Question

Calculate the power of an engine required to lift 10⁵ kg of coal per hour from a mine 360 m deep.
(taking g = 10 ms-²)​

Answer 1

m = 10^5 kg, g = 10ms^−2 ,h = 360m, t = 1h =3600s

The work needed in lifting a mass m to a height h against the force due to gravity is

W=mg×h

And, power, P= W / t = mgh / t

∴P= ( 10^5 × 10 × 360 / 3600 ) W

=> 10^5 = 100kW

Answer 2

Solution:

Values which had been provided with us are as follows,

→ Mass of coal = 10⁵ kg

→ Depth of mine = 360 m

→ Time which had been taken is one hour

We are asked to find out the power of engine.

But before that would convert the time which is in hours into seconds.

Let's convert it:-

As we know that 1 hour contains 60 minutes.

→ 1 hour = 60 minutes

Thus, 60 minutes would contain seconds be,

→ 60 × 60

→ 3600 s

Always remember that the power spent by a source is measured as amount of work done by a particular source in a quantity that is seconds.

Now we would be using the formula of calculating the Power (P).

→ Power (P) = Work done / Time taken

Thus,

→ Power (P) = mgh / t

Substituting the required values:

Again where,

• g = 10 ms-²
• m = 10⁵ kg
• h = 360 m
• t = 3600 s

Calculating:

→ 10⁵ × 10 × 360 / 3600

→ 10⁵ × 10 × 36 / 360

→ 10⁵ × 1 × 36 / 36

→ 10⁵ × 36 / 36

→ 10⁵

Thus, power of an engine required to lift 10⁵ kg of coal per hour from a mine 360 m deep is 100 W

Additional Information:

• The rate of doing work is called power

• Power is a scalar quantity

• S.I. unit of power is watt ( W )

• 1 watt = 1 Joule / 1 second = 1 J s-¹

• Bigger units of power = kilowatt, megawatt, gigawatt.

• Smaller units of power = milliwatt , microwatt