Chapter 30 : Work, Power and Energy 607
Solution. Given : Mass of the locomotive i.e. mass of the train + own mass (m) = 400 t ;
Velocity acquired (v) = 54 km.p.h. = 15 m/s ; Time (t) = 5 min = 300 s and frictional resistance
= 40 N/t = 40 × 400 = 16 000 N = 16 kN
Let a = Acceleration of the locomotive train.
We know that final velocity of the locomotive after 300 seconds (v)
15 = 0 + a × 300 = 300 a ...(Q v = u + at)∴^215
0·05 m/s
300a==Force required for this acceleration
= ma = 400 × 0·05 = 20 kN ...(i)
As the air resistance varies with the square of the velocity, therefore air resistance at
54 km.p.h.
2
54
500 4500 N 4·5 kN
18⎛⎞
===⎜⎟
⎝⎠...(ii)∴ Total resistance = 16 + 20 + 4·5 = 40·5 kNand work done in one second = Total resistance × Distance = 40·5 × 15
= 607·5 kN-m/s = 607·5 kJ/s
∴ Power = 607·5 kW Ans.EXERCISE 30.1
- A trolley of mass 200 kg moves on a level track for a distance of 500 metres. If the
resistance of the track is 100 N, find the work done in moving the trolley. (Ans. 50 kJ) - What is the power of an engine, which can do a work of 5 kJ in 10 s? (Ans. 500 W)
- An army truck of mass 8 tonnes has a resistance of 75 N/t. Find the power of the truck for
moving with a constant speed of 45 km.p.h. (Ans. 7.5 kW) - A train of mass 200 tonnes moves on a level track having a track resistance of 85 newtons
per tonne. Find the maximum speed of the engine, when the power developed is 320 kW.
(Ans. 67.75 km.p.h.) - A train of mass 150 tonnes moves on a level track with a speed of 20 m/s. The tractive
resistance is 100 newtons per tonne. Determine the power of the engine to maintain this
speed.
Also determine the power of the engine, when the train is to move with an acceleration of
0·3 m/s^2 on a level track. (Ans. 300 kW ; 1200 kW)
30.14.MOTION ON INCLINED PLANE
In the previous articles, we have been discussing the motion of bodies on level surface. But
sometimes, the motion of a body takes place up or down an inclined plane as shown in Fig 30·6 (a)
and (b).