Work, Power and Energy
WorkWork is said to be done when a force causes a displacement to a body on which it acts.
Work is a scalar quantity. It is the dot product of applied force F and displacement d.
Diagram Coming Soon
W = F . d
W = F d cos θ ………………………… (1)
Where θ is the angle between F and d.
Equation (1) can be written as
W = (F cos θ) d
i.e., work done is the product of the component of force (F cos θ) in the direction of displacement and the magnitude of displacement d.
equation (1) can also be written as
W = F (d cos θ)
i.e., work done is the product of magnitude of force F and the component of the displacement (d cos θ) in the direction of force.
Unit of Work
M.K.S system → Joule, BTU, eV
C.G.S system → Erg
F.P.S system → Foot Pound
1 BTU = 1055 joule
1 eV = 1.60 x 10(-19)
Important Cases
Work can be positive or negative depending upon the angle θ between F and d.
Case I
When θ = 0º i.e., when F and d have same direction.
W = F . d
W = F d cos 0º ………….. {since θ = 0º}
W = F d …………………….. {since cos 0º = 1}
Work is positive in this case.
Case II
When θ = 180º i.e., when F and d have opposite direction.
W = F . d
W = F d cos 180º ………………………. {since θ = 180º}
W = – F d ………………………………….. {since cos 180º = -1}
Work is negative in this case
Case III
When θ = 90º i.e, when F and d are mutually perpendicular.
W = F . d
W = F d cos 90º ………………………. {since θ = 90º}
W = 0 ……………………………………. {since cos 90º = 0}
Work Done Against Gravitational Force
Consider a body of mass ‘m’ placed initially at a height h(i), from the surface of the earth. We displaces it to a height h(f) from the surface of the earth. Here work is done on the body of mass ‘m’ by displacing it to a height ‘h’ against the gravitational force.
W = F . d = F d cos θ
Here,
F = W = m g
d = h(r) – h(i) = h
θ = 180º
{since mg and h are in opposite direction}
Since,
W = m g h cos 180º
W = m g h (-1)
W = – m g h
Since this work is done against gravitational force, therefore, it is stored in the body as its potential energy (F.E)
Therefore,
P . E = m g h
Power
Power is defined as the rate of doing work.
If work ΔW is done in time Δt by a body, then its average power is given by P(av) = ΔW / Δt
Power of an agency at a certain instant is called instantaneous power.
Relation Between Power and Velocity
Suppose a constant force F moves a body through a displacement Δd in time Δt, then
P = ΔW / Δt
P = F . Δd / Δt ………………… {since ΔW = F.Δd}
P = F . Δd / Δt
P = F . V …………………………… {since Δd / Δt = V}
Thus power is the dot product of force and velocity.
Units of Power
The unit of power in S.I system is watt.
P = ΔW / Δt = joule / sec = watt
1 watt is defined as the power of an agency which does work at the rate of 1 joule per second.
Bigger Units → Mwatt = 10(6) watt
Gwatt = 10(9) watt
Kilowatt = 10(3) watt
In B.E.S system, the unit of power is horse-power (hp).
1 hp = 550 ft-lb/sec = 746 watt
Energy
The ability of a body to perform work is called its energy. The unit of energy in S.I system is joule.
Kinetic Energy
The energy possessed by a body by virtue of its motion is called it kinetic energy.
K.E = 1/2 mv2
m = mass,
v = velocity
Prove K.E = 1/2 mv2
Proof
Kinetic energy of a moving body is measured by the amount of work that a moving body can do against an unbalanced force before coming to rest.
Consider a body of mas ‘m’ thrown upward in the gravitational field with velocity v. It comes to rest after attaining height ‘h’. We are interested in finding ‘h’.
Therefore, we use
2 a S = vf2 – vi2 ………………………. (1)
Here a = -g
S = h = ?
vi = v (magnitude of v)
vf = 0
Therefore,
(1) => 2(-g) = (0)2 – (v)2
-2 g h = -v2
2 g h = v2
h = v2/2g
Therefore, Work done by the body due to its motion = F . d
= F d cos θ
Here
F = m g
d = h = v2 / 2g
θ = 0º
Therefore, Work done by the body due to its motion = (mg) (v2/2g) cos0º
= mg x v2 / 2g
= 1/2 m v2
And we know that this work done by the body due to its motion.
Therefore,
K.E = 1/2 m v2
Potential Energy
When a body is moved against a field of force, the energy stored in it is called its potential energy.
If a body of mass ‘m’ is lifted to a height ‘h’ by applying a force equal to its weight then its potential energy is given by
P.E = m g h
Potential energy is possessed by
1. A spring when it is compressed
2. A charge when it is moved against electrostatic force.
Prove P.E = m g h OR Ug = m g h
Proof
Consider a ball of mass ‘m’ taken very slowly to the height ‘h’. Therefore, work done by external force is
Wex = Fex . S = Fex S cos θ ……………………………. (1)
Since ball is lifted very slowly, therefore, external force in this case must be equal to the weight of the body i.e., mg.
Therefore,
Fex = m g
S = h
θ = 0º ……………………. {since Fex and h have same direction}
Therefore,
(1) => Wex = m g h cos 0º
Wex = m g h …………………………………………………………. (2)
Work done by the gravitational force is
Wg = Fg . S = Fg S cosθ …………………………………………. (3)
Since,
Fg = m g
S = h
θ = 180º …………………. {since Fg and h have opposite direction}
Therefore,
(3) => Wg = m g h cos 180º
Wg = m g h (-1)
Wg = – m g h …………………………………………………………. (4)
Comparing (2) and (4), we get
Wg = -Wex
Or
Wex = – Wg
The work done on a body by an external force against the gravitational force is stored in it as its gravitational potential energy (Ug).
Therefore,
Ug = Wex
Ug = – Wg ………………………… {since Wex = -Wg}
Ug = -(-m g h) ………………… {since Wg = – m g h}
Ug = m g h ……………………………………….. Proved
Absolute Potential Energy
In gravitational field, absolute potential energy of a body at a point is defined as the amount of work done in moving it from that point to a point where the gravitational field is zero.
Determination of Absolute Potential Energy
Consider a body of mass ‘m’ which is lifted from point 1 to point N in the gravitational field. The distance between 1 and N is so large that the value of g is not constant between the two points. Hence to calculate the work done against the force of gravity, the simple formula W = F .d cannot be applied.
Therefore, in order to calculate work done from 1 to N, we divide the entire displacement into a large number of small displacement intervals of equal length Δr. The interval Δr is taken so small that the value of g remains constant during this interval.
Diagram Coming Soon
Now we calculate the work done in moving the body from point 1 to point 2. For this work the value of constant force F may be taken as the average of the forces acting at the ends of interval Δr. At point 1 force is F1 and at point 2, force is F2.
Law of Conservation of Energy
Statement
Energy can neither be created nor be destroyed, however, it can be transformed from one form to another.
Explanation
According to this law energy may change its form within the system but we cannot get one form of energy without spending some other form of energy. A loss in one form of energy is accompanied by an increase in other forms of energy. The total energy remains constant.
Proof
For the verification of this law in case of mechanical energy (Kinetic and potential energy). Let us consider a body of mass ‘m’ placed at a point P which is at a height ‘h’ from the surface of the earth. We find total energy at point P, point O and point Q. Point Q is at a distance of (h-x) from the surface of earth.
