WORK, ENERGY AND POWER 161
(b) Power output =force×velocity,
from which, velocity=
power output
force
.
Force acting on the 300 kg load due to
gravity=300 kg× 9 .81 m/s^2
=2943 N
Hence,
velocity=
1500
2943
= 0 .510 m/s or 510 mm/s.
Problem 20. A lorry is travelling at a
constant velocity of 72 km/h. The force
resisting motion is 800 N. Calculate the
tractive power necessary to keep the lorry
moving at this speed.
Power=force×velocity.
The force necessary to keep the lorry moving at
constant speed is equal and opposite to the force
resisting motion, i.e. 800 N.
Velocity=72 km/h=
72 × 1000
60 × 60
m/s
=20 m/s.
Hence, power=800 N×20 m/s
=16 000 N m/s=16 000 J/s
=16 000 W or 16 kW.
Thus the tractive power needed to keep the lorry
moving at a constant speed of 72 km/h is 16 kW.
Problem 21. The variation of tractive force
with distance for a vehicle which is
accelerating from rest is:
force (kN) 8.0 7.4 5.8 4.5 3.7 3.0
distance (m) 0 10 20 30 40 50
Determine the average power necessary if
the time taken to travel the 50 m from rest is
25 s.
8.0
6.0
4.0
Force (kN)
2.0
01020
Distance (m)
30 40 50
y 1 y 2 y 3 y 4 y 5
Figure 14.9
The force/distance diagram is shown in Figure 14.9.
The work done is determined from the area under
the curve. Using the mid-ordinate rule with five
intervals gives:
area=
(
width of
interval
)(
sum of
mid-ordinate
)
=( 10 )[y 1 +y 2 +y 3 +y 4 +y 5 ]
=( 10 )[7. 8 + 6. 6 + 5. 1
+ 4. 0 + 3 .3]
=( 10 )[26.8]=268 kN m,
i.e. work done=268 kJ
Average power=
work done
time taken
=
268000 J
25 s
=10720 Wor 10 .72 kW.
Now try the following exercise
Exercise 70 Further problems on power
- The output power of a motor is 10 kW.
How much work does it do in 1 minute?
[600 kJ] - Determine the power required to lift a
load through a height of 20 m in 12.5 s
if the force required is 2.5 kN. [4 kW] - 25 kJ of work is done by a force in mov-
ing an object uniformly through 50 m in
40 s. Calculate (a) the value of the force,
and (b) the power.
[(a) 500 N (b) 625 W]