Question

An engine of mass 50 tonnes pulls a train of mass 300 tonnes up an incline of 30 0 . The train
starts from rest and moves with a constant acceleration against a total resistance of 150
newtons per tonnes. If the train attains a speed of 60 km.p.h. in a distance of 2 kilometer, find
a) The power of the engine.
b) The tension in the coupling between the engine and train.

Answer 1

Answer:

this is a physics question....post it in physics to get answers and not computer

I hope this helps

Answer 2

The power of the engine is 20406.68 HP

The tension in the coupling between the engine and train is 1565.84 kN

Explanation:

Inclination of the plane $\theta=30^\circ$

Initial speed $u=0$

Final speed $v=60$ km/h

Distance travelled $s=2$ km

Let the acceleration be a

Then by the third equation of motion

$v^2=u^2+2as$

$60^2=0^2+2a\times 2$

$\implies a=\frac{60\times 60}{4}$

$\implies a=900$ km/h²

$\implies a=900\times 1000/(3600)^2$ m/s²

$\implies a=\frac{5}{72}$ m/s²

Mass of the system m = 50 + 300 = 350 tonnes = $350\times 1000$ kg

Therefore, force applied on the system

$F-(350\times 10^3)g\sin 30^\circ-350\times 150=(350\times 10^3)a$

$\implies F=350\times 1000\times\frac{5}{72}+1750\times 1000+52500$ N

$\implies F=1826.81$ kN

Time taken by the engine be t then

$v=u+at$

$\impllies 60=0+900\times t$

$\implies t=\frac{60}{900}$ hour

$\implies t=\frac{60}{900}\times 3600$ s

$\implies t=240$ s

Therefore, power

$P=F\times s/t$

$\implies P=\frac{1826.81\times 1000\times 2000}{240}$ J/s

$\impies P=15223.38\times 1000$ J/s

$\implies P=\frac{15223.38\times 1000}{746}$ HP

$\implies P=20406.68$ HP

If the tension in the coupling be T then, for the engine

$F-T-50\times 10^3g\sin30^\circ-(150\times 50)=50\times 10^3a$

$\implies 1826.81\times 1000-50000\times 10\times\frac{1}{2}-7500=50000\times 0.0694+T$

$\implies 1826.81\times 1000-250\times 1000-7.5\times 1000=3.47\times 1000+T$

$\implies T=1000(1826.81-250-7.5-3.47)$ N

$\implies T=1565.84$ kN

Hope this answer is helpful.

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