The chapter fundamentally explores the concepts of work, energy, and power, and the intrinsic relationship between them. In physics, work is said to be done only when a force applied on an object causes it to displace in the direction of the force. Its calculation is given by the formula W = F × s × cosθ, where the angle θ is crucial. For instance, no work is done if the force is perpendicular to the displacement. Energy is the capacity to do work, and it exists in various forms, with kinetic energy (energy of motion, KE = ½mv²) and potential energy (stored energy due to position, PE = mgh) being the primary focuses. A key principle is the Law of Conservation of Energy, which states that energy cannot be created or destroyed, only transformed from one form to another. Finally, power is the rate at which work is done or energy is transferred, calculated as P = W/t. It measures how quickly a task is accomplished, distinguishing it from the total amount of work or energy itself. These three concepts are deeply interconnected, as energy is transferred through the doing of work, and power quantifies the speed of this energy transfer.
EXERCISE- 2 (A)
Question 1: Define work. Is work a scalar or a vector?
Ans: Work is the product of force and the displacement in the force’s direction.
It is calculated as:
Work = Force × Displacement × cos θ
Here, θ is the angle between the force vector and the displacement vector.
Work is a scalar quantity. Although force and displacement are vectors, work is computed from their dot product. The result of a dot product is always a scalar, meaning work has only magnitude and no direction.
Question 2: How is the work done by a force measured when (i) force is in direction of displacement, (ii) force is at an angle to the direction of displacement?
Ans: (i) Force is in the direction of displacement:
When the force acts in the same direction as the displacement of the object, the work done is measured as the product of the magnitude of the force and the displacement.
Formula: W=F×s
Here,
W is work,
F is force, and
s is displacement.
(ii)θ to the direction of displacement, we only consider the component of force in the direction of displacement (Fcosθ)
Formula:
W=F×s×cosθ
Here,
θ is the angle between the force vector and the displacement vector.
Question 3: A force F acts on a body and displaces it by a distance S in a direction at an angle θ with the direction of force. (a) Write the expression for the work done by the force. (b) what should be the angle between the force and displacement to get the (i) zero work (ii) maximum work?
Ans: (a) Expression for Work Done
The work done
W by a constant force is given by:
W=FScosθ
where:
F = magnitude of force,
S = magnitude of displacement,
θ = angle between force and displacement vectors.
(b) Angle for Different Conditions
(i) Zero Work
Work done is zero when
θ=90 ∘ .
Here,
cos90 ∘ =0, so force is perpendicular to displacement and does no work.
(ii) Maximum Work
Work done is maximum when
θ=0 ∘ .
Here,
cos0 ∘ =1, so force and displacement are in the same direction, giving maximum work.
Question 4: A body is acted upon by a force. State two condition when the work done is zero.
Ans: The work done by a force is zero in the following two conditions:
Zero Displacement: When there is no displacement of the body, i.e.,
s=0.
(Since Work = Force × Displacement)
Force Perpendicular to Displacement: When the force acts in a direction perpendicular to the direction of the body’s displacement.
(For example, the gravitational force on a horizontally moving object).
Question 5: State the condition when the work done by a force is (i) positive, (ii) negative. Explain with the help of examples.
Ans: The work done by a force is given by
W=F⋅s⋅cosθ, where
F is the force,
s is the displacement, and
θ is the angle between the force vector and the displacement vector.
(i) Positive Work:
Work done is positive when the angle θ between the force and displacement is less than 90° (0°≤θ<90°).
This happens when the force has a component in the direction of displacement.
Example: When you lift a box upwards, the force you apply is upwards and the displacement is also upwards. Here,
θ=0°, so work done by you is positive.
(ii) Negative Work:
Work done is negative when the angle θ is greater than 90° but less than or equal to 180° (90°<θ≤180°).
This happens when the force has a component opposite to the direction of displacement.
Example: When a ball is thrown upwards, the force of gravity acts downwards while the displacement is upwards. Here,
θ=180°, so work done by gravity is negative.
Question 6: A body is moved in a direction opposite to the direction of force acting on it. State whether the work is done by the force or work is done against the force.
