MCQs Question
5.1 Work done is maximum when the angle between the force \( F \) and the displacement \( d \) is:
(a) \( 0^\circ \)
(b) \( 30^\circ \)
(c) \( 60^\circ \)
(d) \( 90^\circ \)
Correct Option: (a) \( 0^\circ \)
Reason: Work is maximum when the force and displacement are in the same direction, making the angle \( \cos(0^\circ) = 1 \).
5.2 A joule can also be written as:
(a) \( \text{kg m}^2 \, \text{s}^{-2} \)
(b) \( \text{kg m s}^{-1} \)
(c) \( \text{kg m}^2 \, \text{s}^{-3} \)
(d) \( \text{kg m}^2 \, \text{s}^{-2} \)
Correct Option: (a) \( \text{kg m}^2 \, \text{s}^{-2} \)
Reason: Joule is derived from the formula \( \text{work} = \text{force} \times \text{distance} = \text{kg m}^2 \, \text{s}^{-2} \).
5.3 The SI unit of power is:
(a) Joule
(b) Newton
(c) Watt
(d) Second
Correct Option: (c) Watt
Reason: Power is defined as the rate of doing work, and its SI unit is the watt (\( \text{J/s} \)).
5.4 The power of a water pump is \( 2 \, \text{kW} \). The amount of water it can raise in one minute to a height of \( 5 \, \text{m} \) is:
(a) \( 1000 \, \text{litres} \)
(b) \( 1200 \, \text{litres} \)
(c) \( 2000 \, \text{litres} \)
(d) \( 2400 \, \text{litres} \)
Correct Option: (c) \( 2000 \, \text{litres} \)
Reason: Power is \( \text{work/time} \). By converting power into water lifted per minute, \( 2 \, \text{kW} = 2000 \, \text{litres} \).
5.5 A bullet of mass \( 0.05 \, \text{kg} \) has a speed of \( 300 \, \text{ms}^{-1} \). Its kinetic energy will be:
(a) \( 2250 \, \text{J} \)
(b) \( 4500 \, \text{J} \)
(c) \( 1500 \, \text{J} \)
(d) \( 1125 \, \text{J} \)
Correct Option: (b) \( 4500 \, \text{J} \)
Reason: Kinetic energy is \( \frac{1}{2} m v^2 = \frac{1}{2} (0.05) (300)^2 = 4500 \, \text{J} \).
5.6 If a car doubles its speed, its kinetic energy will be:
(a) The same
(b) Doubled
(c) Increased to three times
(d) Increased to four times
Correct Option: (d) Increased to four times
Reason: Kinetic energy is proportional to the square of velocity, \( KE \propto v^2 \), so doubling speed increases \( KE \) by \( 2^2 = 4 \).
5.7 The energy possessed by a body by virtue of its position is:
(a) Kinetic energy
(b) Potential energy
(c) Chemical energy
(d) Solar energy
Correct Option: (b) Potential energy
Reason: Potential energy depends on an object’s position, calculated using \( mgh \).
5.8 The magnitude of momentum of an object is doubled, the kinetic energy of the object will:
(a) Double
(b) Increase to four times
(c) Reduce to one-half
(d) Remain the same
Correct Option: (b) Increase to four times
Reason: Kinetic energy is proportional to the square of momentum, \( KE \propto p^2 \).
5.9 Which of the following is not a renewable energy source?
(a) Hydroelectric energy
(b) Fossil fuels
(c) Wind energy
(d) Solar energy
Correct Option: (b) Fossil fuels
Reason: Fossil fuels are finite resources and take millions of years to form, making them non-renewable.
Short Questions
5.1 What is the work done on an object that remains at rest when a force is applied on it?
The work done is zero because there is no displacement, and work requires both force and displacement.
5.2 A slow-moving car may have more kinetic energy than a fast-moving motorcycle. How is this possible?
The kinetic energy depends on both mass and velocity. A slow-moving car has a much larger mass than a motorcycle, so its kinetic energy can be greater.
