hence, Energy consumed by four 500 W devices in 10 hours
= 4 × 500 × 10
= 20000 Wh
= 20 kWh

Q22: Free-falling objects will eventually stop when they hit the ground. What will happen to their kinetic energy?
Ans: The object will eventually stop after it hits the ground in free fall, because its kinetic energy will be transferred to the ground when it hits the ground.

Q23: A large force acting on an object, and the displacement of that object is zero, what will be the work done?  
Ans: The work done on the body is defined as the force exerted on the body that causes a net displacement of the body.  
Work done = Force x Displacement  
If the force does not cause any displacement, the work done to the object is zero.

Q24: Write some differences between kinetic and potential energy.
Ans: Differences between kinetic and potential energy:

Q25: Describe the law of conservation of energy.
Ans: The law of conservation of energy says that:

• Energy cannot be produced or destroyed. It can only be transformed from one form to another.  
• The energy of the universe is constant.

Q26: A person weighing 50 kg climbs the stairs with a height difference of 5 meters, within 4 seconds.  
(a) What kind of work is done by that person?  
(b) What is the average power of that person?
Ans: Mass of the man = 50 Kg
Distance moved by that man = 5 meter
Time taken to cover the given distance = 4s
(a) Work Done = Force  Acceleration
In this case, the increase in Potential energy = Work done =Mgh
=50×10×5
=2500 J
(b) Power = work Done  Time Taken  =25004=625 Watts

Q27: Write differences between power and energy.
Ans: Differences between power and energy are given below:


Q28: Write down the expressions for
(a) Potential energy of an object
(b) Kinetic energy of an object
Ans: 
(a) The expression for Potential energy of an object = P.E = mgh
Where, m = Mass of Body
g = Acceleration due to gravity
h = Height
(b) The expression for Kinetic energy of an object = 12mv2
Where, m = Mass of body
v = Velocity of body

Q29: If a force of 12.5 N is applied to complete a work of 100 J, what is the distance covered by the force?
Ans:  W = Work = 100 J
F = Force = 12.5 N
And S is the distance moved or displacement
Since, Work done = Force  Displacement
W = FS
100 =12.5 × S
100 × 1012.5 = S
1000125 = S
8 m=S (Displacement)

Q30: A car weighing 1800 kg is moving at a speed of 30 m/s when braking. If the average braking force is 6000 N, it is determined that the vehicle has traveled to a standstill distance. What is the distance at which it becomes stable?
Ans: M = Mass of the car = 1800 Kg
V = Velocity of the car = 30 m/s
F = Force applied while braking = 6000 N
KE=1/2mv²
KE =121800×900
KE=810000 J
KE of car = Work done by the car = Force  Displacement
810000=6000× Displacement
8100006000= Displacement
135 m= Displacement

Q31: What do you understand about average power?
Ans: The agent may not always be able to complete the same amount of work in a given time period. In other words, the power of this work will change over time. Therefore, in this case, we can take the average power of the work done by the body per unit time (that is, the total energy consumed divided by the total time).

Q32: Take a look at the steps below. Based on your understanding of the word "work", prove whether the work will proceed.  

• Suma swims in the pond.  
• The donkey carries a heavy load.  
• The windmill draws water from the well.  
• Green plants perform photosynthesis.  
• The trains are pulled by engines. 
• Drying food grains in the sun.  
• Sailing boats are powered by wind.

Ans: The work is said to be done when a force acts on an object and moves in the direction of the force. According to this explanation, the following activities were taken in which work will be proceeded:

• Suma swims in the pond.  
• The donkey carries a heavy load.  
• The windmill draws water from the well.  
• The trains are pulled by engines. 
• Sailing boats are powered by wind.

Q33: The law of conservation of energy is explained by discussing the energy changes that occur when we move the pendulum laterally and swing it. Why does the pendulum eventually stop? What happens to the energy and does it violate energy conservation law?
Ans: Bob will eventually stop due to the friction created by the air and the rigid support that holds the thread in place. This does not violate the law of conservation of energy, because mechanical energy can be converted into another unusable form of energy for some useful work. This energy loss is called energy dissipation.

Q34: Get the expression of the potential energy of an object. Calculate PE for a body of 10 kg which is resting at a height of 10 m.
Ans: The potential energy of an object with mass = mkg, at height above the ground =h m
Gravitational force of attraction on that body = mgN
To lift that body to B height at h  m above the ground.
Force applied to lift this body with a constant velocity =mgN
Distance moved by the body after applying force = hm
Work done in lifting the body from a to B distance = Force × Distance
Energy cannot be destroyed, hence, this energy is stored as potential energy in the stone.
m = 10Kg
g = 10 m/s²
h = 10 m
PE = mgh
PE = mgh
= 10 × 10 × 10
= 1000Joules