Ans: When a body is moved in a direction opposite to the direction of the force acting on it, the work is done against the force.
This is because the applied force (by the person or agent moving the body) is opposite to the given force (like friction or gravity), so the displacement occurs in the opposite direction to that force. In such cases, energy is spent to overcome the force, meaning work is done against it.
Question 7: When a body moves in a circular path, how much work is done by the body? Give reason.
Ans: When a body moves in a circular path, the work done by the centripetal force is zero.
Reason: The centripetal force acts towards the center of the circle. This force is always perpendicular to the direction of the body’s instantaneous displacement (which is along the tangent). Since work done is the product of force and displacement in the direction of force (W = F s cosθ), and the angle θ between the force and displacement is 90°, the value of cos90° is zero. Therefore, the work done is zero.
Question 8: A satellite revolves around the earth in a circular orbit. What is the work done by the force of gravity? Give reason.
Ans: The gravitational force acts towards the center of the Earth, along the radius of the circular orbit. At every point in the orbit, the displacement of the satellite is tangential (perpendicular to the radius). Since the angle between the gravitational force and the displacement is 90°, work done is given by:
W=F⋅s⋅cos90 ∘
=F⋅s⋅0=0
Thus, no work is done by gravity in a circular orbit.
Question 9: In which of the following cases, is work being done? (i) A man pushing a wall. (ii) a coolie standing with a load of 12 kgf on his head. (iii) A boy climbing up a staircase.
Ans: (i) A man pushing a wall → No work done, as the wall does not move.
(ii) A coolie standing with a load on his head → No work done, as there is no displacement.
(iii) A boy climbing up a staircase → Work is done, as force is applied and displacement occurs in the direction of the force.
Question 10: A coolie carrying a load on his head and moving on a frictionless horizontal platform does no work. Explain the reason.
Ans: The coolie does no work because the force he applies to carry the load is vertically upwards (to support the weight), while his displacement is horizontal.
Since the angle between the force and the displacement is 90 degrees, and work is calculated as Force × Displacement × cosθ, the work done becomes zero (as cos 90° = 0).
Question 11: The work done by a fielder when he takes a catch in a cricket match, is negative Explain
Ans: When a fielder catches a ball, the force he applies is opposite to the ball’s motion. Since the angle between his force and the ball’s displacement is 180 degrees, and cos 180° is -1, the work done by the fielder is negative. This negative work reduces the ball’s kinetic energy, stopping it.
Question 12: Give an example when work done by the force of gravity acting on a body is zero even though the body gets displaces from its initial position.
Ans: When a body moves horizontally on level ground, the force of gravity acts vertically downward while the displacement is perpendicular to it.
In this case, θ = 90° between force and displacement, so
Work done = F s cos 90° = 0,
even though the body is displaced.
Example: Pushing a box on a flat floor — gravity does no work.
Question 13: What are the S.I. and C.G.S units of work? How are they related? Establish the relationship.
Ans: The S.I. unit of work is the joule (J). The C.G.S unit of work is the erg.
They are related because both units measure the same physical quantity, work, which is force multiplied by displacement. The relationship between them is established through the conversion of their base units. One newton (the S.I. unit of force) is equal to 10^5 dynes (the C.G.S. unit of force), and one meter (S.I. unit of length) is equal to 100 centimeters.
Therefore, 1 joule = 1 newton × 1 meter = (10^5 dyne) × (100 cm) = 10^7 dyne-cm.
Question 14: State and define the S.I. unit of work.
Ans: The joule (J) is the SI unit for measuring work.One joule is defined as the work done when a one newton force displaces an object by one meter in the force’s direction. Essentially, it is the product of force and displacement, expressed as 1 J = 1 N × 1 m. For instance, pushing a box with a force of one newton over a one-meter distance means you have performed one joule of work.
Question 15: Express joule in terms of erg.
Ans: 1 joule is equal to 10⁷ ergs.