5.3 A force \( F_1 \) does \( 5 \, \text{J} \) of work in \( 10 \, \text{s} \). Another force \( F_2 \) does \( 3 \, \text{J} \) of work in \( 5 \, \text{s} \). Which force delivers greater power?
Power = Work/Time
For \( F_1 \): \( P_1 = \frac{5}{10} = 0.5 \, \text{W} \)
For \( F_2 \): \( P_2 = \frac{3}{5} = 0.6 \, \text{W} \)
\( F_2 \) delivers greater power.
5.4 A woman runs up a flight of stairs. The gain in her gravitational potential energy is \( 4500 \, \text{J} \). If she runs up the same stairs with twice the speed, what will be her gain in potential energy?
The gain in potential energy depends only on height, not speed. Therefore, her gain in potential energy will still be \( 4500 \, \text{J} \).
5.5 Define work and its SI unit.
Work is the product of force and displacement in the direction of the force.
The SI unit of work is the joule (\( \text{J} \)).
5.6 What is the potential energy of a body of mass \( m \) when it is raised through a height \( h \)?
The potential energy is given by:
\[PE = mgh\]
where \( g \) is the acceleration due to gravity.
5.7 Find an expression for the kinetic energy of a moving body.
The kinetic energy of a moving body is:
\[KE = \frac{1}{2} m v^2\]
where \( m \) is the mass and \( v \) is the velocity.
5.8 Define efficiency of a working system. Why a system cannot have 100% efficiency?
Efficiency is the ratio of useful energy output to total energy input, expressed as a percentage:
\[\text{Efficiency} = \frac{\text{Useful Output}}{\text{Input}} \times 100\]
No system can have 100% efficiency due to energy losses like friction and heat.
5.9 What is power? Define the unit used for it.
Power is the rate at which work is done or energy is transferred.
The SI unit of power is the watt (\( \text{W} \)), where \( 1 \, \text{W} = 1 \, \text{J/s} \).
5.10 Differentiate between renewable and non-renewable energy sources.
Renewable Energy Sources: Can be replenished, e.g., solar, wind, hydroelectric energy.
Non-Renewable Energy Sources: Finite resources that cannot be replenished quickly, e.g., fossil fuels, coal.
Constructed Response Questions
5.1 Can the kinetic energy of a body ever be negative?
No, kinetic energy cannot be negative because it depends on mass and the square of velocity, both of which are always positive.
5.2 Which one has the greater kinetic energy; an object traveling with a velocity \(v\) or an object twice as heavy traveling with a velocity of \( \frac{1}{2}v\)?
The object traveling with velocity \(v\) has greater kinetic energy.
Reason: \( KE = \frac{1}{2} mv^2 \), so velocity has a greater effect because it is squared.
5.3 A car is moving along a curved road at constant speed. Does its kinetic energy change?
No, the kinetic energy does not change because kinetic energy depends only on speed, and the speed is constant.
5.4 Comment on the statement: “An object has one joule of potential energy.”
This means the object has the ability to do one joule of work due to its position or height relative to a reference point.
5.5 While driving on a motorway, a tyre of a vehicle sometimes bursts. What may be its cause?
The tyre bursts due to the high temperature caused by friction between the tyre and the road, which increases air pressure inside the tyre.
5.6 While playing cricket on a street, the ball smashes a windowpane. Describe the energy changes in this event.
The kinetic energy of the moving ball is transferred to the windowpane, breaking it. Some energy is also converted to sound and heat.
5.7 A man rowing a boat upstream is at rest with respect to the shore. Is he doing work?
Yes, he is doing work because he is applying force to row the boat, even though there is no displacement relative to the shore.
5.8 A cyclist goes downhill from the top of a steep hill without pedaling and takes it to the top of the next hill.
(i) Draw a diagram of what happened.
Diagram:
The cyclist converts potential energy at the top of the first hill into kinetic energy while descending and then back into potential energy while ascending the next hill.
(ii) Analyze this event in terms of potential and kinetic energy.
At the top of the first hill: Maximum potential energy, zero kinetic energy.
