Calculate the power of an engine required to lift 10^5 Kg of Coal per hour from a mine 360m deep, take g as 10 m/s^2.
Answers
Answered by
36
EXPLANATION.
Power of an engine required to lift 10⁵ kg of coal.
Per hour from a mine 360m deep.
⇒ g = 10 m/s².
As we know that,
Formula of :
⇒ Work done = m x g x h.
⇒ Mass = 10⁵ kg.
⇒ g = 10 m/s².
⇒ h = 360m deep.
Work done = 10⁵ x 10 x 360.
Work done = 10⁵ x 3600 joule.
As we know that,
Formula of :
⇒ Power = work done/Time.
⇒ Work done = 10⁵ x 3600 joule.
⇒ Time = 1 hours = 3600 seconds.
⇒ Power = [10⁵ x 3600]/(3600).
⇒ Power = 10⁵ watt.
Answered by
43
Answer :-
Given: m = 10⁵ Kg, g = 10 ms⁻² , h = 360 m,
t = 1 h = 60 × 60s = 3600 s
The work need in lifting a mass m to a height h against the force due to gravity is:
⇒ W = mg × h = mgh
and Power P = Work done/Time taken
- mgh/t
∴ P = 10⁵ × 10 × 360/3600 = 10⁵ W = 100 kW
Note:
- In fact, the actual power of the engine will be much more than 100 kW because (i) some energy will get wasted in overcoming the force of friction and (ii) the efficiency of the engine will always be less than 100%.
Similar questions