This conversion comes from the relationship between their base units. The joule is the SI unit of energy, defined as 1 newton × 1 meter. The erg is the CGS unit, defined as 1 dyne × 1 centimeter. Since 1 newton = 10⁵ dynes and 1 meter = 10² centimeters, we multiply them: 1 joule = (10⁵ dyne) × (10² cm) = 10⁷ dyne-cm = 10⁷ erg.
Question 16: A body of mass m falls down through a height h. Obtain an expression for the work done by the force of gravity.
Ans: When a body of mass m falls vertically through a height h, the force acting on it is gravity, given by
F=mg, where
g is the acceleration due to gravity.
The displacement of the body is in the same direction as the gravitational force. Therefore, the work done by gravity is calculated as:
Work=Force×Displacement
W=(mg)×h
W=mgh
Thus, the expression for the work done by the force of gravity is
W=mgh.
Question 17: A boy of mass m climbs up a staircase of vertical height h. (a) What is the work done by the boy against the force of gravity? (b) What would have been the work done if he uses a lift in climbing the same vertical height?
Ans: When climbing the stairs, the boy uses his muscles to exert a force equal to his weight, moving himself upward against gravity. The total work he performs is calculated as mgh, which represents the gain in gravitational potential energy. This amount of work depends only on the net vertical height ‘h’ he ascends, not on the path taken or the number of steps. If he takes the lift instead, he still gains the same height and the same amount of potential energy, mgh. The key difference is that in the lift, this necessary work against gravity is done by the motor, not by the boy’s own body.
Question 18: Define the term energy and state its S.I. unit.
Ans: When climbing the stairs, the boy uses his muscles to exert a force equal to his weight, moving himself upward against gravity. The total work he performs is calculated as mgh, which represents the gain in gravitational potential energy. This amount of work depends only on the net vertical height ‘h’ he ascends, not on the path taken or the number of steps. If he takes the lift instead, he still gains the same height and the same amount of potential energy, mgh. The key difference is that in the lift, this necessary work against gravity is done by the motor, not by the boy’s own body.
Question 19: What physical quantity does the electron volt (eV) measure? How is it related to the S.I. unit of that quality?
Ans: An electron volt (eV) measures energy.
Specifically, it is defined as the amount of kinetic energy gained or lost by a single electron when it is accelerated through an electric potential difference of one volt. It is a very convenient unit for measuring the tiny energies of atomic and subatomic particles. Its relation to the S.I. unit of energy, the joule (J), is given by the conversion:
1 eV = 1.602 × 10⁻¹⁹ joules.
Question 20: Complete the following sentence: 1 J = Calorie Ans: The statement “1 J = Calorie” is incomplete and, in its current form, incorrect. To complete it accurately, we need to specify which type of calorie is being referenced.Therefore, one dietary Calorie is equal to 1,000 of the smaller calories used in physics and chemistry. When we use the standard small ‘c’ calorie (cal), which is defined as the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius, the correct conversion is approximately 1 calorie = 4.184 joules. This means one joule is equal to a much smaller fraction of a calorie, specifically about 0.239 calories. So, to make the sentence accurate, you could write it as “1 Calorie (kcal) = 4184 J” or, for the smaller physics calorie, “1 calorie (cal) = 4.184 J.” Understanding this distinction is crucial for correctly converting energy units between the metric system (joules) and the historical unit of energy (calories).
Question 21: Name the physical quantity which is measured in calorie. How is it related to the S.I. unit of the quality?
Ans:The physical quantity measured using the calorie is heat energy.
This unit connects to the official SI unit for energy, the joule, through a direct conversion. The relationship is:
1 calorie ≈ 4.186 joules (in many practical calculations, this is often rounded to 4.2 joules).
What this means in simple terms is that the amount of heat energy needed to raise the temperature of 1 gram of water by 1 degree Celsius—which is the definition of a calorie—is equivalent to expending about 4.186 joules of energy. So, while the calorie is a practical unit from the context of heat and chemistry, the joule is the universal unit for all forms of energy, whether it’s mechanical, electrical, or heat.
Question 22: Define a kilowatt hour. How is it related to joule?
Ans: A kilowatt hour (kWh) is a commercial unit of electrical energy. It is defined as the amount of electrical energy consumed when an electrical appliance with a power rating of 1 kilowatt (1000 watts) is used for 1 hour.
