**WORK, ENERGY AND POWER**

**Work**

Work is said to be done whenever a force moves a body through a certain distance in the direction of force, and is equal to the product of the force and the distance moved.

If W = work done

F = force applied

and S distance covered.

Then W = F x S

The SI unit of work is the joule (J). The joule is defined as the work done when a force of 1 Newton moves a body through a distance of

1 metre in the direction of the applied force.

Other units of work are:

kilojoule KJ = 10^{3}J

megajoule MJ = 10^{6}J.

**Work Done By An Inclined Force**

If a force (F) responsible for moving a body (B) along a horizontal plane, is applied to the body at an angle of θ to the horizontal, the component force, FCosθ along the plane is obtained by resolution of the applied force.

Assuming the body (B) is moved through a distance d. The work done (w) by the applied force can be obtained by the relation.

W = FCosθ x d

**Example**:

When a force of 200N is applied on a body at an angle of 50^{o} to the horizontal direction, the body covers a distance of 5m. Calculate the work done.

**Solution**:

Horizontal component force, F = 200 x cos50^{o }

Distance covered, s = 5m

Work done = F x s

= 200 x cos50^{o} x 5 = 642.79J

**Work done in moving a load up an inclined plane**.

If a crate of 500N is pulled up an inclined plane against friction, unto a platform ‘h’ metre high as shown in the figure above. The work done would be that against

(i) the weight of the crate and (ii) friction.

This implies that,

**Work done in a force field**

The earth’s gravitational field is an example of a force field. In gravitational field, there is always a force pulling a body towards the earth’s centre. We defined the weight of a body as the force of attraction on the body due to the earth’s gravity. This weight acts downwards. To lift a load through a height h, a pulling force must be applied to overcome the weight of the body. Therefore, when an object is lifted vertically upwards, work is done against the force of gravity. The magnitude of the work done is given by:

Work = force x distance = mg x h

W = mgh

Where m = mass of the body (kg),

g = acceleration due to gravity (g = 10 ms^{-2}),

h = height (m), and

W = Work done in joules

**Example**

A man of mass 80kg carries a load of bricks of mass 20kg up a vertical ladder of length 6m. What work has he done? (Take g as 10ms^{-2})

**Solution:**

A total mass of (80 + 20)kg = 100kg has been lifted vertically through 6m.

Workdone = 100 x 10 x 6

= 6000J

= 6kJ.

**Fall under gravity**

If on the other hand a body of mass m falls from a vertical height h the total work done will also be equal to mgh.

**Example**

A stone of mass 10kg falls from a height of 2m. Calculate the work done.

**Solution**

Work done = mgh

= 10 x 10 x 2

= 200J.

**Exercise**

A load of mass m slides down a smooth plane, with inclination, θ from point A to B. The same mass is then dropped from A to the ground C a vertical distance h. In which of the two situations is more work done?

**Solution: **

In each case the force used is mg. Consider the work

done along AC and AB.

Workdone from A to C = mgh

Workdone from A to B = mg x ABsinθ = mgh

The same work is done in each case. Hence work done by gravitational force depends only on vertical distance moved and not on the path taken.

**Force/Distance Graph**

If a graph is plotted of experimental values of force (on the vertical axis) against distance moved (on the horizontal axis) a force/distance graph or work diagram is produced. The area under the graph represents the work done.

For example, a constant force of 20 N used to raise a load a height of 8 m may be represented on a force/distance graph as shown below.

The area under the graph (rectangle = length x breadth) showed shaded represents the work done. Hence

work done = 20N x 8m = 160J

Similarly, a spring extended by 20mm by a force of 500N may be represented by the work diagram shown below,

where

work done = area of shaded portion

work done = ½ base x height

work done = ½ x (20 x 10^{-3})m x 500N = 5J

**Exercise**:

A motor supplies a constant force of 1 kN which is used to move a load a distance of 5m. The force is then changed to a constant 500N and the load is moved a further 15m. Draw the force/distance graph for the operation and from the graph determine the work done by the motor.

**Solution:**

The force/distance graph or work diagram is shown in the figure below.

Between points A and B a constant force of 1000N moves the load 5m; between points C and D a constant force of 500N moves the load from 5m to 20m

Total work done = area under the force/distance graph

= (1000N x 5 m) + (500N x 15 m)

= 5000J + 7500J = 12,500J

= 12.5 KJ

**ENERGY**

Energy is a measure of the change imparted to a system. It is given to an object when a force does work on the object. The amount of energy transferred to the object equals the work done.

Further, when an object does work, it loses an amount of energy equal to the work it does. Energy and work have the same units. joules. Energy, like work, is a scalar quantity. An object that is capable of doing work possesses energy.

**Forms of energy**

Energy occurs in different forms. Examples are mechanical energy, heat energy, nuclear energy, sound energy, light energy etc.

However, because of total dependence of energy for industrial, scientific agricultural and technological development in large scale, world energy resources are grouped into two depending on mode of generation and regeneration.

**(i) Renewable energy:** are derived from sources that do not have need for transformation nor require additive or chemical changes or temperature and pressure effects. The sources are natural raw sources that provide continuous requirement for energy generating plant at no cost, example, the wind, solar source, ocean waves etc.

**(ii) Non-renewable energy:** sources are sources that are exhaustible but provide also the much needed energy requirement. The energy derived are possible after certain chemical changes or that through atomic fission or fusion example, petroleum, biomass, nuclear material etc.

