3. Q. What is work?
A. Work is said to be done when a force acts on a
body and moves it through a certain
displacement.
WORK
d
Work done by the Force F
mm
F
5. If force is applied on a body and it moves the
body through a displacement ‘d ’, then the work 'W' is
defined by the relation
W = F. d
We have already studied scalars and vectors.
Q. What is the nature of work?
A. Work is a scalar quantity.
WORK
6. Unit of work:
The SI unit of work is joule (J).
One Joule
When a force of one Newton moves a body
through a distance of one meter in the direction of
force, then the work done is equal to one joule.
WORK
7. Positive Work
CASES OF WORK
When force and displacement are in the same
direction
Then = 0°
W = F d cos 0°
= F d x 1
= F d
11. When force and displacement are in the opposite
direction
Then = 180°
W = F d cos 180°
= F d x (-1) cos 180° = -1
W = - F d
CASES OF WORK
Negative Work
13. Q. Give an example of positive work?
A. Pushing something horizontally is an example of
positive work.
Q. Give an example of negative work?
A. Lifting something vertically upwards is an example
of negative work.
WORK
16. • Power is measured by the amount of Work done in
One Second.
• If Work W is done in ‘t’ seconds, then Power
Thus,
• Smaller the time in which Work is done, the greater is
the Power.
• Power is a Scalar Quantity.
• Unit of Power is Watt.
• Watt is equal to Joule/second.
• Watt is equal to Kgm2/s3 .
• Dimension of power is M1 L2 T-3.
P=W/
T
18. • Energy is the Capacity of a body to
do Work.
• Energy represents the total amount
of Work that a Body can do.
• Unit of Energy is Joule.
• Joule = Kgm2/s2.
19. Numerical
The two springs for reversing the motion of a
heald shaft each have to be stretched
15cm to put them in position with the heald
shaft down. If the stiffness of each spring
is 1.5N/cm find the work done in putting
the springs in position.
Stiffness∞ force
20. Solution :
Total force required to stretch each spring =
1.5×15 = 22.5N
Total force = 22.5×2
= 45N
Work done = f.d
=45×15 Ncm
=675Ncm
22. Numerical on Power
A ringframe traveller, moving in a circle of
5cm in diameter at 9000rev/min, offers a
resistance to movement of 0.15N. If the
frame has 240 spindles, calculate the
power expended in moving the travellers.
23. Solution:
Distance moved by traveller in one
revolution = 5¶
= 5(3.14)
= 15.70cm
Distance moved per second =
15.70×9000÷60
= 23.55m
24. Solution:
W = f.d
Work done per second on each traveller =
23.55×0.15N
W = 3.53J
Total work done = 3.53 on each of 240
spindles per second
Power expended = 847.2W