Relationship with joule:
We know that,
1 kilowatt = 1000 watts
1 hour = 3600 seconds
Also, Electrical Energy = Power × Time
So,
1 kWh = 1 kW × 1 hour
= 1000 W × 3600 s
= 3,600,000 Ws
Since 1 watt-second (Ws) = 1 joule (J),
1 kWh = 3,600,000 J or 3.6 × 10⁶ J
Thus, 1 kilowatt hour is equal to 3.6 million joules.
Question 23: Define the term power. State its S.I. unit.
Ans:Think of it as a measure of the lens’s “strength.” A lens with a high power will bend light rays very sharply, while a lens with a low power will bend them more gently.
The power of a lens is directly related to its focal length. The focal length is the distance between the lens and the point where it brings parallel light rays to a focus. We can find the power using a simple formula:
Power (P) = 1 / Focal Length (f)
However, for this formula to work correctly, the focal length must be measured in metres.
The unit we use for measuring power is the dioptre, which is represented by the letter D.
What is one dioptre?
If a lens has a focal length of exactly 1 metre, its power is 1 dioptre (1 D). So, one dioptre is simply the power of a lens that has a one-metre focal length.
Question 24: State two factors on which power spent by a source depends. Explain your answer with examples.
Ans:The amount of power a source, like a battery or a mains outlet, uses isn’t just down to one thing—it’s a combination of two key players working together.
1. The Push (Voltage):
Think of voltage as the electrical push or pressure from the source. It’s the amount of energy given to each unit of electric charge. A higher voltage means a stronger push, so each bit of charge carries more energy. If you compare a small battery to a wall socket, the socket has a much higher voltage. This is why a lamp plugged into the wall shines brighter than one connected to a battery; the stronger push from the wall socket makes the source spend power at a faster rate. Simply put, more voltage means more power spent.
2. The Flow (Current):
Now, think of current as the amount of flow—how much electric charge is moving through the circuit every second. Even with a strong push (high voltage), if only a tiny trickle of charge is moving (low current), the total power spent won’t be very high. It’s like a wide, slow-moving river versus a narrow, fast-moving hose. When you turn on more appliances in your home, you’re asking for more current to flow. The source has to work harder to supply this increased flow, and as a result, it ends up spending more power.
To put it simply, the power spent comes from the combination of how hard the source pushes (Voltage) and how much it is pushing (Current). This relationship is neatly captured by the formula: Power = Voltage × Current.
So, if either the push (voltage) or the flow (current) increases, the total power the source has to deliver goes up.
Question 25: Differentiate between work and power.
Ans:Work and power are two fundamental concepts in physics, but they describe different things.
Work refers to the amount of energy transferred when a force is applied to an object and causes it to move. It is calculated as the product of force and displacement in the direction of the force (Work = Force × Displacement). Its SI unit is the joule (J). Essentially, work tells you how much total energy was used in a task, regardless of how long it took. For example, lifting a box vertically upwards does a certain amount of work on it.
Power, on the other hand, is the rate at which work is done or energy is transferred. It measures how fast work is completed. It is calculated as work divided by time (Power = Work / Time). Its SI unit is the watt (W), which is equal to one joule per second. For instance, a more powerful motor can lift the same box to the same height faster than a less powerful one; both do the same amount of work, but the more powerful motor has a higher power output because it does the work in less time.
Question 26: Differentiate between energy and power.
Ans:Energy is the total capacity to do work. It is the amount of work that can be performed, and it is measured in Joules (J). For example, the food we eat contains chemical energy, which our bodies use throughout the day. Energy is like the total amount of money you have in your bank account.
Power, on the other hand, is the rate at which energy is used or work is done. It measures how fast energy is transferred or converted. Its unit is Watt (W), which is equal to one Joule per second. For instance, a 100-watt light bulb uses electrical energy ten times faster than a 10-watt bulb. Power is like the rate at which you spend that money—whether you spend it slowly or very quickly.
Question 27: State and define the S.I. unit of power.
Ans:S.I. Unit of Power:
The S.I. unit of power is the watt, denoted by the symbol W.
