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Question: What would be the power of an engine required to lift 9 metric tonnes of coal per hour for a mine 20...

What would be the power of an engine required to lift 9 metric tonnes of coal per hour for a mine 200m deep. (g = 9.8 ms-2)

Explanation

Solution

To solve this problem one must be aware of the concepts like the definition of power, work-energy theorem etc. According to the work-energy theorem work and energy are mutually interconvertible however the exact statement of the work-energy theorem is that the net work done by the forces acting on an object is equal to its change in kinetic energy.

Complete step by step solution:
The data given in the problem required to solve this problem is given below.

  1. Mass of the coal to be lifted, m = 9 metric tonnes = 9000 kg
  2. Time duration in which the lifting operation is to be performed = 1 hr = 3600 s
  3. Depth of the mine, h = 200 m
  4. Acceleration due to gravity, g = 9.8 m/s2

    We know that the work done is the product of force and displacement.
    Work done= force × displacement{\mathbf{Work}}{\text{ }}{\mathbf{done}} = {\text{ }}{\mathbf{force}}{\text{ }} \times {\text{ }}{\mathbf{displacement}}….(1)
    The force required to move the object up should be equal to the weight of the coal to be lifted and it is given by, F=mgF = mg
    The coal is lifted to a height of h = 200 m. Hence the displacement will be equal to h.
    Putting the values of force and displacement in equation (1), we get,
    W=mgh=9.8×9000×200W = mgh = 9.8 \times 9000 \times 200
    W=1.764×107JW = 1.764 \times {10^7}J
    We know that work done per unit time is known as power and it is given by,
    P=WtP = \dfrac{W}{t}
    Where P is the power in watt or joule per sec.
    P=1.764×1073600P = \dfrac{{1.764 \times {{10}^7}}}{{3600}}
    P=4900WP = 4900W

Hence, in order to lift coal of mass 9 metric tonnes per hour from a mine that is 200 m deep, we need an engine of 4900 w of power.

Note: Power is defined as the rate of change of work done with respect to time. To understand this more clearly let us take an example from this question itself. So in this problem, the work is to lift 9 tonnes of coal to a height of 200m. So if we need to do this work in 1 hour we will require an engine of capacity 4900 W. If we want the same amount of work in half our we will require a bigger engine precisely an engine with twice the capacity of the previous engine as the time has been halved.