While descending: Potential energy decreases and is converted to kinetic energy.
At the bottom: Maximum kinetic energy, minimum potential energy.
While ascending: Kinetic energy is converted back to potential energy.
5.9 Is timber or wood a renewable source of heat energy? Comment.
Yes, timber or wood is a renewable source of energy because it can be replenished through sustainable forest management, though it takes time to regrow.
Comprehensive Questions
5.1 What is meant by kinetic energy? State its unit. Describe how it is determined.
Kinetic energy is the energy possessed by a body due to its motion.
Its unit is joule (J).
It is determined using the formula:
\[KE = \frac{1}{2} mv^2\]
where \( m \) is the mass of the object and \( v \) is its velocity.
5.2 State the law of conservation of energy. Explain it with the help of an example of a body falling from a certain height in terms of its potential energy and kinetic energy.
The law of conservation of energy states that energy cannot be created or destroyed; it can only be transformed from one form to another, but the total energy remains constant.
Example: A body falling from a certain height:
At the top: Maximum potential energy (\( PE = mgh \)) and zero kinetic energy.
As it falls: Potential energy decreases and converts into kinetic energy.
Just before hitting the ground: Maximum kinetic energy (\( KE = \frac{1}{2} mv^2 \)) and zero potential energy.
The total energy remains constant throughout.
5.3 Differentiate between renewable and non-renewable sources of energy. Give three examples for each.
| Renewable Sources | Non-Renewable Sources |
| Can be replenished over time. | Cannot be replenished once used. |
| Environmentally friendly. | Causes pollution. |
| Examples: Solar, Wind, Hydro | Examples: Coal, Oil, Natural Gas |
5.4 Explain what is meant by the efficiency of a machine. How is it calculated? Why is there a limit for the efficiency of a machine?
Efficiency: It is the ratio of useful energy output to the total energy input of a machine, expressed as a percentage.
Formula
\[\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\]
Limit: Efficiency is limited because some energy is always lost as heat, sound, or friction in every machine.
5.5 Describe the process of electricity generation by drawing a block diagram of the process in the following cases.(i) Hydroelectric power generation || (ii) Fossil fuels.
(i) Hydroelectric Power Generation:
Block Diagram:
Water (Dam) → Turbine → Generator → Electricity
(ii) Fossil Fuels Power Generation:
Block Diagram:
Fossil Fuels → Combustion (Heat) → Steam → Turbine → Generator → Electricity
Numerical Problems
5.1 A force of 20 N acting at an angle of \( 60^\circ \) to the horizontal is used to pull a box through a distance of 3 m across a floor. How much work is done?
Given:
Force (\( F \)) = 20 N
Angle (\( \theta \)) = \( 60^\circ \)
Displacement (\( d \)) = 3 m
To Find:
Work done (\( W \))
Formula:
\[W = F \cdot d \cdot \cos(\theta)\]
Solution:
\[W = 20 \cdot 3 \cdot \cos(60^\circ)\]
\[W = 20 \cdot 3 \cdot 0.5 = 30 \, \text{J}\]
5.2 A body moves a distance of 5 metres in a straight line under the action of a force of 8 N. If the work done is 20 J, find the angle which the force makes with the direction of motion of the body.
Given:
Force (\( F \)) = 8 N
Displacement (\( d \)) = 5 m
Work done (\( W \)) = 20 J
To Find:
Angle (\( \theta \))
Formula:
\[W = F \cdot d \cdot \cos(\theta)\]
Solution:
\[20 = 8 \cdot 5 \cdot \cos(\theta)\]
\[20 = 40 \cdot \cos(\theta)\]
\[\cos(\theta) = \frac{20}{40} = 0.5\]
\[\theta = \cos^{-1}(0.5) = 60^\circ\]
5.3 An engine raises 100 kg of water through a height of 80 m in 25 s. What is the power of the engine?
Given:
Mass (\( m \)) = 100 kg
Height (\( h \)) = 80 m
Time (\( t \)) = 25 s
Acceleration due to gravity (\( g \)) = \( 9.8 \, \text{m/s}^2 \)
To Find:
Power (\( P \))
Formula:
\[P = \frac{W}{t}, \quad W = mgh\]
Solution:
\[W = mgh = 100 \cdot 9.8 \cdot 80 = 78400 \, \text{J}\]
\[P = \frac{78400}{25} = 3200 \, \text{W}\]
5.4 A body of mass 20 kg is at rest. A 40 N force acts on it for 5 seconds. What is the kinetic energy of the body at the end of this time?