Definition:
One watt is defined as the rate of doing work or consuming energy equal to one joule per second.
Mathematically,
1watt= 1second1joule
In simpler terms, if an appliance consumes or produces energy at the rate of 1 joule every second, its power is said to be 1 watt.
Question 28: What is horse power (H.P)? How is it related to the S.I. unit of power? Ans: Horsepower (H.P.) is a traditional unit of power that was originally developed by engineer James Watt. He created it to compare the output of steam engines with the power of draft horses. In simple terms, one horsepower (1 H.P.) is defined as the power needed to lift 550 pounds (lb) of weight by one foot (ft) in one second (s).
Its relationship with the Standard International (S.I.) unit of power, the watt (W), is fixed and defined as follows:
1 Horsepower (H.P.) = 746 Watts (W)
This means that one horsepower is exactly equal to 746 joules of work done per second. For easier calculation, especially in the automotive and electrical industries, this value is often rounded to approximately 750 watts.
So, to summarize, while horsepower is a older, non-SI unit, it is officially and precisely related to the SI unit of power, the watt, through this conversion factor.
Question 29: Differentiate between watt and watt hour.
Ans: While “watt” and “watt-hour” sound similar, they measure two completely different things: power and energy.
Watt (W): This is the unit of power. Power is the rate at which energy is used or generated. Think of it as how fast an electrical device is consuming energy. For example, a 60-watt light bulb is drawing energy at a rate of 60 joules every second. A higher wattage means a faster rate of energy consumption.
Watt-hour (Wh): This is the unit of energy. It represents the total amount of energy used or produced over time. It is calculated by multiplying power (in watts) by time (in hours). For instance, if that 60-watt bulb is left on for 5 hours, it has consumed 60 W × 5 h = 300 watt-hours of electrical energy. Your electricity bill is based on the total kilowatt-hours (kWh) you have used in a month.
Watt is like the rate of flow of the water (e.g., litres per second).
Watt-hour is like the total volume of water that has passed through the pipe after a certain time (e.g., litres).
Question 30: Name the quality which is measured in (a) kWh (b) kW (c) Wh (d) eV
Ans:(a) kWh (kilowatt-hour)
The kilowatt-hour (kWh) is the unit used to measure electrical energy. It represents the amount of energy consumed by a 1 kilowatt (kW) appliance running for 1 hour. Electricity bills are calculated based on the number of kWh used.
(b) kW (kilowatt)
The kilowatt (kW) is the unit that measures power. It tells us the rate at which energy is used or generated. For example, a powerful electric heater might have a rating of 2 kW, meaning it converts electrical energy into heat at that rate.
(c) Wh (watt-hour)
The watt-hour (Wh) is another unit for measuring energy, just like the kWh, but on a smaller scale. One kWh is equal to 1000 Wh. It is the energy consumed by a 1-watt device in one hour.
(d) eV (electronvolt)
The electronvolt (eV) is a very tiny unit used to measure energy, especially in atomic and nuclear physics. It is defined as the kinetic energy gained by a single electron when it is accelerated by an electric potential difference of 1 volt.
MULTIPLE CHOICE TYPE:
Question 1: One horse power is equal to: (a) 1000 W (b) 500 W (c) 764 W (d) 746 W
Ans: 746 W
Question 2: kWh is the unit of: (a) power (b) force (c) energy (d) none of these
Ans: The unit kWh is the unit of energy.
NUMERICALS:
Question 1: A body, when acted upon by a force of 10 kgf, gets displaced by 0.5 m. Calculate the work done by the force, when the displacement is (i) in the direction of force, (ii) at an angle of 60° with the force, and (iii) normal to the force. (g = 10 N kg-1)
Ans:(i) When displacement is in the direction of force:
Work done = 50 J
(ii) When displacement is at 60° to the force:
Work done = 25 J
(iii) When displacement is normal (perpendicular) to the force:
Work done = 0 J
Question 2: A boy of mass kg runs upstairs and reaches the 8 m high floor in 5 s Calculate: the force of gravity acting on the boy. (i) the work done by him against gravity. (ii) the power spent by boy. (Take g = 10 m s-2)
Ans: (i) Force of gravity acting on the boy
The force of gravity acting on a body is its weight.
