# Mechanical power: what is it, applications, examples

The **mechanical power** is the rate at which work is done, expressed in mathematical form by the amount of work performed per unit time. And since work is done at the expense of the energy absorbed, it can also be considered as energy per unit of time.

Calling *P* power, *W* work, *E* energy, *and* time, all these can be summarized in easy-to-use mathematical expressions:

Figure 1. The Gossamer Albatross, the “flying bicycle”, crossed the English Channel in the late 1970s, using only human energy. Source: Wikimedia Commons. Albatross Gossamer. Guroadrunner on Wikipedia in English [Public domain]

The good:

It was named after the Scottish engineer James Watt (1736-1819), known for creating the condenser steam engine, an invention that started the Industrial Revolution.

Other units of power used in industries are hp ( *power* or power) and CV (horsepower). The origin of these units also dates back to James Watt and the Industrial Revolution, when the measurement standard was the pace at which a horse was working.

Both hp and CV are approximately equivalent to ¾ of a kilo-W, and are still widely used, especially in mechanical engineering, for example, in the designation of engines.

Watt multiples, such as the aforementioned kilo-W = 1000 W, are also frequently used in electrical energy. It’s because joule is a relatively small unit of energy. The British system uses foot-pound/second.

__What is it and applications in industry and energy__

__What is it and applications in industry and energy__

The concept of power is applicable to all types of energy, whether mechanical, electrical, chemical, wind, sonic or any other type. Time is very important in the industry because processes must be executed as quickly as possible.

Any engine will do the work necessary to have enough time, but the important thing is to do it in the shortest possible time to increase efficiency.

A very simple application is immediately described to clarify the distinction between work and power.

Suppose a heavy object is pulled by a rope. To do this, some external agent is needed to do the necessary work. Let’s say that this agent transfers 90 J of energy to the object chain system, so that it fires for 10 seconds.

In this case, the energy transfer rate is 90 J / 10 or 9 J / s. So we can say that this agent, a person or an engine, has an output power of 9 W.

If another external agent is able to obtain the same displacement, in less time or transferring less energy, it is capable of developing greater power.

Another example: suppose an energy transfer of 90 J, which can put the system in motion for 4 seconds. The output power will be 22.5 W.

**machine performance**

Power is very much about performance. The energy supplied to a machine is never completely transformed into useful work. An important part usually dissipates in heat, which depends on many factors, for example the design of the machine.

Therefore, it is important to know the performance of machines, which is defined as the ratio between the work delivered and the energy supplied:

*η = work delivered by machine / energy supplied*

Where the Greek letter *η* denotes performance, a dimensionless quantity always less than 1. If it is also multiplied by 100, the income is given in percentage terms.

__Examples__

__Examples__

– Humans and animals develop potency during locomotion. For example, when climbing stairs, it is necessary to work against gravity. Comparing two people climbing a ladder, the person who goes up all the steps first will have developed more strength than the other, but both have done the same job.

– Appliances and machines have specified output power. An incandescent bulb suitable for lighting a room well has a power output of 100 W. This means that the bulb transforms electrical energy into light and heat (mostly) at a rate of 100 J/s.

– The motor of a lawn mower can consume about 250 W and that of a car is on the order of 70 kW.

– A home water pump typically delivers 0.5 hp.

– The sun generates 3.6 x 10 ^{26} W of power.

**power and speed**

The instantaneous power is obtained by taking an infinitesimal time: *P* = *dW / dt* . The force that does the work that causes the infinitesimal small displacement *d x* is

**(both are vectors), so**

*F**dW =*

**F**

*●**d*. Replacing everything in the expression for power, it remains:

**x****human power**

People are able to generate wattages of around 1500 W or 2 horsepower, at least for a short period of time, like lifting weights.

On average, daily power (8 hours) is 0.1 hp per person. Much of which translates to heat, roughly the same amount a 75W incandescent bulb generates.

