Work, energy and power

(A) concept of 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. 

For example, work is done when you drag a lump of stone a certain distance or when a car moves a certain distance etc.

Thus,

       Work= force×distance in the direction of force.    W= f×s

                 W= f×s cos O

  The SI unit of work is the joule (J).

One joule is the work done when a force of 1 Newton moves a distance of 1 metre.

Larger units are the kilojoule (kJ) and the megajoule (MJ).

      1kJ = 10³J,  1MJ = 10⁶J.

(B) concept of energy

Energy can be defined as the capacity to do work. Thus the unit of energy is the same as that of work, i.e. the SI unit of energy is the joule, J. Energy can appear in any form.

(C)work done in a force field 

In the gravitational field, there is always a force pulling a body towards the Earth’s centre. The weight of a body is the force of attraction on a body as a result of the Earth’s gravity acting on its. The magnitude of the work done is given by:

   Work = force × distance

           = mg × h

    W = mgh

Where m = mass of the body (kg),

            g = acceleration due to gravity

                   (g = 10ms^-2)

           h = height (m), and

           W = work done in joules (J).

(D) mechanical energy

There are two kinds of mechanical energy:

Kinetic and potential energy.

Kinetic energy is the energy a body possesses because it is in motion. The symbol for kinetic energy is E_k. Kinetic energy is the energy due to motion.

Examples of kinetic energy

1.  A rolling ball
2. An object falling under gravity
3. Wind or air in motion
4. An athlete running a race
5. A bullet movement
6. A plane flying, etc. 

The work done in bringing the body to rest from initial speed v to final speed zero should be equal to the kinetic energy of the body when its speed is v.

     Work done = force × distance

                       = Fs

But s = average speed × time = ( v + 0)t/2 = 

½ vt and F = ma

                 = m v/t

Work done = ½ vt × mv/t

                  = ½ mv²

            E_k = ½ m²

Where m is the mass in kg, v is the speed in ms^-1.i.e. work done = change in kinetic energy of the body.

Potential energy is the energy a body possesses because of its position. The symbol for potential energy is E_p.

Potential energy = work done

E_p = force × distance

E_p = mg × h

Gravitational potential energy

Source: www.commons.wikimedia.org 

Note: 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 ke² 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.

(E) conservation law of mechanical energy

Energy can be transformed from one form  to another, the total energy of the system remains the same. Potential energy can be transformed to kinetic energy and vice versa but in all cases the sum remains constant. Let K.E. = E_k and P.E. = E_p

          E_k + E_p = constant

Or    E_k  + E_p.     at any point

 =    E_k + E_p    at another point

A typical example of alternating potential and kinetic energy is that of simple pendulum.

Energy in simple pendulum

Source: www.commons.wikimedia.org 

Note: E_p + E_k = constant

          E_p at furthest point, E_k at rest position.

V = √(2gh) 

(F)power

Power is defined as the time rate of doing work or the time rate of transfer of energy.

Average power(P)

             = Work done or energy expended

                            Time taken

             =.   Energy or work per second

             = W/t

            =  F×v

Where v is Velocity.

The SI unit of power 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 = 10³W

1MW = 10⁶W

Sometimes the horse power (h.p) unit is used.    1h.p = 746watts

                     = 0.75kW. 

Note: 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 × 3600s

          = 3.6 × 10⁶J

          = 3.6MJ 

(G) mechanical advantage (M.A.) and Velocity ratio (V.R.)

Mechanical advantage is also known as force ratio (F.R). Mechanical advantage of a machine is defined as the ratio of the load to the effort.

i.e M.A = L/E

General principles of a machine

Source: www.en.wikipedia.org 

The V.R of a machine is defined as the ratio of the distance moved by the effort to the distance moved by the load.

