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PHYSICAL DEFINITION
When the point of application of force moves in the direction of the applied
force under its effect then work is said to be done.
MATHEMATICAL DEFINITION OF WORK
F
s
Work is defined as the product of force and displacement in the direction of force
W=Fxs
FSinθ F
θ
FCosθ
s
If force and displacement are not parallel to each other rather
they are inclined at an angle, then in the evaluation of work component of force (F)
in the direction of displacement (s) will be considered.
W = (Fcosθ) x s
or, W = FsCosθ
VECTOR DEFINITION OF WORK
F
θ
s
83
,K.V. Silchar
Force and displacement both are vector quantities but their product, work
is a scalar quantity, hence work must be scalar product or dot product of force and
displacement vector.
W=F .s
WORK DONE BY VARIABLE FORCE
Force varying with displacement
In this condition we consider the force to be constant for any
elementary displacement and work done in that elementary displacement is
evaluated. Total work is obtained by integrating the elementary work from initial to
final limits.
dW = F . ds
s2
W = ∫ F . ds
s1
Force varying with time
In this condition we consider the force to be constant for any
elementary displacement and work done in that elementary displacement is
evaluated.
dW = F . ds
Multiplying and dividing by dt,
dW = F . ds dt
dt
or, dW = F . v dt (v=ds/dt)
Total work is obtained by integrating the elementary work from initial to
final limits.
t2
W = ∫ F . v dt
t1
WORK DONE BY VARIABLE FORCE FROM GRAPH
Let force be the function of displacement & its graph be as shown.
aF B
a F M N
a A
s1 ds s2
84
, K.V. Silchar
To find work done from s1 to s2 we consider two points M & N very
close on the graph such that magnitude of force (F) is almost same at both the
points. If elementary displacement from M to N is ds, then elementary work done
from M to N is.
dW = F.ds
dW = (length x breadth)of strip MNds
dW = Area of strip MNds
Thus work done in any part of the graph is equal to area
under that part. Hence total work done from s1 to s2 will be given by the area
enclosed under the graph from s1 to s2.
W = Area (ABS2S1A)
DIFFERENT CASES OF WORK DONE BY CONSTANT FORCE
Case i) Force and displacement are in same direction
θ=0
Since, W = Fs Cos θ
Therefore W = Fs Cos 0
or, W = Fs
Ex - Coolie pushing a load horizontally
a F
s
Case ii) Force and displacement are mutually perpendicular to each other
θ = 90
Since, W = Fs Cos θ
Therefore W = Fs Cos 90
or, W=0
Ex - coolie carrying a load on his head & moving horizontally with constant velocity.
Then he applies force vertically to balance weight of body & its displacement is
horizontal.
F
a
s
mg
85
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