It is calculated using the formula:
Weight = mass × acceleration due to gravity
Given:
Mass of the boy = 60 kg
Acceleration due to gravity = 10 m/s²
So,
Force of gravity = 60 × 10
= 600 N
Answer: 600 N
(ii) Work done by him against gravity
Work is done when a force causes displacement.
Here, the boy works against gravity to climb up.
The formula used is:
Work = force × displacement
Since the force here is equal to the weight (mg), and displacement is the height climbed,
Work = m × g × h
Given:
Mass, m = 60 kg
g = 10 m/s²
Height, h = 8 m
So,
Work done = 60 × 10 × 8
= 4800 J
Answer: 4800 J
(iii) Power spent by the boy
It is calculated as:
Power = Work done / Time taken
Given:
Work done = 4800 J
Time taken = 5 s
So,
Power = 4800 / 5
= 960 W
Answer: 960 W
Final Answers:
- Force of gravity = 600 N
- Work done against gravity = 4800 J
- Power spent = 960 W
Question 3: It takes 20 s for a person A to climb up the stairs, while another person B does the same in 15 s. Compare the (i) Work done and (ii) power developed by the persons A and B.
Ans: To compare the work done and power developed by persons A and B, we need to make a few logical assumptions:
Both persons have the same mass (m).
They are climbing the same set of stairs, meaning they are being raised through the same height (h) against gravity.
(i) Comparison of Work Done
The work done (W) against gravity to climb the stairs is given by the formula:
W=m×g×h
where:
m is the mass of the person,
g is the acceleration due to gravity,
h is the height of the stairs.
Since both persons have the same mass and climb the same height against the same gravitational force, the work done by both is the same.
∴ The ratio of work done by A to work done by B is 1 : 1.
(ii) Comparison of Power Developed
Power (P) is defined as the rate of doing work, or Work done per unit time:
P=W/t
where
t is the time taken.
For person A:
PA=W/20
For person B:
PB=W/15
Now, let’s find the ratio of their power:
PA/PB=W20/W15=15/20=3/4
∴ The ratio of power developed by A to power developed by B is 3 : 4.
Question 4: A boy weighing 350 N runs up a flight of 30 steps, each 20 cm high in 1 minute, Calculate: (i) the work done and (ii) power spent.
Ans:Step 1: Understanding the problem
A boy weighing 350 N runs up a flight of 30 steps, each 20 cm high, in 1 minute.
We need to calculate:
(i) Work done
(ii) Power spent
Step 2: Find total height climbed
Height of each step = 20 cm = 0.20 m
Number of steps = 30
Total height,
h=30×0.20 m
h=6.0 m
Step 3: Calculate work done
Work done = Force × displacement in the direction of force
Here, force = weight of the boy = 350 N
Displacement = vertical height climbed = 6.0 m
W=350×6.0
W=2100 J
Step 4: Calculate power spent
Time taken = 1 minute = 60 seconds
Power = Work done ÷ Time taken
P=2100/60
P=35 W
Final Answer: 2100 J, 35 W
Question 5: A man spends 6.4 KJ energy in displacing a body by 64 m in the direction in which he applies force, in 2.5 s Calculate: (i) the force applied and (ii) the power Spent (in H.P) by the man.
Ans: Step 1: Understand the given data
Energy spent E=6.4 kJ=6400 J
Displacement s=64 m (in the direction of force)
Time t=2.5 s
Step 2: Find the force applied
Work done = Force × displacement
E=F×s
6400=F×64
F=6400/64 =100 N
So,
(i) Force applied=100 N
Step 3: Find power in watts
P=Work done/time
=6400/2.5
=2560 W
Step 4: Convert watts to horsepower (H.P.)
We know
1 H.P.=746 W
P (in H.P.)= 746/2560 ≈3.43 H.P.
So,
(ii) Power spent≈3.43 H.P.
Final Answer: 100 N, 3.43 H.P.