       V.R = x/y

Note: efficiency of a machine is formulated as follows;£= work output  × 100%

                      work input

Since work = force × distance 

Work output.= Load ×distance moved by L 

Work input.    Effort×distance moved by E

Where L= load, E= effort

             = Ly/Ex

             = L/E / x/y

             =  M.A

                 V.R

  £ = M.A  ×  100%

        V.R

(H) classification, uses and applications of machines

(I)The lever

The lever is a simple form of machine. The term lever is applied to a rigid body pivoted about a point called a fulcrum (F). An effort (E) is applied at one point on the lever and this overcomes a load (L) at some other point.

The lever

Source: www.commons.wikimedia.org 

First order levers 

The first order lever type is one in which F is between L and E. Simple examples of first order levers are the crowbar, the claw hammer and pliers.

The claw hammer

Source: www.flickr.com

Crowbar

Source: www.pixabay.com

Pliers

Source: www.commons.wikimedia.org 

Note: if y and x represent the distances of load L and effort E from the fulcrum F in each of the above levers, then taking moments about F gives y×L = x × E 

       L/E = x/y > M.A = V.R

Second order levers

This is the class of levers in which the load L is between the fulcrum F and effort E. Examples include nut crackers and wheel barrows.

Nut cracker

Source: www.commons.wikimedia.org

Wheelbarrow

Source: www.pikist.com 

Third order levers

In levers of this class, the effort E is between L and F. Examples include the forearm of a human body and a pair of laboratory tongs.

Forearm 

Source: www.pxfuel.com

Tongs

Source: www.commons.wikimedia.org 

(II) pulleys

A simple pulley is a fixed wheel with a rope passing round a groove in its rim. A load, L, is attached at one end of the rope while the effort, E, is applied at the other end. If we neglect friction at the wheels and the weight of the rope, then tension, T, in the rope will be the same throughout. 

Therefore, when there is no friction, L=T=E.

The mechanical advantage (M.A)=1=V.R. if there is no friction present.

Systems of pulleys

Source: www.en.wikipedia.org 

Note: pulleys can be combined to form machines used for lifting loads. They are used by builders for hauling heavy loads to high floors or in loading and unloading ships. The aim of combining pulleys is to achieve a larger Velocity ratio and thus also higher mechanical advantage.

       V.R = d/d/4= 4

(III)the inclined plane

Heavy loads such as drums of oil and engine blocks can be raised with the aid of a sloping plank with little applied effort. The arrangement is called an inclined plane and it is said to be a machine.

Inclined plane

Source: www.commons.wikimedia.org 

Then the Velocity ratio for inclined plane

= Distance moved by E

   Distance moved by L

i.e V.R = AB

               BC

    = x/h

V.R = 1/sin O

In general, V.R and hence the M.A increases when O decreases.

(IV)the hydraulic press

A small effort is required to lift a large load. If there is no friction (the liquid inside the press is usually an oil).

                 M.A = L/E

                        = A₂ / A₁ 

                        = R²/ r² 

Where R and r are the radii of large and small pistons respectively.

Hydraulic press

Source: www.commons.wikimedia.org 

Note: also if x and y are the distances moved by E and L respectively, we have A₁x = A₂y

x/y = A₂ /A₁ = V.R = R²/ r²

(V)the screw 

The screw can be thought of as an inclined plane wrapped round a cylinder to form a thread. The simplest example of a screw is a nut and bolt. As the nut is turned, it moves along the thread of the bolt as though travelling up an inclined plane.

The screw as an inclined plane

Source: www.flickr.com 

Thread of a screw

Source: www.commons.wikimedia.org

Screw jack

Source: www.pikist.com 

      V.R = 2πa

                 P

(VI)the wheel and axle

The wheel and axle

Source: www.commons.wikimedia.org 

Let a and b be the radii of the wheel and axle respectively. Then Velocity ratio

       V.R = distance moved by effort

                distance moved by load 

            =.  2 πa

                 2πb

            =.  a/b

Thus velocity ratio of the wheel and axle is given by:  V.R = a/b

(Vii) gears 

Gears work on the principle of wheel and axles. Thus the Velocity ratio of gear system is:

V.R. =

 Number of teeth ratio on the driven gear

   Number of teeth on the driving gear

Note: other machines that work on wheel and axle principle are: the brace used by Carpenters to bore holes on wood, screw driver, box spanner, etc.

