## Energy

This is the ability to do work.#### Forms of energy

a.**Chemical energy:**- this is found in foods, oils charcoal firewood etc.

b.

**Mechanical energy:**- there are four types;

**Potential energy**– a body possesses potential energy due to its relative position or state**Kinetic energy**– energy possessed by a body due to its motion i.e. wind, water**Wave energy**– wave energy may be produced by vibrating objects or particles i.e. light, sound or tidal waves.**Electrical energy**– this is energy formed by conversion of other forms of energy i.e. generators.

#### Transformation and conservation of energy

- Any device that facilitates energy transformations is called transducer.- The law of conservation of energy states that

*energy cannot be created or destroyed; it can only be transformed from one form to another.*### Work

Work is done when a force acts on a body and the body moves in the direction of the force.Work done = force × distance moved by object.

**W = F × d**

Work is measured in

**Nm. 1 Nm = 1 Joule (J)**

**Example 1**

Calculate the work done by a stone mason lifting a stone of mass 15 kg through a height of 2.0 m. (take g=10N/kg)

*Solution*

Work done = force × distance

= (15 × 10) × 2 = 300 Nm or 300 J

**Example 2**

A girl of mass 50 kg walks up a flight of 12 steps. If each step is 30 cm high, calculate the work done by the girl climbing the stairs.

*Solution*

Work done = force × distance

= (50 × 10) × (12 × 30) ÷ 100 = 500 × 3.6 = 1,800 J

**Example 3**

A force of 7.5 N stretches a certain spring by 5 cm. How much work is done in stretching this spring by 8.0 cm?

*Solution*

A force of 7.5 produces an extension of 5.0 cm.

Hence 8.0 cm = (7.5 × 8)/ 5 = 12.0 N

Work done = ½ × force × extension

= ½ × 12.0 × 0.08 = 0.48 J

**Example 4**

A car travelling at a speed of 72 km/h is uniformly retarded by an application of brakes and comes to rest after 8 seconds. If the car with its occupants has a mass of 1,250 kg. Calculate;

a) The breaking force

b) The work done in bringing it to rest

*Solution*

a) F = ma and a = v – u/t

But 72 km/h = 20m/

a = 0 - 20/8 = - 2.5 m/s

Retardation = 2.5 m/s

Braking force F = 1,250 × 2.5

= 3,125 N

b) Work done = kinetic energy lost by the car

= ½ mv^{2} – ½ mu^{2}

= ½ × 1250 × 0

2 – ½ × 1250 × 20^{2}

= - 2.5 × 10^{5}J

**Example 5**

A spring constant k = 100 Nm is stretched to a distance of 20 cm. calculate the work done by the spring

*Solution*

Solution

Work = ½ ks^{2}

= ½ × 100 × 0.2^{2}

= 2 J

### Power

Power is the time rate of doing work or the rate of energy conversion.Power

**(P) = work done / time**

**P = W / t**

The SI unit for power is the watt (W) or joules per second (J/s).

**Example 6**

A person weighing 500 N takes 4 seconds to climb upstairs to a height of 3.0 m. what is the average power in climbing up the height?

*Solution*

Power = work done / time = (force × distance) / time

= (500 × 3) / 4 = 375 W

**Example 7**

A box of mass 500 kg is dragged along a level ground at a speed of 12 m/s. If the force of friction between the box and floor is 1200 N. Calculate the power developed

*Solution*

Power = Fv

= 2,000 × 12

= 24,000 W = 24 kW.

### Machine

A machine is any device that uses a force applied at one point to overcome a force at another point.Force applied is called the effort while the resisting force overcome is called

**load.**

Machines makes work easier or convenient to be done. Three quantities dealing with machines are;-

**Mechanical advantage (M.A.)**- this is defined as the ratio of the load**(L)**to the effort**(E).**It has no units.

M.A = load (L) / effort (E)-
**Velocity ratio**– this is the ratio of the distance moved by the effort to the distance moved by the load.

**V.R**= distance moved by effort/ distance moved by the load. **Efficiency**– is obtained by dividing the work output by the work input and the getting percentage

Efficiency = (work output/work input) × 100

= (M.A / V.R) × 100

= (work done on load / work done on effort) × 100

**Example 8**

A machine; the load moves 2 m when the effort moves 8 m. If an effort of 20 N is used to raise a load of 60 N, what is the efficiency of the machine?

