Power

We hear the word “power” quite frequently in day to day life. Cricket commentators talk about the Power behind a batsman’s shot while broadcasting a cricket match, and football commentators may talk about a particularly powerful free kick by a player.  We among ourselves have, at one point or another, arm-wrestled with a friend to decide who is more powerful!

Browse more Topics Under Work Energy And Power

• Collisions
• Concepts of Potential Energy
• Conservation of Mechanical Energy
• Potential Energy of a Spring
• Power
• The Scalar Product
• Work and Kinetic Energy
• Work-Energy Theorem
• Various Forms of Energy: The Law of Conservation of Energy

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Power

Simply put is the rate at which work is done. Power, like work, is a scalar quantity.  It is denoted by P and its SI unit is watt (W). This unit is named after Scottish scientist James Watt, who is credited with the invention of the steam engine. Thus, the average power formula over a time period is:

For example, a wrecking ball of mass 500 kg is dropped from a height of 50 metres on the ground. Let us calculate the power of its impact on the ground. From the law of conservation of mechanical energy, we know that Δ(K + V) = 0. Let us try and solve this.

To find the Time

Here, ΔV = m g h  = 500 * 9.8 * 50 or,  ΔV = ΔK = 2,45,000J
Since, it is a free fall, W = ΔK = 2,45,000J. Therefore K = (mv²)/2
Hence, 2,45,000 = (500 * v²)/2 or, (2,45,000 * 2)/500 = v²
Which gives v = √98o = 31.3 m/s.
Since, for a free fall, v = gt, we have: 31.3 = 9.8 × t
or t = 31.3/9.8 =  3.19s

To find the Power

By applying the power formula, P = W/t
we have, here, P = 2,45,000/3.19 = 76,802.5 W
Hence, P = 76.8 KW. Hence, the power of the wrecking ball’s impact on the surface is 76.8 kilowatt.

Power in Terms of Force and Velocity

We already know that power is the time rate of Work done.  So, at any given point in time, the power can be defined as:

dW is the work done in the time period of dt.  Since, we know that W = F.d,  we can rewrite the above equation as:

Here, F is the force, dx is the displacement of the object and dt is the time period of that displacement. Since, we know that v = dx/t, we can rewrite the above equation as:

F is the force and v is the instantaneous velocity of the object.  Force and velocity are the vector quantities and power is the scalar product of these two vectors.  Hence, we have derived the power formula in terms of force and velocity. So,

P = F v cosθ

Let us consider another example. While testing a newly developed engine, the engineers of Honda see that their 125 kg testing motorcycle propels from standstill to a top speed of 100 km per hour in 5 seconds. If the same engine is used in their racing motorcycle of mass 85 kg, how many seconds will elapse before the racing motorcycle achieves a speed of 100 km per hour?

Solution: We have: v = 100 × (1000/3600) = 27.78 m/s. We know that, a = v/t or average acceleration of the test motorcycle a = 27.78/5 = 5.56 m/s². We also know that, F = ma. Hence Force applied by engine is F = 125 × 5.56 = 695 N. By applying the power formula, P = F v, the peak power of the engine is P = 695 × 27.78 = 19,307 W

For the Racing Motorcycle

Since, the engine is the same, its peak power will remain constant. Therefore P = 19,307W. At peak power, the speed of racing motorcycle is same, or v = 27.78 m/s. By applying the power formula, P = F v or, 19307 = F × 27.78, we get, F = 19307/27.78 = 695 N. Since, F = m a, we have 695 = 85 × a.

Therefore the acceleration of the racing motorcycle is a = 695/85 = 8.18 m/s². Hence the time required for the racing motorcycle to reach 100 kmph is, t = v/a = 27.78/8.18 or 3.4 seconds.

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Power, Horsepower and Kilo-Watt-Hour

Other than the Watt, we encounter “horsepower” or hp as the most common unit to describe power. All cars and motorcycles use hp to describe the power of their engines.  Also, most electric motors also use h.p. to describe their power rating.

1 hp = 746 W

Kilo-watt-hour or KWH is the unit of energy.  It is used most commonly in our electric bills to denote the number of electrical units consumed.  One kWh is the application of 1000 W of power for 1 hour.  So,

1 KWH = 1000 W x 3600 seconds = 3600000 J

Solved Examples for You

Q: A builder calls an engineer to install an elevator in his building.  When the engineer inquires about the requirements, builder says that he wants the elevator to be able to carry a maximum of eight people at an average speed of 3 m/s.  If the average load of a person is assumed to be 70 kg, and the elevator itself weighs 150 kg, find out the minimum power rating of the required motor in hp, if there is no friction involved.

If the total height of the top floor of the building is 50 m, find out the cost of transporting eight people to the top floor, if the cost of electricity is ₹11.65 per unit.

Solution: Mass of elevator with eight people = 8 × 70 + 150 = 710 kg. Hence, F = 710 × 9.8 = 6,958N. Thus the minimum force required by the motor to overcome the gravity is F = 6,958N. By applying the power formula, P = F v, the minimum power rating of the required motor is P = 6958 × 2 = 13,916 W.

Also, 1 hp = 746 W, therefore required hp rating of motor is = 13916/746 or, 18.65 hp. The height of the top floor is 50 m and the average speed of the elevator is 2 m/s. Also v = d/t or, t = 50/2

The time taken by elevator to the top floor is 25 seconds. Electrical consumption in kWh = motor rating in kW × time in hours. Electrical units consumed in one trip = 13.916 × (25/3600). Energy Units consumed in one trip = 0.097 kWh. Since the cost of one unit is ₹11.65 therefore the total cost of one trip in terms of electricity is, 11.65 × 0.097 or, ₹1.13