Fig. (2
Illustration:
1) A block of mass M is pulled along a horizontal surface by applying a force at an angle
with horizontal. Co-efficient of friction between block and surface is
. If the block travels with uniform velocity, find the work done by this applied force during a displacement d of the block.
Solution: The force acting on the block is shown in figure.
Fig (3)
As the block moves with uniform velocity the force add up to zero.
FCosq =
N ---------------- (1)
FSinq+N= Mg ---------------- (2)
Solving (1) and (2)
FCosq =
(Mg-FSinq)
F =
Work done by this force during a displacement d is
W = F.dCosq =
WORK DONE BY VARYING FORCE:If the force is a variable (i.e. changes) then work done is
= (Where
remains constant for displacement Ds)
(If
S
0 )
Graphical interpretation of above result:
Work done = Area under F vs. S graph.
Area need to add with sign.
Fig (4)
SPRING:The spring force varies as a lines function of compression or elongation.
F = -KX, where K is stiffness constant of spring.
Work done in on spring:
(Initial compression/elongation is X0 and final compression/elongation is X1)
Proof:
(Assuming force is parallel to displacement)
=
Illustration:
A block of mass m released from rest onto an ideal non-deformed spring of spring constant ‘K’ from a negligible height. Neglecting the air resistance, find the compression ‘d’ of the spring.
Solution:
Fig (6)
Here there are two forces involved, weight and spring force.
(Work done by weight)
Net work done must be equal to change in kinetic energy (theorem to be discussed in next section).
K= 0-0 = 0
W
mg + W
spring = 0
isq. Then work done is 