*Solution*

Efficiency = (M.A / V.R) × 100 M.A = load/effort = 60/20 = 3

V.R = DE/ DL = 8/2 = 4

Efficiency = ¾ × 100 = 75%

**Some simple machines**

#### a) Levers

– this is a simple machine whose operation relies on the principle of moments.#### b) Pulleys

– this is a wheel with a grooved rim used for lifting heavy loads to high levels.The can be used as a single fixed pulley, or as a block-and-tackle system.

**M.A = Load/ Effort**

V.R = no. of pulleys/ no. of strings supporting the load

**Example 9**

A block and tackle system has 3 pulleys in the upper fixed block and two in the lower moveable block. What load can be lifted by an effort of 200 N if the efficiency of the machine is 60%?

*Solution*

V.R = total number of pulleys = 5

Efficiency = (M.A/V.R) × 100 = 60%

0.6 = M.A/5 = 3, but M.A = Load/Effort

Therefore, load = 3 × 200 = 600 N

#### c) Wheel and Axle

– consists of a large wheel of big radius attached to an axle of smaller radius.

**V.R = R/r and M.A = R/r**

**Example 10**

A wheel and axle is used to raise a load of 280 N by a force of 40 N applied to the rim of the wheel. If the radii of the wheel and axle are 70 cm and 5 cm respectively. Calculate the M.A, V.R and efficiency.

*Solution*

M.A = 280 / 40 = 7

V.R = R/r = 70/5 = 14

Efficiency = (M.A/V.R) × 100 = 7/14 × 100 = 50 %

#### d) Inclined plane

**V.R = 1/ sin Θ**

M.A = Load/ Effort

M.A = Load/ Effort

**Example 11**

A man uses an inclined plane to lift a 50 kg load through a vertical height of 4.0 m. the inclined plane makes an angle of 300 with the horizontal. If the efficiency of the inclined plane is 72%, calculate;

a) The effort needed to move the load up the inclined plane at a constant velocity.

b) The work done against friction in raising the load through the height of 4.0 m. (take g= 10 N/kg)

*Solution*

a) V.R = 1 / sin C = 1/ sin 300 = 2 M.A = efficiency × V.R = (72/100)× 2 = 1.44

Effort = load (mg) / effort (50 × 10)/ 1.44 = 347.2 N

b) Work done against friction = work input – work output

Work output = m g h = 50 × 10 × 4 = 2,000 J

Work input = effort × distance moved by effort

347.2 × (4 × sin 300) = 2,777.6 J

Therefore work done against friction = 2,777.6 – 2,000 = 777.6 J

#### e) The screw:

- the distance between two successive threads is called the pitch**V.R of screw = circumference of screw head / pitch P**

= 2πr / P

= 2πr / P

**Example 12**

A car weighing 1,600 kg is lifted with a jack-screw of 11 mm pitch. If the handle is 28 cm from the screw, find the force applied.

*Solution*

Neglecting friction M.A = V.R

V.R = 2Πr /P = M.A = L / E

1,600 / E = (2Π × 0.28) / 0.011

E = (1,600 × 0.011 × 7) / 22 × 2 × 0.28 =10 N

#### f) Gears:

- the wheel in which effort is applied is called the driver while the load wheel is**V.R = revolutions of driver wheel / revolutions of driven**

Or

**V.R = no. of teeth in the driven wheel/ no. of teeth in the driving wheel**

#### g) Pulley belts:

-these are used in bicycles and other industrial machines**V.R = radius of the driven pulley / radius of the driving pulley**

#### h) Hydraulic machines

**V.R = R**

^{2}/ r^{2}where R- radius of the load piston and r- radius of the effort piston**Example 13**

The radius of the effort piston of a hydraulic lift is 1.4 cm while that of the load piston is 7.0 cm. This machine is used to raise a load of 120 kg at a constant velocity through a height of 2.5 cm. given that the machine is 80% efficient, calculate;

a) The effort needed

b) The energy wasted using the machine

*Solution*

a) V.R = R^{2}/ r^{2} = (7 × 7) / 1.4 × 1.4 = 25

Efficiency = M.A / V.R = (80 /100) × 25 = 20

But M.A = Load / Effort = (120 × 10) / 20 = 60 N

b) Efficiency = work output / work input = work done on load (m g h) /80

= (120 × 10 × 2.5) / work input

80 / 100 = 3,000 / work input

Work input = (3,000 × 100) /80 = 3,750 J

Energy wasted = work input – work output

= 3,750 – 3,000 = 750 J