Mechanical Engineering Principles

WORK, ENERGY AND POWER 155

1000

500

0 5 10 15

Distance / m

Force / N

20

AB

CD

E F G

Figure 14.4

pointsCandDa constant force of 500 N moves
theloadfrom5mto20m

Total work done=area under the force/distance
graph

=areaABFE+areaCDGF

=(1000 N×5m)

+(500 N×15 m)

=5000 J+7500 J

=12500 J= 12 .5kJ

Problem 4. A spring, initially in a relaxed

state, is extended by 100 mm. Determine the

work done by using a work diagram if the

spring requires a force of 0.6 N per mm of

stretch.

Force required for a 100 mm extension
=100 mm× 0 .6N/mm=60 N.
Figure 14.5 shows the force/extension graph or
work diagram representing the increase in extension
in proportion to the force, as the force is increased

60

Force / N

30

050

Extension / mm

100

Figure 14.5

from 0 to 60 N The work done is the area under the

graph, hence

work done=^12 ×base×height

=^12 ×100 mm×60 N

=^12 × 100 × 10 −^3 m×60 N

=3J

(Alternatively, average force during

extension=

( 60 − 0 )

2

=30 N

and total

extension=100 mm= 0 .1m,

hence

work done=average force×extension

=30 N× 0 .1m=3J)

Problem 5. A spring requires a force of

10 N to cause an extension of 50 mm.

Determine the work done in extending the

spring (a) from zero to 30 mm, and (b) from

30 mm to 50 mm.

Figure 14.6 shows the force/extension graph for the

spring.

(a) Work done in extending the spring from zero to

30 mm is given by areaABOof Figure 14.6,

i.e.

10

8

6

4

2

01020

Extension / mm

Force / N

30 40 50

A

B C

D

E

Figure 14.6