An engine draws a train up an incline of 1 in 100 at the rate of 36 km

1.	An engine draws a train up an incline of 1 in 100 at the rate of 36 kmh-1. If the resistance due to friction is 5 kg wt per ton, find out the power of the engine. Mass of train and engine is 100 metric ton.
Answer» Given : m = 100 metric ton = 100 × 1000 kg Total force of friction, f₁ = 100 × 5 = 500 kg wt = 500 × 9.8 N = 4900 N \(sin\,θ=\frac{1}{100}\) Suppose, f₂ = Downward force on the train = component of its weight acting in downward direction parallel to the inclined plane = mg sin θ = 100 × 1000 × 9.8 × \(\frac{1}{100}\) = 9800 N When F be the total force against which engine has to work, then F = f₁ + f₂ = 4900 + 9800 = 14700 N v = velocity of train = 36 kmh^-1 = 36 × \(\frac{5}{18}\) ms^-1 = 10 ms^-1 ∴ Power of the engine, is given by P = F × v = 14700 × 10 = 147000 watt = 147 kW.

An engine draws a train up an incline of 1 in 100 at the rate of 36 kmh-1. If the resistance due to friction is 5 kg wt per ton, find out the power of the engine. Mass of train and engine is 100 metric ton.

Answer»

Given : m = 100 metric ton

= 100 × 1000 kg

Total force of friction,

f₁ = 100 × 5 = 500 kg wt

= 500 × 9.8 N = 4900 N

\(sin\,θ=\frac{1}{100}\)

Suppose, f₂ = Downward force on the train = component of its weight acting in downward direction parallel to the inclined plane = mg sin θ

= 100 × 1000 × 9.8 × \(\frac{1}{100}\)

= 9800 N

When F be the total force against which engine has to work, then

F = f₁ + f₂

= 4900 + 9800 = 14700 N

v = velocity of train

= 36 kmh^-1

= 36 × \(\frac{5}{18}\) ms^-1

= 10 ms^-1

∴ Power of the engine, is given by

P = F × v = 14700 × 10

= 147000 watt

= 147 kW.

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