Question

A train of mass 2 x 10^5 kg moves at a constant speed of 72 kmh-¹ up a straight inclined against a frictional force of 1.28 × 10^4N. The incline is such that the train rises vertically 1.0 m for every 100 m travelled along the incline. Calculate the necessary power developed by the train. ​

Answer 1

100×1.28

Step-by-step explanation:

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Answer 2

Approximately
`6.5 * 10^(5)\; {\rm W}` (assuming that
`g = 9.81\; {\rm N\cdot kg^(-1)}`.)

Step-by-step explanation:

Refer to the diagram attached (not to scale.) Let
`\theta` denote the angle of elevation of the incline. Sine the incline rises
`1.0\; {\rm m}` (opposite) for every
`100\; {\rm m}` along the incline (hypotenuse):

`\displaystyle \sin(\theta) = \frac{(\text{opposite})}{(\text{hypotenuse})} = (1.0)/(100)`.

Let
`m = 2* 10^(5)\; {\rm kg}` denote the mass of the train. Decompose the weight
`m\, g` of the train into two components: along the incline and perpendicular to the incline. Refer to the diagram attached (not to scale):

• Weight along the incline:
  `m\, g\, \sin(\theta)`.
• Weight perpendicular to the incline:
  `m\, g\, \cos(\theta)`.

Hence, forces on the train along the incline are:

• Weight along the incline,
  `m\, g\, \sin(\theta)`,
• Friction, and
• Force driving the train forward.

Since the train is moving at a constant velocity, forces on the train should be balanced- both along the incline and perpendicular to the incline.

For forces to be balanced in the component along the incline, the force driving the train upward should be equal to
`m\, g\, \sin(\theta) + (\text{friction})`.

Since
`\sin(\theta) = (1.0 / 100)` and
`(\text{friction}) = 1.28 * 10^(4)\; {\rm N}`:

`\begin{aligned} & m\, g\, \sin(\theta) + (\text{friction}) \\ =\; & (2 * 10^(5))\, (9.81)\, (1.0 / 100) + (1.28 * 10^(4)) \\ \approx\; & 32420\; {\rm N}\end{aligned}`.

Apply unit conversion and ensure that velocity of the train is in standard units:

`\begin{aligned} v &= 72\; {\rm km\cdot h^(-1)} * \frac{1000\; {\rm m}}{1\; {\rm km}} * \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &= 20\; {\rm m\cdot s^(-1)}\end{aligned}`.

Power
`P` is the dot product of force
`F` and velocity
`v`. Since the force driving the train forward along the slope is in the same direction as velocity, the power of this force would be:

`\begin{aligned} P &= F\, v \\ &= (32420 \; {\rm N})\, (20\; {\rm m\cdot s^(-1)}) \\ &\approx 6.5 * 10^(5)\; {\rm W}\end{aligned}`.