**Principles of Conservation of Energy**

The principle of conservation of energy states that although energy can be changed from one form to another, the total energy of a given system remains unchanged, i.e. energy can neither be created nor destroyed during a transformation but can be transformed from one form to another.

In summary, in an isolated or closed system, the total amount of energy is always constant.

**Mechanical Energy**

Mechanical energy is made up of (i) Potential energy and (ii) Kinetic energy

**Kinetic Energy (K.E.)**

Kinetic energy is the energy possessed by a body due to motion. The kinetic energy of a body of mass, m moving with a velocity, v is given by

K.E. = ½ mv^{2}

change in kinetic energy = ½ mv^{2} – ½ mu^{2} = ½ m(v^{2} – u^{2})

recall the third equation of motion, v^{2} – u^{2} = 2as

therefore, change in Kinetic energy = ½ m (2as)

= mas

= FxS (Newton’s 2^{nd} Law of motion; F=ma)

= work done.

**Potential Energy (PE)**

Potential energy is the energy possessed by a body by virtue of its position.

If a body of mass, m is held at a height ‘h’ metres above the ground.

then potential energy P.E. = weight of body x height = mg x h

i.e P.E. = mgh

This is equal to the work done when the mass m falls a distance h under gravity. The potential energy in this case is called **gravitational potential energy** because it arises from the force of gravity.

If a spiral spring is stretched a distance e, it is made to acquire an energy referred to as **elastic potential energy**. The magnitude of this energy can be shown to be ½ ke^{2} where k is the elastic constant of the spring. A stretched catapult gives the stone potential energy, so that when released, the stone flies through the air with considerable speed.

**Conservation of Mechanical Energy**

This is the principle in which the energy changes by virtue of the position of a body and when in motion are conserved.

The principle of conservation of mechanical energy can be illustrated when a body is dropped from a given height as shown below:

At a height, h_{A}, the potential energy of a body at A maximum. (= mgh)

At a height, h_{B}, the total energy possessed is the sum of the potential energy at height h_{B} and kinetic energy of body at B.

i.e Total Energy at h_{B} = K.E. + P.E.

= ½ mV_{B}^{2} + mgh_{B}

At C, before reaching the ground, the energy possessed by the body is Kinetic Energy which is equal to that at A.

Hence by the principle of conversation of Mechanical Energy

Potential Energy at A = Total Energy at B = Kinetic Energy at C

i.e. mgh = ½ mv_{B}^{2} + mgh_{B} = ½ mv_{C}^{2}

However, ½ mvB2 = mgh_{A} – mgh_{B}

= mg(h_{A} – h_{B})

**Application to problems**

**Example 1**

A body of mass 100kg is released from a height of 200m. With what energy does the body strike the ground?

Solution:

Gravitational potential energy is:

Ep = mgh = 100 x 10 x 200

= 200000J

= 200kJ

**Example 2**

A bullet of mass 40 g is moving with a speed of 216 km/hr. Calculate its kinetic energy

**Solution**

We must convert all units to SI units before using the formula.

40g = 0.04kg

216km/hr = __216 x 1000__

3600s

= 60 ms^{-1}

K.E. = ½ mv^{2}

= ½ x 0.04 x 60

= 72J

**Example 3**

A 5 kg mass is dropped from a height of 30 m above the ground. Determine the velocity of the mass when it is 18 m above the ground. (Take g = 10 ms^{-2})

**Solution
**

**Exercises**

1. A 2 kg mass is dropped from a 12 m height. At what height would the velocity be 5 ms^{-1}?

2. A stone of mass 6 kg is released from a height of 30 m, Calculate the

(i) potential energy and

(ii) the velocity before Impact. (Take g = 10 ms^{-2})

3. A ball of mass 2 kg falls from rest from a height of 200 m. Calculate its kinetic energy after falling a distance of 50 m. (Neglect air resistance and take g = 10 ms^{-2})

4. A stone of mass 10 kg was released from a height of 50 m. Calculate its kinetic energy after falling through a distance of 20 m. [g = 10 ms^{-2}] (NECO 2018)

A. 200 J B. 500 J C. 2000 J D. 3000 J E. 5000 J

**Power**

Power is defined as the time rate of doing work.

Average Power = __workdone by a force
__ time taken to do this work

Average power = __force x distance__ = force x = force x speed__
__ time

The SI unit of is called the watt (W) which is the rate of transfer of energy of one joule per second.

1W = 1Js-1

Large units are the kilowatt (kW) and megawatt (MW)

1Kw = 10

^{3}W

1MW = 10

^{6}W

Sometimes the horse power (h.p.) unit is used.

1h.p. = 746 watts

= 0.75Kw

A unit of energy that is sometimes used for electrical energy is the kilowatt-hour, kWh. This is the energy used by an appliance with a power of one kilowatt in one hour.

1kWh = 1000W x 3600s

= 3.6 x 10^{6}J

= 3.6MJ

**Example:**

Calculate the power of a pump which lifts 500 kg of water through a vertical height of 4 m in 5 s

(g = 10 ms^{-2})

**Solution:**

Power = = = = = 4000W = 4Kw

**Exercise**

A man pushes a wheel barrow filled with load of mass 30 kg at a velocity of 5 ms^{-1} for 3 s. Calculate the power expended. [g = 10 ms^{-2}] (NECO 2018)

A. 5000 J B. 4500 J C. 1500 J D. 500 J E. 180 J

**Scroll Down to Select Page 9 for the next topic – Heat Energy
**

## 0 Comments