A force of 7 N acts on an object. The displacement is, say, 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Given: Force F = 7 N, Displacement s = 8 m, angle θ = 0° (force and displacement in same direction).
The work done is 56 joules.
When do we say that work is done?
Work is said to be done when:
(i) A force is applied on an object.
(ii) The object moves (displacement occurs) in the direction of the applied force.
where θ is the angle between the force and displacement.
Work is NOT done when:
(a) Force is perpendicular to displacement (θ = 90°, cos 90° = 0) — e.g., a satellite moving in circular orbit.
(b) There is no displacement (s = 0) — e.g., pushing a wall.
Write an expression for the work done when a force is acting on an object in the direction of its displacement.
When force and displacement are in the same direction (θ = 0°):
where W = work done (in joules, J), F = force applied (in newtons, N), s = displacement in the direction of force (in metres, m).
In the general case when the angle between force and displacement is θ:
The SI unit of work is the joule (J). Work is a scalar quantity.
Define 1 joule of work.
1 joule of work is said to be done when a force of 1 newton displaces an object by 1 metre in the direction of the applied force.
The joule is named after the physicist James Prescott Joule. It is the SI unit of both work and energy.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
Given: Force F = 140 N, Displacement s = 15 m, θ = 0°
The work done in ploughing the field is 2100 joules.
What is the kinetic energy of an object?
The kinetic energy of an object is the energy possessed by it due to its motion. An object of mass m moving with velocity v has kinetic energy given by:
where m = mass of the object (in kg) and v = velocity (in m/s). KE is measured in joules (J).
Key properties: (i) Kinetic energy is always positive (since m and v² are always positive). (ii) It is a scalar quantity. (iii) KE depends on the square of velocity — doubling velocity quadruples kinetic energy.
Write an expression for the kinetic energy of an object.
The kinetic energy (KE) of an object of mass m moving with speed v is:
where m is the mass in kilograms (kg) and v is the speed in metres per second (m/s). The kinetic energy is measured in joules (J).
This expression is derived from the work-energy theorem: the net work done on an object equals the change in its kinetic energy.
The kinetic energy of an object of mass m moving with a velocity of 5 m/s is 25 J. What will be its kinetic energy when its velocity is doubled? Halved?
Given: v = 5 m/s, KE = 25 J. So ½mv² = 25 J.
(a) When velocity is doubled (v′ = 2v = 10 m/s):
Kinetic energy becomes 100 J (four times the original).
(b) When velocity is halved (v′′ = v/2 = 2.5 m/s):
Kinetic energy becomes 6.25 J (one-quarter of the original).
What is power?
Power is defined as the rate of doing work (or the rate of transfer of energy).
where P = power, W = work done (in joules), t = time taken (in seconds).
Power can also be written as:
The SI unit of power is the watt (W): 1 W = 1 J/s. Larger units: 1 kilowatt (kW) = 1000 W; 1 megawatt (MW) = 106 W; 1 horsepower (hp) = 746 W. Power is a scalar quantity.
Define 1 watt of power.
One watt is the power of an agent that does work at the rate of 1 joule per second.
In other words, if a machine does 1 joule of work every second, its power is 1 watt. The watt is named after the Scottish inventor James Watt.
A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Given: Energy consumed W = 1000 J, Time t = 10 s.
The power of the lamp is 100 watts.
Define the average power.
Average power is the total work done divided by the total time taken. It is used when the rate of doing work (power) varies over time.
For example, if a person does 600 J of work in 60 seconds (not necessarily at a constant rate), the average power is Pavg = 600/60 = 10 W. Average power gives the mean rate of energy expenditure over the entire time interval, regardless of fluctuations.
An object of mass 15 kg is moving with a uniform velocity of 4 m/s. What is the kinetic energy possessed by the object?
Given: Mass m = 15 kg, Velocity v = 4 m/s.
The kinetic energy of the object is 120 joules.
What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
The work done by the force of gravity on a satellite moving in a circular orbit is zero.
Justification: A satellite moves in a circular orbit. The gravitational force always acts toward the centre of the earth (centripetal direction — radially inward). The displacement of the satellite at every instant is tangential (along the orbit — perpendicular to the radius).
Since force and displacement are perpendicular to each other, θ = 90°.
Therefore, the work done by the gravitational force on a satellite is zero.
Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this in your friends' group.
Yes, there can be displacement of an object in the absence of a net force acting on it.
According to Newton's first law of motion, an object in a state of uniform motion continues to move in a straight line with constant velocity unless acted upon by an external force. Such an object undergoes displacement without any (net) force.
Example: An object moving in outer space far from any gravitational or other forces will continue to move and will have displacement even though no force is acting on it. However, in this case, while there is displacement, no work is being done on the object by any force (since there is no force). This scenario is different from saying no work is done when force exists but is perpendicular to displacement.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Just before hitting the ground, a freely falling object has maximum kinetic energy. When it stops on striking the ground, the kinetic energy is not destroyed (it cannot be, according to the law of conservation of energy). It is transformed into other forms of energy:
(i) Heat energy — generated due to the impact and friction with the ground.
(ii) Sound energy — the thud/sound produced on impact.
(iii) Deformation energy — if the object or ground deforms (permanent deformation stores energy).
The total energy before and after impact remains the same; it is merely converted from kinetic energy into these other forms.
An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Given: Power P = 1500 W, Time t = 10 hours = 10 × 3600 = 36,000 s.
In commercial units (kilowatt-hours):
The heater uses 5.4 × 107 J (or 15 kWh) of energy in 10 hours.
Note: 1 kWh = 3.6 × 106 J (commercial unit of electrical energy, also called 1 unit).
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?
Law of Conservation of Energy — Pendulum Example:
Position A (extreme position, bob pulled to side): The bob is at maximum height. It has maximum Potential Energy (PE = mgh) and zero Kinetic Energy (velocity = 0).
Position B (lowest/mean position): As the bob swings down, PE converts to KE. At the lowest point, height = 0, so PE = 0 and KE is maximum.
Position C (other extreme): The bob rises to the same height on the other side. KE converts back to PE. KE = 0, PE = maximum again.
Why does the bob come to rest? In practice, air resistance and friction at the pivot point continuously convert a small amount of mechanical energy into heat energy. Over many oscillations, the total mechanical energy decreases gradually, the amplitude decreases, and eventually the bob comes to rest at position B (lowest point). The total energy (mechanical + heat) is still conserved.
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?
Using the work-energy theorem: Work done = Change in Kinetic Energy.
The negative sign indicates that the work must be done against the direction of motion (opposing force must be applied). The magnitude of work required to bring the object to rest is ½mv2 joules.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h.
Given: Mass m = 1500 kg, Velocity v = 60 km/h.
Convert velocity to m/s:
Work required to stop = Initial KE of the car:
The work required to stop the car is approximately 2.08 × 105 J (i.e., about 208,333 joules). This work is done by the braking force against the direction of motion.
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