Mechanical Engineering Principles

160 MECHANICAL ENGINEERING PRINCIPLES

Power=

work done

time taken

=

39 240 J

20 s

=1962 W or 1 .962 kW

Problem 15. 10 kJ of work is done by a

force in moving a body uniformly through

125 m in 50 s. Determine (a) the value of

the force, and (b) the power.

(a) Work done=force×distance,

hence 10 000 J=force×125 m,

from which, force=

10 000 J

125 m

=80 N

(b) Power=

work done

time taken

=

10 000 J

50 s

=200 W

Problem 16. A car hauls a trailer at

90 km/h when exerting a steady pull of

600 N. Calculate (a) the work done in 30

minutes and (b) the power required.

(a) Work done = force×distance moved. The

distance moved in 30 min, i.e.^12 h, at

90 km/h=45 km.

Hence,work done = 600 N×45 000 m=

27 000 kJor27 MJ

(b) Power required

=

work done

time taken

=

27 × 106 J

30 ×60 s

=15 000 W or 15 kW

Problem 17. To what height will a mass of

weight 981 N be raised in 40 s by a machine

using a power of 2 kW?

Work done=force×distance.

Hence, work done=981 N×height.

Power=

work done

time taken

,

from which, work done=power×time taken

=2000 W×40 s

=80 000 J

Hence 80 000=981 N×height,

from which, height=

80 000 J

981 N

= 81 .55 m

Problem 18. A planing machine has a

cutting stroke of 2 m and the stroke takes

4 seconds. If the constant resistance to the

cutting tool is 900 N, calculate for each

cutting stroke (a) the power consumed at the

tool point, and (b) the power input to the

system if the efficiency of the system is 75%.

(a) Work done in each cutting

stroke=force×distance

=900 N×2m=1800 J

Power consumedat tool point

=

work done

time taken

=

1800 J

4s

=450 W

(b) Efficiency=

output energy

input energy =

output power

input power

Hence

75

100

=

450

input power

from which, input power= 450 ×

100

75

=600 W

Problem 19. An electric motor provides

power to a winding machine. The input

power to the motor is 2.5 kW and the overall

efficiency is 60%. Calculate (a) the output

power of the machine, (b) the rate at which

it can raise a 300 kg load vertically upwards.

(a) Efficiency,

η=

power output

power input

i.e.

60

100

=

power output

2500

from which,

power output=

60

100

× 2500

=1500 W or 1 .5kW.