Engineering Mechanics

Chapter 30 : Work, Power and Energy „„„„„ 611

(b) Power required to propel the truck on a level track

Since the truck is to move on a level track, therefore it has no resistance due to inclination.
Thus it has to overcome tractive resistance only.

∴Work done in one second = Tractive resistance × Distance = 750 × 10 N- m/s

= 7500 N-m/s = 7.5 kN-m/s = 7.5 kJ/s

∴ Power = 7.5 kW Ans.

(c) Power required to propel the truck down the incline

Since the truck is propelled down the incline, therefore, it has to overcome tractive resistance
minus gravitational pull i.e. resistance due to inclination.

∴ Net resisting force = Tractive resistance – Resistance due to inclination
= 750 – 490 = 260 N
and work done in one second = Net resisting force × Distance = 260 × 10 N-m/s
= 2600 N-m/s = 2.6 kN-m/s = 2.6 kJ/s
∴ Power = 2.6 kW Ans.
Example 30.15. An engine of mass 50 tonnes pulls a train of mass 300 tonnes up an incline
of 1 in 100. The train starts from rest and moves with a constant acceleration against a total resis-
tance of 50 newtons per tonnes. If the train attains a speed of 36 km.p.h. in a distance of 1 kilometre,
find power of the engine. Also find tension in the coupling between the engine and train.

Solution. Given : Mass of the engine (m 1 ) = 50 t ; Mass of the train (m 2 ) = 300 t or total mass

(m) = 50 + 300 = 350 t ; Slope

1

(sin ) 0.01

100

θ= = ; Initial velocity (u) = 0 (because, it starts from

rest); Tractive resistance = 50 N/t = 50 × 350 = 17 500 N = 17.5 kN ; Final velocity (v) = 36 km.p.h.

= 10 m/s and distance (s) = 1 km = 1000 m

Fig. 30.7.

Power of the engine

Let a = Acceleration of the train.

We know that resistance due to inclination

= mg sin α = 350 × 9.8 × 0.01 = 34.3 kN

We also know that relation for accelerationm,

v^2 = u^2 + 2 as

(10)^2 = (0)^2 + 2a × 1000 = 2000 a

or

(^100) 0.05 m/s 2
2000
a==