Mechanical Engineering Principles

WORK, ENERGY AND POWER 159

For the gearing,

efficiency =

output energy

input energy

i.e.

75

100

=

10000

input energy

from which, the input energy to the gears

= 10000 ×

100

75

=13333 J.

The input energy to the gears is the same as the
output energy of the motor. Thus, for the motor,

efficiency =

output energy

input energy

i.e.

80

100

=

13333

input energy

Henceinput energy to the hoist

= 13333 ×

100

80

=16667 J= 16 .67 kJ

Now try the following exercise

Exercise 69 Further problems on energy

1. A machine lifts a mass of weight 490.5 N
   through a height of 12 m when 7.85 kJ
   of energy is supplied to it. Determine the
   efficiency of the machine. [75%]


2. Determine the output energy of an electric
   motor which is 60% efficient if it uses
   2 kJ of electrical energy. [1.2 kJ]


3. A machine that is used for lifting a partic-
   ular mass is supplied with 5 kJ of energy.
   If the machine has an efficiency of 65%
   and exerts a force of 812.5 N to what
   height will it lift the mass? [4 m]


4. A load is hoisted 42 m and requires a
   force of 100 N. The efficiency of the hoist
   gear is 60% and that of the motor is 70%.
   Determine the input energy to the hoist.
   [10 kJ]

14.3 Power

Poweris a measure of the rate at which work is
done or at which energy is converted from one form

to another.

PowerP=

energy used

time taken

or P=

work done

time taken

The unit of power is thewatt, W,where1watt

is equal to 1 joule per second. The watt is a small

unit for many purposes and a larger unit called the

kilowatt, kW, is used, where 1 kW=1000 W.

The power output of a motor, which does 120 kJ of

work in 30 s, is thus given by

P=

120 kJ

30 s

=4kW

Since work done=force×distance,

then power=

work done

time taken

=

force×distance

time taken

=force×

distance

time taken

However,

distance

time taken

=velocity

Hence power=force×velocity

Problem 13. The output power of a motor

is 8 kW. How much work does it do in 30 s?

Power=

work done

time taken

,

from which, work done=power×time

=8000 W×30 s

=2 40 000 J=240 kJ

Problem 14. Calculate the power required

to lift a mass through a height of 10 m in

20 s if the force required is 3924 N.

Work done=force×distance moved

=3924 N×10 m=39 240 J