Question 6: A weight lifter a load of 200 kgf to a height of 2.5 m in 5 s. Calculate: (i) the work done, and (ii) the power developed by him. Take g = 10 N kg
Ans:(i) Work Done:
First, we need to find the force, which is the weight of the load.
Load,
m=200 kgf
Since 1 kgf is the force exerted by a 1 kg mass under gravity (g=10 N/kg), we can directly calculate the force in Newtons.
Force,
F=m×g=200kg×10N/kg=2000N
Height,
h=2.5m
Work is done against gravity, so the formula is:
Work Done (W)=Force×Displacement
W=F×h
W=2000N×2.5m
W=5000J
So, the work done by the weight lifter is 5000 Joules.
(ii) Power Developed:
Power is the rate at which work is done.
Work Done,
W=5000J
Time Taken,
t=5s
Power (P)=Work Done/Time Taken
P=W/t=5000J/5
P=1000W
Therefore, the power developed by him is 1000 Watts.
Question 7: A machine raises a load of 750 N through a height of 16 m in 5 s. calculate: (i) energy spent by machine, (ii) power at which the machine works.
Ans:(i) Energy spent by the machine
The energy spent by the machine is used to lift the load against gravity. This energy is stored in the load as Gravitational Potential Energy.
We know the formula for potential energy is:
Distance moved in the direction of force
Energy=Force×Distance moved in the direction of force
Here, the force is the weight of the load, which is 750 N.
The distance moved in the direction of the force (upwards) is the height, which is 16 m.
So,
Energy Spent=750N×16m
Energy Spent=12000J
(ii) Power at which the machine works
Power is the rate at which work is done or energy is spent. The formula for power is:
Power=Energy Spent/Time Taken
We have the Energy spent = 12000 J
The Time taken = 5 s
So,
Power=2400W
Final Answers:
(i) The energy spent by the machine is 12000 Joules.
(ii) The power at which the machine works is 2400 Watts.
Question 8: An electric heater of power 3 KW is used for 10 h. How much energy does it consume? Express your answer in (i) kWh, (ii) joule.
Ans:Energy = Power × Time
Given Data:
Power of the heater, P = 3 KW
Time of operation, t = 10 hours
(i) Energy in kilowatt-hour (kWh):
Since the power is already in kilowatts (KW) and the time is in hours (h), the calculation is direct. The unit “kilowatt-hour” (kWh) is literally a product of kilowatts and hours.
Energy (in kWh) = Power (in KW) × Time (in h)
= 3 KW × 10 h
= 30 kWh
In everyday terms, we say the heater has consumed 30 units of electrical energy.
(ii) Energy in joules (J):
We know that 1 kilowatt-hour (kWh) is a standard unit of energy defined as the energy consumed by a 1 KW device running for 1 hour. Its equivalence in joules is a fixed value.
1 kWh = 3.6 × 10⁶ J
Therefore, to find the energy in joules, we multiply the energy in kWh by this conversion factor.
Energy (in J) = Energy (in kWh) × 3.6 × 10⁶ J/kWh
= 30 kWh × 3.6 × 10⁶ J
= (30 × 3.6) × 10⁶ J
= 108 × 10⁶ J
= 1.08 × 10⁸ J
Final Answers:
(i) The energy consumed is 30 kWh.
(ii) The energy consumed is 1.08 × 10⁸ J.
Question 9: A boy of mass 40 kg runs up a flight of 15 steps each 15 cm high in 10 s. Find: (i) the work done and (ii) the power developed by him Take g = 10 N kg -1
Ans:Step 1: Understanding the given data
Mass of the boy,
m=40 kg
Number of steps climbed =15
Height of each step =15 cm=0.15 m
Time taken, t=10 s
Acceleration due to gravity,
g=10 N/kg
Step 2: Calculate total height climbed
Total height,
Total height, h=15×0.15=2.25 m
Step 3: Calculate work done
Work done against gravity is given by:
W=m×g×h
Substituting the values:
W=40×10×2.25=900 J
(i) Work done = 900 J
Step 4: Calculate power developed
Power is the work done per unit time:
P=W/t=900/10=90 W
(ii) Power developed = 90 W
Final Answer: 900 J and90 W
Question 10: A water pump raises 50 litres of water through a height of 25 m in 5 s. Calculate the power which the pump supplies. (Take g = 10 N kg-1 and density of water = 1000 kg m-3)
Ans:Step 1: Find the mass of water raised
We are given that the volume of water raised is 50 litres.