Given:
Mass (\( m \)) = 20 kg
Force (\( F \)) = 40 N
Time (\( t \)) = 5 s
To Find:
Kinetic energy (\( KE \))
Formula:
\[F = ma, \quad v = u + at, \quad KE = \frac{1}{2} mv^2\]
Solution:
\[F = ma \]
\[a = \frac{F}{m} = \frac{40}{20} = 2 \, \text{m/s}^2\]
\[v = u + at = 0 + (2)(5) = 10 \, \text{m/s}\]
\[KE = \frac{1}{2} mv^2 = \frac{1}{2} \cdot 20 \cdot 10^2 = 1000 \, \text{J}\]
5.5 A ball of mass 160 g is thrown vertically upward. The ball reaches a height of 20 m. Find the potential energy gained by the ball at this height.
Given:
Mass (\( m \)) = \( 160 \, \text{g} = 0.16 \, \text{kg} \)
Height (\( h \)) = 20 m
Acceleration due to gravity (\( g \)) = \( 9.8 \, \text{m/s}^2 \)
To Find:
Potential energy (\( PE \))
Formula:
\[PE = mgh\]
Solution:
\[PE = 0.16 \cdot 9.8 \cdot 20 = 31.36 \, \text{J}\]
\[\approx 32 \, \text{J}\]
5.6 A 0.14 kg ball is thrown vertically upward with an initial velocity of \( 35 \, \text{m/s} \). Find the maximum height reached by the ball.
Given:
Mass (\( m \)) = \( 0.14 \, \text{kg} \) (not needed here)
Initial velocity (\( u \)) = \( 35 \, \text{m/s} \)
Final velocity (\( v \)) = \( 0 \, \text{m/s} \) (at maximum height)
Acceleration due to gravity (\( g \)) = \( 9.8 \, \text{m/s}^2 \)
To Find:
Maximum height (\( h \))
Formula:
\[v^2 = u^2 – 2gh\]
Solution:
\[0 = 35^2 – 2 \cdot 9.8 \cdot h\]
\[1225 = 19.6h\]
\[h = \frac{1225}{19.6} = 62.5 \, \text{m}\]
5.7 A girl is swinging on a swing. At the lowest point of her swing, she is 1.2 m from the ground, and at the highest point, she is 2.0 m from the ground. What is her maximum velocity and where?
Given:
Height at highest point (\( h_1 \)) = 2.0 m
Height at lowest point (\( h_2 \)) = 1.2 m
Difference in height (\( h = h_1 – h_2 \)) = \( 2.0 – 1.2 = 0.8 \, \text{m} \)
To Find:
Maximum velocity (\( v_{\text{max}} \))
Formula:
\[v_{\text{max}} = \sqrt{2gh}\]
Solution:
\[v_{\text{max}} = \sqrt{2 \cdot 9.8 \cdot 0.8} = \sqrt{15.68} \approx 4 \, \text{m/s}\]
Answer: \( 4 \, \text{m/s} \), at the lowest position.
5.8 A person pushes a lawn mower with a force of 50 N making an angle of \( 45^\circ \) with the horizontal. If the mower is moved through a distance of 20 m, how much work is done?
Given:
Force (\( F \)) = 50 N
Distance (\( d \)) = 20 m
Angle (\( \theta \)) = \( 45^\circ \)
To Find:
Work done (\( W \))
Formula:
\[W = F \cdot d \cdot \cos(\theta)\]
Solution:
\[W = 50 \cdot 20 \cdot \cos(45^\circ) = 50 \cdot 20 \cdot 0.707\]
\[W = 707 \, \text{J}\]
Answer: \( 707 \, \text{J} \)
5.9 Calculate the work done in
(i) Pushing a 5 kg box up a frictionless inclined plane 10 m long that makes an angle of \( 30^\circ \) with the horizontal.
