Work
Work
Work
F • •
If the force and the displacement are not in the same direction then the
component of the force (or the component of the displacement) which is
in the same direction of the displacement must be multiplied.
d
F
• F cos
•
If the box is lifted against the gravitational pull (with a constant speed)
then the same amount of force must be applied upward as the earth is
applying downward.
Fg = mg
The amount of force applied on the box will be same as the weight i.e.
‘mg’ and in case of perpendicular displacement it is the vertical height ‘h’.
Therefore:
W=F.d (here F is the gravitational force)
W = m.g.h (as F = mg and d = h).
X2
So here work done is equal to gravitational potential energy gain by the
box when it is lifted a vertical height ‘h’.
As W = F. d therefore:
WF ……………….. (1)
Wd ……………….. (2)
1
F ………………..(3)
d
Thus if more force is needed to move an object in the same distance then
more work will be done and if the object is pushed a longer distance then
more work will be done. Again in the Earth’s gravitational field work done
due to height gain is W = m g h . Thus:
W m …………………(4)
W h …………………. (5) (g is constant)
Thus if a massive object is lifted more force will be required and more
work will be done, and if the object is lifted a greater height (must be
inside a uniform gravitational field) then more work will be done.
NOTES:
i. An opposing force must change a displacement, to make a work
done. If a force is applied and no displacement then there will be no
work done The applied force and the displacement must be in
parallel to make a work done. If the applied force and the
displacement are in the same direction then the work done is
positive.
ii. If the applied force and the displacement moved are perpendicular
to each other then there will be no work done.
iii. The object must move with a constant speed. So that the applied
force and the frictional forces are same. Work is always done by the
opposing force.
Force applied
Displacement
Displacement Force
Work done in this diagram is ‘Zero’ Work done in this diagram is not ‘Zero’ as
as the force and the displacement is the force and the displacement are in
perpendicular to each other. parallel.
4m
2m 30N
30o
Energy:
Energy is the capacity of doing work. The unit of energy is ‘joule’ (J),
and it is a scalar quantity.
Types of Energy:
iii. Kinetic Energy: Energy gained by an object for its motion. K.E can
be calculated by the equation:
1
E ke = m v ………………….(ii)
2
2
v. Sound energy.
1
x. Elastic potential energy = F . ∆ x ……………………(vi)
2
GPE ∝ h ……………(ii)
2m
m
E/J
EGPE
2E h h 2EGPE
m/kg
m 2m
GPE ∝ m m
E/J m
2h
2E EGPE
h 2EGPE
h/m
h 2h
GPE ∝ h
When this object is released, the energy (GPE) will start converting to
Kinetic energy, just before hitting the ground all of its GPE will be
converted into KE, and the amount of KE will same as GPE. Therefore :
1 2
mgh= m v
2
v=√ ( 2 gh ) …………….. (i)
and as the object is just dropped, in this case the initial velocity u=0
.Thus
2
v =2 gh
Kinetic Energy:
A moving object has kinetic energy due to its motion, and this
amount of energy can be calculated by the equation:
1 2
KE= m v ………………………. (ii)
2
KE = mad
v vt v vt
KE=m a= d=
t 2 t 2
When a moving object applies the brake and comes to a rest then there is
an energy change and work is done.
The amount of work done is equal to the total amount of KE change into
heat and sound produced when the brakes are applied.
As the amount of work done is the KE change for a moving object so :
Work done:
1 2
W= mv
2
1 2
Fd= mv
2
Brak
ing
dist
anc
e/m
Velocity /m/s
When an object is thrown vertically upward then the KE of the object
converts into GPE as it gains height. At its maximum height all of its KE
converts into GPE, and the object starts to fall to the ground then the GPE
again converts into KE and just before the impact all the GPE converts
into KE. While dropping, the object has both KE and GPE.
KE = 0
GPE= max
K G
E/ P
h J E/
J
KE = max GPE = 0
Height/m Height/m
The KE at the beginning is equal to the GPE at the maximm height. Here
it is assumed that no energy has been converted to heat due to air
friction or any other process.
NOTES:
i. The total amount of energy in this universe is constant.
ii. To calculate the energy of a system, calculate the amount of initial
energy (input energy) that should be the final amount of energy.
iii. Find the total amount of energy change that should be the work
done in that system.
iv. Different types of energy and their source must be able to be
identified.
v. In a system some energy will always be lost by mainly ‘heat’ and
‘sound’ that will give error in the final calculation. So when there is
a difference in the results as there is an energy lost.
W
Thus : Power P= ………………..(a)
t
E
P=
t ………………(b)
W =P .t ……………(d)
W F.d d
P= = =F . =F . v
t t t
P=F v …………..(e)
Po
Po
w
w
er
er
/
/
w
w
at
at
Energy/J Time/s
1
P∝E P∝
t
- Power indicates, the machine can do the work (or convert energy)
at a faster rate.
- Power indicates, the machine can convert more energy in a certain
time.
NOTES:
i. To calculate the power of a machine, first calculate the total energy
change or work done.
ii. Find the time taken for this energy change or work done. Convert
this time into second.
iii. Divide the work done or energy change by the time taken.
EFFICIENCY:
Efficiency is a comparison between the input energy and the useful output
energy. Every machine converts energy from one/more form into other
form/s of energy. Therefore by comparing the input energy and the useful
output energy the efficiency of a machine can be calculated.
Q1 Q2
Heat
Heat Source Engine Cold Sink
Work done
The efficiency of a heat engine is the ratio of useful output (Work done).
So:
W
efficiency =
Q1
( Q1−Q2 )
e= …………………(c)
Q1
Q2
e=1− …………………….(d)
Q1
Thus if the density of the air is ρ then the mass of the air hitting the
blades per second is
m=V ρ
This equation is also applicable for the water turbine or steam turbine.