Mechanical Engineering Principles

156 MECHANICAL ENGINEERING PRINCIPLES

work done=^12 ×base×height

=^12 × 30 × 10 −^3 m×6N

= 90 × 10 −^3 J= 0 .09 J

(b) Work done in extending the spring from
30 mm to 50 mm is given by areaABCEof
Figure 14.6, i.e.

work done=areaABCD+areaADE

=( 20 × 10 −^3 m×6N)

+^12 ( 20 × 10 −^3 m)(4N)

= 0 .12 J+ 0 .04 J= 0 .16 J

Problem 6. Calculate the work done when

a mass of 20 kg is lifted vertically through a

distance of 5.0 m. Assume that the

acceleration due to gravity is 9.81 m/s^2.

The force to be overcome when lifting a mass of

20 kg vertically upwards is mg,

i.e. 20× 9. 81 = 196 .2 N (see Chapter 13).

work done=force×distance

= 196. 2 × 5. 0 =981 J

Problem 7. Water is pumped vertically

upwards through a distance of 50.0 m and

the work done is 294.3 kJ. Determine the

number of litres of water pumped. (1 litre of

water has a mass of 1 kg).

Work done=force×distance,

i.e. 2 94 300=force× 50. 0 ,

from which force=

2 94 300

50. 0

=5886 N

The force to be overcome when lifting a mass

m kg vertically upwards is mg, i.e. (m× 9 .81) N

(see Chapter 13).

Thus 5886=m× 9. 81 , from which mass,

m=

5886

9. 81

=600 kg.

Since 1 litre of water has a mass of 1 kg,600 litres

of water are pumped.

Problem 8. The force on a cutting tool of a

shaping machine varies over the length of

cut as follows:

Distance (mm) 0 20 40 60 80 100

Force(kN) 6072655344 50

Determine the work done as the tool moves

through a distance of 100 mm.

The force/distance graph for the given data is shown

in Figure 14.7. The work done is given by the area

under the graph; the area may be determined by an

approximate method. Using the mid-ordinate rule,

with each strip of width 20 mm, mid-ordinatesy 1 ,

y 2 ,y 3 ,y 4 andy 5 are erected as shown, and each is

measured.

Area under curve =

(

width of

interval

)(

sum of

mid-ordinate

)

=( 20 )( 69 + 69. 5 + 59 + 48

+ 45. 5 )

=( 20 )( 291 )

70

60

50

40

Force / kN 30

20

10

(^0204060)
Distance / mm
80 100
y^1
= 69
y^3
= 59
y^4
= 48
y^2
= 69.5
y^5
= 45.5
Figure 14.7