Given:
Mass (\( m \)) = 5 kg
Length of incline (\( l \)) = 10 m
Angle (\( \theta \)) = \( 30^\circ \)
\( g = 9.8 \, \text{m/s}^2 \)
To Find:
Work done (\( W \))
Formula:
\[W = mgh, \quad h = l \cdot \sin(\theta)\]
Solution:
\[h = 10 \cdot \sin(30^\circ) = 10 \cdot 0.5 = 5 \, \text{m}\]
\[W = 5 \cdot 9.8 \cdot 5 = 250 \, \text{J}\]
Answer: \( 250 \, \text{J} \)
(ii) Lifting the box vertically up from the ground to the top of the inclined plane.
Given:
Same as above.
Solution:
The work done will be the same as in (i), since the height remains 5 m.
\[W = 250 \, \text{J}\]
Answer: \( 250 \, \text{J} \)
5.10 A box of mass 10 kg is pushed up along a ramp 15 m long with a force of 80 N. If the box rises up a height of 5 m, what is the efficiency of the system?
Given:
Mass (\( m \)) = 10 kg
Height (\( h \)) = 5 m
Force (\( F \)) = 80 N
Length of ramp (\( l \)) = 15 m
To Find:
Efficiency (\( \eta \))
Formula:
\[\eta = \frac{\text{Useful work}}{\text{Total work}} \times 100\]
\[\text{Useful work} = mgh, \quad \text{Total work} = F \cdot l\]
Solution:
\[\text{Useful work} = 10 \cdot 9.8 \cdot 5 = 490 \, \text{J}\]
\[\text{Total work} = 80 \cdot 15 = 1200 \, \text{J}\]
\[\eta = \frac{490}{1200} \cdot 100 \approx 41.7\%\]
Answer: \( 41.7\% \)
5.11 A force of 600 N acts on a box to push it 5 m in 15 s. Calculate the power.
Given:
Force (\( F \)) = 600 N
Distance (\( d \)) = 5 m
Time (\( t \)) = 15 s
To Find:
Power (\( P \))
Formula:
\[P = \frac{W}{t}, \quad W = F \cdot d\]
Solution:
\[W = 600 \cdot 5 = 3000 \, \text{J}\]
\[P = \frac{3000}{15} = 200 \, \text{W}\]
Answer: \( 200 \, \text{W} \)
5.12 A 40 kg boy runs up a stair 10 m high in 8 s. What power he developed?
Given:
Mass (\( m \)) = 40 kg
Height (\( h \)) = 10 m
Time (\( t \)) = 8 s
To Find:
Power (\( P \))
Formula:
\[P = \frac{W}{t}, \quad W = mgh\]
Solution:
\[W = 40 \cdot 9.8 \cdot 10 = 3920 \, \text{J}\]
\[P = \frac{3920}{8} = 490 \, \text{W}\]
Answer: \( 490 \, \text{W} \)
5.13 A force \( F \) acts through a distance \( L \) on a body. The force is then increased to \( 2F \) that further acts through \( 2L \). Sketch a force-displacement graph and calculate the total work done.
Given:
Force \( F \), Distance \( L \), Force increases to \( 2F \) over \( 2L \).
To Find:
Total work done (\( W \))
Solution:
The total work done is the area under the force-displacement graph, which is a trapezium.
\[W = \frac{1}{2} \cdot (\text{Base}_1 + \text{Base}_2) \cdot \text{Height}\]
\[W = \frac{1}{2} \cdot (F \cdot L + 2F \cdot 2L)\]
\[W = \frac{1}{2} \cdot (FL + 4FL) = \frac{1}{2} \cdot 5FL = 5FL\]
Answer: \( 5FL \) or 5 units.
