## ENERGY WORK POWER AND MACHINES

* ENERGY WORK POWER AND MACHINES:Energy *

*This is the ability to do work. *

Forms of energy.

- Chemical energy: – this is found in foods, oils charcoal firewood etc.
- Mechanical energy: – there are two 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**. Energy can be transformed from one form to another i.e. mechanical – electrical – heat energy. **The law of conservation of energy **states that *“energy cannot be created or destroyed; it can only be transformed from one form to another”.*

ENERGY WORK POWER AND MACHINES

*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)**

*Examples *

*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

*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

*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

*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; *

*The breaking force**The work done in bringing it to rest*

*Solution *

- F = ma and a = v – u/t But 72 km/h = 20m/s a = 0 -20/8 = – 2.5 m/s

Retardation = 2.5 m/s

Braking force F = 1,250 × 2.5

= 3,125 N

- 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

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

*Solution *

Work = ½ ks^{2}

= ½ × 100 × 0.2^{2}

= 2 J

ENERGY WORK POWER AND MACHINES

*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**).

*Examples *

*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

*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 = F v

= 2,000 × 12 = 24,000 W = 24 kW.

*Machines *

** 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;-

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

*M.A = load (L) / effort (E) *

- b)
– this is the ratio of the distance moved by the effort to the distance moved by the load*Velocity ratio*

*V.R = distance moved by effort/ distance moved by the load *

- c)
– is obtained by dividing the work output by the work input and the getting percentage*Efficiency*

Efficiency = (work output/work input) × 100

= *(M.A / V.R) × 100*

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

ENERGY WORK POWER AND MACHINES

*Examples *

*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 =D_{E}/ D_{L} = 8/2 = 4

Efficiency = ¾ × 100 = 75%

*Some simple machines *

*Levers*– this is a simple machine whose operation relies on the principle of moments*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 *

*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 *

*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 **

*Example *

*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 30 ^{0} with the horizontal. If the efficiency of the inclined plane is 72%, calculate; *

*The effort needed to move the load up the inclined plane at a constant velocity.**The work done against friction in raising the load through the height of 4.0 m. (take g= 10 N/kg)*

*Solution *

- R = 1 / sin C = 1/ sin 30
^{0}= 2 M.A = efficiency × V.R = (72/100)× 2 = 1.44

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

- 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 30^{0}) = 2,777.6 J

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

ENERGY WORK POWER AND MACHINES

- 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 *

*Example *

*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 the driven wheel.

*V.R = revolutions of driver wheel / revolutions of driven wheel *

* Or *

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

Example

- Pulley belts: -these are used in bicycles and other industrial machines
*R = radius of the driven pulley / radius of the driving pulley* - Hydraulic machines

*V.R = R ^{2} / r^{2} where R- radius of the load piston and r- radius of the effort piston *

*Example*

*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; *

*The effort needed**The energy wasted using the machine*

*Solution *

- 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

- 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

ENERGY WORK POWER AND MACHINES

ALL PHYSICS NOTES FORM 1-4 WITH TOPICAL QUESTIONS & ANSWERS

PRIMARY NOTES, SCHEMES OF WORK AND EXAMINATIONS