Friction

Friction is defined as a force which acts at the surface of separation of two bodies in contact and tends to oppose the motion of one over the other.

If a body A is placed on another B, the normal reaction of B on A, R is equal and opposite the weight mg thus  R= mg

Experiment shows that if A is increased by adding another weight A₂ on it, the limiting frictional force F or P increases, i.e. as R increases F increases so that F is directly proportional to R i.e. F = R.  F= uR

Features of friction

i. Friction opposed motion

II. It depends on the nature of the surface in contact

III. It depends on normal reaction f= uR

IV. It is independent of area in contact

V. It is independent of relative velocity between the surfaces.

Reducing friction machines

Friction is reduced in machines for two reasons:

1. It causes wear
2. It uses up energy and reduces the efficiency of the machine

Past questions

1.a wooden blocks of mass 1.6kg rests on a rough horizontal surface. If the limiting frictional force between the block and the surface is 8N. Calculate the coefficient of friction. [g= 10ms^-2] (WASSCE 2000)

A. 0.6

B. 0.5

C. 0.3

D 0.2

Answer: B

Solution: m= 1.6kg; F= 8N; g= 10ms²; u=F/R; R= mg; R = 1.6 × 10 = 16N;

Thus u= 8/16 = 0.5

2.a body of mass 1000kg is released from a height of 10m above the ground. Determine its kinetic energy just before it strikes the ground.[g=10ms^-2] (wassce 2015)

A. 10J

B. 10³J

C. 10⁴J

D. 10⁵J

Answer: D

Solution: K.E = mgh = P.E ; 1000×10×10=100000J

= 10⁵J

3.the energy stored in a spring of stiffness constant K 2000Nm^-1 when extended by 4cm is (WASSCE 2017)

A. 0.16J

B. 1.60J

C. 16.00J

D. 160.00J

Answer: B

Solution: the energy stored in a spring of stiffness constant k=200Nm^-1, e= 4cm= 4×1/100 × m = 0.04m

E= ½ ke²; = ½ × 2000 × (0.04)² = 1.6J 

4.a stone of mass 2.0kg is thrown vertically upwards with a Velocity of 2.0ms^-1, calculate the initial kinetic energy of the stone. (WASSCE 2006)

A. 200J

B. 400J

C. 800J

D. 1600J

Answer: B

Solution: k.e = ½ mv² = ½ ×2×20×20 = 20×20=400J

5.

The figure above represents a block and tackle pulley system on which an effort of W newtons supports a load of 120.0N. If the efficiency of the machine is 40°, then the value of W is. (Jamb 1986)

A. 28.0N

B. 48.0

C. 50.0N

D. 288.0

Answer: C 

Solution: V.R = 6, load = 120N

Efficiency = M.A × 1/V.R × 100%

40 = 120/effort × ⅙ × 100/1;

Effort = 50N

6. A stone of mass m kg is held h meters above the floor for 50s. The work done in joules over this period is (jamb 1990)

A. mh

B. mgh

C. mgh/50

D. 0

Answer: D

Solution: Work = mgh the conclusion is no motion 

7.a hydraulic press has a large circular piston of radius 0.8m and a circular plunger of radius 0.2m. A force of 500N is exerted by the plunger. Find the force exerted on the piston. (Jamb 1995)

A.8000N

B. 4000N

C. 2000N

D. 31N

Answer: C

Solution: pressure on piston = pressure on the plunger.

500/π×0.8×0.8 = f/π×0.2×0.2; 

F = 31N

8.

The diagram above shows a solid figure with base PQ and centre of gravity G on an inclined plane. Which of the following statements is correct?(jamb 1999)

A. The solid will fall over if the vertical line through G lies outsides the base

B. The solid will fall over if the vertical line through G lies inside the base 

C. The solid will not fall over if the vertical line through G lies outside the base.

D. The solid can never fall over.

Answer: A