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# Power – Definition, Formula, Unit, Problems

• Last Updated : 13 Sep, 2021

In everyday life, we hear the word “Power” rather often. While broadcasting a cricket match, cricket commentators may discuss the force behind a batsman’s stroke, while football commentators may discuss a particularly strong free-kick by a player. At some point or another, we’ve all arm-wrestled with a friend to see who is the most powerful. Before we discuss power, let’s discuss a few points of work and energy.

### Work

When a body or item moves due to the application of external force, it is said to be doing work. Work may be defined as an activity involving movement and force in the force’s direction.

When we push any object, we exert an external force known as F, and the object moves a given distance (d) as a result of this force. Displacement refers to the movement of a ball from one position to another. The work done here is calculated as W=F×d.

Joule is the SI unit of work. Newton-meter is the other unit of work.

### Energy

The ability to work is defined as energy. There is no way to generate or destroy energy. It can only be changed from one kind to the next. The unit of Energy is the same as the unit of Work, which is Joules. There are several forms of energy since they may be found in a variety of objects.

Kinetic and potential energy are two types of energy. Kinetic Energy is the energy carried in an item during motion and is measured by the quantity of work done, whereas Potential Energy is the energy stored in an item. Joule is the SI unit of Energy.

Some other types of energy are thermal energy, nuclear energy, heat energy, mechanical energy, electrical energy, radiant energy, chemical energy, gravitational energy, and elastic energy.

### Power

Work is defined as a force that causes displacement. The period of time that this force operates to generate the displacement has nothing to do with work. Sometimes the task is completed fast, while other times it is completed slowly. A rock climber, for example, takes an unusually long time to lift her body up a few meters along the edge of a cliff.

A trail hiker, on the other hand, who takes the simpler way up the mountain, may raise her body a few meters in a short period of time. Although the two persons perform the same amount of labor, the hiker completes it in significantly less time than the rock climber. The power is the quantity that has to do with the rate at which a specific amount of work is completed. The hiker surpasses the rock climber in terms of power. The rate at which work is completed is referred to as power.

Power is defined as the rate with which any work is done. Alternatively, work divided by time is known as power.

It is a scalar quantity and is denoted by P.

The formula for power is given as:

P = W ⁄ t

where,

• W is the work done
• t is the time for which work is done
• P is the power gain or loss.

Units of Power:

The SI unit of Power is Watt (W) whose multiples are: KW, MW, GW…

A Watt is defined as, when a body does work of one joule in one second it is called One-Watt Power.

1 watt = 1 J / s

Another unit of power is Horsepower (hp), where 1 hp = 746 W.

Dimensional Formula for Power is [ML2T3].

### Power in Terms of Force and Velocity

We already know that power is the rate at which work is completed in a given amount of time. As a result, the power may be defined as follows at any given time:

P = dW ⁄ dt

Here, dW is the work done in time interval dt.

As we know, W = F d, substitute W in the above equation. (Assume d = x)

P = F dx ⁄ dt

Here, dx is the displacement in time interval dt.

We know, v = dx ⁄ dt; substitute v in the above equation.

P = F v

Here, velocity and force are vector quantities and power is the scalar product of these vectors.

Therefore, the derived formula of power is given as:

P = F v cosθ

Here, θ is the angle between force and velocity vector.

### Sample Problems

Problem 1: A boy pushes a box of 20 kg up to a distance of 5 m for 10 seconds. Calculate the power delivered to the box.

Solution:

Given,

Mass of the box, m = 20 kg

Displacement covered, d = 5 m

Time of displacement, t = 10 s

Weight of the box, F = mg = 20 ×10 N = 200 N

Work done by the boy, W = F d = 200 N ×5 J = 1000 J

Power delivered, P = W ⁄ t = 1000 / (10 J/s) = 100 J/s

Hence, power delivered to the box is 100 J/s.

Problem 2: A pump is required to lift 500 kg of water per minute from an 8 m deep well and eject it with a speed of 25 m/s. Calculate the power of the pump.

Solution:

Given,

Mass of the water, m = 500 kg

Height covered, h = 8 m

Eject velocity of water, v = 25 m/s

Delivery time, t = 1 min = 60 s

Total energy is converted into work, W = E = m g h+(1/2) m v2

= (500×10×8)+(500×25×25)/2

= (40000+156250) J

=196250 J

Power delivered, P = W / t

=196250 J / 60 J

= 3271 W

Hence, power delivered by pump is 3271 W.

Problem 3: An elevator is designed to lift a load of 500 kg through 5 floors of a building averaging 3 m per floor in 5 seconds. Calculate the power of the elevator.

Solution:

Given:

Mass of the load, m = 500 kg

Total height covered, h = 5 × 3 m =15 m

Time taken, t = 5 s

Power delivered by elevator, P = W ⁄ t = mgh ⁄ t

= (500 × 10 × 15) / 5 W

=15000 W

=1.5×104 W

Hence, power of the elevator is 1.5×104 W.

Problem 4: A force of 5 N is required to move a body on frictional floor with constant velocity 5 m/s. Find the power generated by the force.

Solution:

Given:

Velocity of body, v = 5 m/s

Force required to maintain the velocity, F = 5 N

Power generated, P = 5 × 5 W

= 25 W

Hence, power generated by the force is 25 W.

Problem 5: What happens to a body on which work is done?