Page 9 - TRIAL 2

Solution 4.6 Efficiency
Work
Power 
Time 1.. The efficiency is a ratio between useful output
Force Displaceme nt energy to the total energy put into the

Time machine.
Displaceme nt 2.. Usually a production of a machine is measure
 Force
Time by using efficiency, which normally expressed
 Force Velocity as a percentage.

P  FV Efficiency  energy output x 100 %
3 energy input
km 10 m 1hr
velocity  70    work output
hr km 3600s work input x 100 %
 19.44m/s
Example 1
P  FV
 1050  19.44 A water pump is running at an efficiency of 82%
with power output 6.7kW to deliver clean water to
 20416.7W water supply tank. Calculate the power input of the

water pump.
Example 4

A construction worker lifts a bag of cement with Solution
force of 418.30 N in 1 minute up a flight of stairs. E  82%
The bag of cement is lifted at 3m. Output  6.7kW

i. How much work is done on the bag of E  output  100%
cement by the construction worker? input
ii. How much is the energy required to walk up 0.82  6.7
the stairs? input
iii. How much power is exerted by the input  6.7
construction worker? 0.82
 8.17kW
Solution
i) How much work is done on the bag of cement by Example 2
the construction worker? A crane’s efficiency to lift a load of 4000kg through
W=mgh
a vertical height of 25m in 15s is 85%. Determine
= (418.30)(3) the input power of the crane?
= 1254.9 J

ii) Energy needed=work done Solution
=1254.9 J
E  85%
iii) How much power is exerted by the construction 0.82 6.7
worker? input
Convert 1 min to s input  6.7
1min=60s 0.82
 8.17kW

= h  25m

m  4000kg
1254.9 t 
= 15s
60
=20.915
8