An athlete in training can generate, on average, 0.5 hp equivalent to approximately 350 J / s, transforming chemical energy (glucose and fat) into mechanical energy.

When it comes to human power, it is generally preferred to measure in kilocalories/hour rather than watts. The required equivalence is:

*1 kilocalorie = 1 nutritional calorie = 4186 J*

A power of 0.5 hp sounds like a very small amount and is for many applications.

However, in 1979, a human-powered bicycle that could fly was created. Paul MacCready designed the *Gossamer Albatross* , which crossed the English Channel generating 190 W of average output (figure 1).

__Electric power distribution__

__Electric power distribution__

An important application is the distribution of electrical energy among users. Companies that supply electricity bill for the energy consumed, not the rate at which it is consumed. That’s why anyone reading your bill carefully will find a very specific unit: the kilowatt-hour or kW-h.

However, when Watt’s name is included on this unit, it refers to energy and not energy.

The kilowatt-hour is used to indicate the consumption of electrical energy, since the joule, as mentioned earlier, is a very small unit: *1 watt-hour or Wh* is the work done in 1 hour for a power of 1 watt.

So 1kW *-h* is the work done in one hour, working with a power of 1kW or 1000W. Let’s put the numbers to pass these quantities into joules:

*1 Wh = 1 L x 3600 s = 3600 J*

*1 kW-h = 1000 W x 3600 s = 3.6 x 10 ^{6} J*

It is estimated that around 200 kWh per month can be consumed in a home.

__Exercises__

__Exercises__

**Exercise 1**

A farmer uses a tractor to pull a hay bale of M = 150 kg on an incline of 15° and bring it to the barn at a constant speed of 5.0 km/h. The coefficient of kinetic friction between the hay bale and the ramp is 0.45. Find the power of the tractor.

**Solution**

For this problem, it is necessary to draw a free-body diagram for the hay bale that rises on the inclined plane. Let ** F be** the force applied by the tractor to lift the load, α = 15° is the inclination angle.

Also involved are kinetic frictional force *f ** _{friction}* that opposes motion, plus normal

**N**and weight

**W**(not to be confused with work weight W).

Newton’s second law offers the following equations:

*∑ Fx = F –W _{x} –f _{rubbing} = 0* (since the bale rises at constant speed.

*∑Fy = N – W _{y} = 0* (no movement along the x axis)

The kinetic friction force is calculated by:

*f _{effr} = coefficient of kinetic friction x magnitude of normal*

*f _{rub} = 0.45. Wy = 0.45 x 150 kg x9.8 m / s2 x cos 15º = 639 N*

*F = W _{x} + f _{rub} = Mg sin α *

*= 150 kg. 9.8 m / s*

^{2}. sin 15 + 639 N =*1019.42 N*

Speed and force have the same direction and meaning, therefore:

*P = F *

*●*

**v**= F. vIt is necessary to transform the speed units:

*v = 5.0 km / h = 1.39 m / s*

By substituting values, you finally get:

*P = 1019.42 N x 1.39 m / s = 1417 W = 1.4 kW*

**Exercise 2**

The engine shown in the figure will raise the block by 2 kg, starting from rest, with an acceleration of 2 m/s ^{2} and in 2 seconds.

Calculate:

a) The height reached by the block at that moment.

b) The power that the engine must develop to achieve it.

**Solution**

a) It is a uniformly varied rectilinear movement; therefore, the corresponding equations will be used, with initial velocity 0. The height reached is given by:

*y = ½ to ^{2} = ½. 2 m / s ^{2} . (2 s) ^{2} = 4 m.*

b) To find the power developed by the engine, the equation can be used:

*P = Δ **W / Δ **t*

And since the force exerted on the block is caused by the tension in the string, which is constant in magnitude:

*P = (ma) .y / Δt **= 2 kg x* 2 m / s ^{2} x 4 m / 2 s = 8 W