We know that 1 litre = 0.001 m³,
So,
Volume = 50×0.001=0.05m 3
Density of water = 1000kg/m 3
Mass = Density × Volume
m=1000×0.05=50kg
Step 2: Find the weight of water
Weight = mass × gravitational acceleration
Using
g=10m/s
W=50×10=500N
Step 3: Work done by the pump
Work = Force × vertical height
Here, height = 25 m
Work =
500×25=12500J
Step 4: Power supplied by the pump
Power = Work/Time
Time given = 5 seconds
Power = 12500/5
=2500W
Final Answer:
The power supplied by the pump is 2500 W.
Question 11: A man raises a box of mass 50 kg to a height of 2 m in 2 minutes, while another man raises the same box to the same height in 5 minutes. Compare: (i) the work done and (ii) the power developed by them.
Ans:(i) Comparison of Work Done
Work = Force × Displacement
Here, the force is equal to the weight of the box, which is:
Weight = mass × acceleration due to gravity
Since both men are lifting boxes of the same mass (50 kg) to the same height (2 m) against gravity, the force applied and the vertical displacement are identical in both cases.
Therefore, the work done by both men is the same.
So, the ratio of work done by the first man to the second man is:
1 : 1
(ii) Comparison of Power Developed
Power is defined as the rate of doing work, given by:
Power = Work done / Time taken
Let the total work done by each man be W.
First man completes the work in 2 minutes = 120 seconds
Power of first man, P₁ = W / 120
Second man completes the same work in 5 minutes = 300 seconds
Power of second man, P₂ = W / 300
Now, comparing their power:
P₁ / P₂ = (W / 120) / (W / 300) = 300 / 120 = 5 / 2
Therefore, the ratio of the power developed by the first man to the second man is:
5 : 2
Question 12: A pump is used to lift 500 kg of water from a depth of 80 m in 10 s. calculate: (a) the work done by the pump (b) the power a which the pump works, (c) the power rating of the pump if its efficiency is 40% (Take g = 10 m s-2)
Ans:
Given:
Mass of water,
m=500 kg
Depth,
h=80 m
Time,
t=10 s
Acceleration due to gravity,
g=10 m/s 2
Efficiency,
η=40%=0.4
(a) Work done by the pump
The work done in lifting the water is equal to the gravitational potential energy gained by the water:
W=m⋅g⋅h
W=500×10×80
W=500×800
W=400,000 J
(b) Power at which the pump works
Power is the work done per unit time:
Poutput=Wt
Poutput
=400,000/10
P output=40,000 W
So, the pump works at a power of 40 kW.
(c) Power rating of the pump if its efficiency is 40%
Efficiency is given by:
η=Output power/Input power
0.4=40,000/Pinput
P input=40,000/0.4
P input =100,000 W=100 kW
So, the power rating of the pump is 100 kW.
Question 13 : An ox can apply a maximum force of 1000 N. It is taking part in a cart race and is able to pull the cart at a constant speed of 30 M S-1 while making its best effort. Calculate the power developed by the ox.
Ans:The power developed by the ox is calculated using the formula:
Power = Force × Velocity
Given:
Force = 1000 N
Velocity = 30 m/s
Power = 1000 × 30 = 30,000 W
Thus, the power developed is 30,000 watts or 30 kilowatts.
Question 14: If the power of a motor is 40 kw, at what speed can it raise a load of 20,000 N?
Ans: We are given:
Power supplied by the motor,
P=40 kW=40,000 W
Force exerted (weight of the load),
F=20,000 N
We know that power can be expressed as:
P=F×v
where
v is the constant speed at which the load is raised (in m/s).
Rearranging the formula to solve for
v=P/F
Substitute the given values:
v= 20,000/40,000
=2 m/s
