Let’s consider two men A and B. A is able to do 10 J of work in 1 hour, but B is able to do only 5 J of work in the same amount of time. Who is more powerful? We say that person A is more powerful because he is able to do a higher amount of work in the same time, compared to B. Let us take two electrical bulbs B_{1} and B_{2}. If you look closely at both bulbs, they have a unique “power rating”. If bulb B_{1} is rated at 25 watt and B_{2} at 50 watt, we say that B_{2} consumes more power.
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Power is defined as the amount of work done or energy converted in a given amount of time. Mathematically, it can be expressed as
where W is the work done in Joule and t is the time taken in seconds. From the above equation, it is clear that more power is consumed or delivered when a large amount of work is done in a short period of time.
The SI unit of power is J/s or watt. Its symbol is W.
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Instantaneous power is defined as the amount of work dW in a time interval dt. Mathematically, we can say that
P = ; where dW is the amount of work done in time dt. Since dW=F.dx; where F is the force applied and dx is the displacement.
Since power is a vector dot product , it is a scalar quantity.
Average power is defined as the total work done W in a total amount of time interval t
Efficiency(n) is the outpower power delivered divided by the input power supplied. Mathematically, it can also be expressed in percentage
Note: Efficiency of a machine is always less than 100 %
For such a case, velocity v=0. Therefore power delivered = zero.
For a given magnitude of force:
is constant in this case.
is variable in this case.
Question.1 A body of mass 1 kg begins to move under the action of a time dependent force, F = (2t i + 3t2 j ) N, where i and j are unit vectors along the x and y axes. What power will be developed by the force at a time t ?
2) An engine pumps 400 kg of water through a height of 10 m in 40 s. Find the power of the engine if its efficiency is 80%(Take g =10 ms^{-2} ).
Question.2 1 kW (b) 2 kW (c) 2.5 kW (d) 1.25 kW
Solution (d)
Work done in raising the water against gravity W = mgh ;
W = 400 × 10 × 10 = 40 kJ
Power used by the engine =
The efficiency of the engine is n=
now to find power input we will use,
Input power = 1 kW0.8 = 1.25 kW.
Question.3 Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10% of energy. How much power is generated by the turbine ? (g =10 ms^{-2})
Solution) c
Power available in falling water = 15 × 10 × 60 = 9000 W; since 10% of energy is lost
Remaining Energy = 90 % of 9000 = 8100= 8.1 kW.
Question.4 If ms^{-1}, then find the instantaneous power
(a)195 W (b)45 W (c)75 W (d) 100 W
Solution) b
Question.1 What is the unit of power?
Answer) The unit of power is J/s or watt.
Question.2 Is Power a scalar or vector quantity?
Answer) Power is a Scalar quantity.
Question.3 When the angle between the force(F) and velocity (v) is 90^{0} ; then the power delivered is
Answer) Since P= F .v = Fvcos ; =90^{0}
∴ P= zero.
Question.4 What is efficiency?
Answer) It is the ratio of the output power delivered to the input power supplied
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