ENERGY, WORK, AND POWER

ENERGY, WORK, AND POWER

STANDARD COMPETENCY :

Student can apply the concept of Energy, Work, and Power.

BASIC COMPETENCY :

  1. Identifying energy
  2. Understanding the concept of work
  3. Interpreting power

LESSON MATTER

A. ENERGY

Energy is capability to do effort. There are three types of energy:

1)      Kinetic Energy (Ek)

2)      Potential Energy (Ep)

3)      Mechanic Energy (Em)

1. Kinetic Energy (Ek)

Kinetic energy is energy that caused by movement of something, it is called energy of movement.

Formulated by:

Ek = 1/2 mv2

m = mass (kg)

v = velocity (ms-1)

Ek = Energy of kinetic (joule)

Example and solution:

A car of mass 2 ton moves with velocity 36 km/hour. Find the kinetic energy?

We have:         m = 2 ton = 2000 kg

v = 36 km/hour = 10 ms-1

Will find:         Ek = ?

Solution:          Ek = ½ mv2

Ek = ½ . 2000 . 102

Ek = 1000.100   = 100.000 joule

Ek = 100 Kilo joule

2. Potential Energy (Ep)

Potential energy also called place energy. Potential energy depends on mass, gravity, and position of the object from ground.

EP = mgh

m  = mass (kg)

g   = gravity (ms-2)

h   = altitude (m)

Example and solution:

A tile of mass 100 gr falls from roof of house at altitude of 6 m. The potential energy is ….

Ep = mgh

Ep = 0,1 .9,8. 6 = 5,88 joule

3. Mechanic Energy (Em)

Mechanic energy is total of kinetic energy and potential energy.

Em = Ek + Ep

Em = 1/2 mv2 + mgh

Example and solution:

Someone of mass 40 kg walking upward with angle 30° and velocity 2 ms-1. Find the mechanic energy if he walks 6 m.

We have:         m = 40 kg

a = 30°

v = 2 ms-1

s = 6m

Will find:         Em = ?

Solution:

sin a    = h/s

½         = h/6

h          = 6 . ½

h = 3 m

Em = Ek + Ep

Ek = ½ mv2 + mgh

Ek = ½ . 40 . 22 + 40 . 9,8 . 2

Ek = 80 + 1176

Ek = 1256 joule

4. Law of Energy Constancy

The law of energy constancy is:

Large of mechanic energy is constant.

Em1 = Em2

Ek1 + Ep1 = Ek2 + Ep2

½ mv12+ mgh1 = ½ mv22 + mgh2

Free fall things, v1 = v0 = 0

Example and solution:

Hammer falls at altitude of 10 m from ground. If g = 10 ms-2, find velocity at it position of 2 m from ground?

We have:

h = 8 m

g  =10 ms-2

Will find:        v = ?

Solution:

v = √2gh

v = √2 . 10 . 8

v = √160

v = 4√10   m/s

B. WORK(W)

Work is cross product between force and distance of movement that caused of force.

If force F works at object so the object moves as far as s, so the value of work is formulated as:

W = F . s

W = work (J)

F  = force (N)

s   = distance (m)

  • If force F makes angle α with respect horizontal plane, then :     W = F . cos α . s
  • If horizontal forces work at some objects, then :  W = Σ F . s
  • If object at coarse plane with scour force (Fg), therefore :  W = (F – Fg). s

Examples and solution:

1)      Two men push the car at horizontal direction until 75 m with forces 120 N and 80 N respectively. What is the works?

We have:

F1 = 120 N                  Solution:     W             = ΣF . s

F2 = 80 N                                           W = (F1 + F2) . s

s   = 75 m                                            W = 200 . 75

Will find    W = ?                               W  = 15000 joule

2)      Someone pulls block with force 120 N and makes angle 600 with horizontal plane. How much work is needed to move block 100 m?

We have:

F   = 120 N                  Solution: W     = ΣF . cos a . s

a   = 60°                                            W  = 120 . cos 600 . 100

s    = 100 m                                         W = 120 . 1/2 . 100

Will find:  W = ?                                 W = 6000 joule

  • The Relation of Work and Kinetic Energy:

The value of work is equal to change of the kinetic energy.

W = ΔEk

W = Ek end – Ek start

W = ½ mvt2 – ½ mv02

Example and solution:

The motor cycle of mass 100 kg moves with velocity 18 km/hour, then gives force 150 N. How much its velocity at 50 m?

We have:

m = 100 kg

F   = 150N

s   = 50 m

v0   = 18 km/hour = 5 m/s

Will find          vt   = ?

Solution

F . s                 = ½ . m . (vt2– v02)

150.50             = ½ .100.(vt2 – 52)

7.500               = 50 . (vt2 – 25)

7.500               = 50 . vt2 – 1250

7.500 + 1250   = 50 vt2

vt2 = 175

v =  5√7 ms-1 = 5 √7 m/s

  • The Relation of Work and Potential Energy

W   = F . s

W = m . g (hA – hB)

W = m . g hA – m.g.hB

W  = EpA – EpB

W  = Δ Ep

Example and solution:

The brick of mass 400 gr falls from altitude 10 m. If g = 9.8 ms-1, how much work at altitude of 2 m?

We have:

m = 400 gr = 0,4kg     Solution:    W = m.g.hA – m.g.hB

hA = 10m                                             W = m.g.(hA – hB)

hB = 2 m                                              W = 0,4 . 9,8 (10-2)

g   = 9,8 ms-2                                                  W =  0,4 x 9,8 x 8

Will find          W = ?

  • The Relation of Effort with Spiral spring potential energy

Spiral spring which length of Xo if pulled by style F will increase length equal Δx.

F = k . Δx

Ep = ½ . k . (Δx)2

W  = ½ . k . (Δx)2

F = force ( N)

k = stiffness constant of spiral spring (N/m)

Δx = wall accretion of spiral spring length (m)

Δx = x1 – xo

Ep = spiral spring potential energy ( J)

W = effort spiral spring (J)

So spring potential energy equal to work spiral spring.

Example of problem and solving:

Long spiral spring 25 cm given by force of 10 N so that long become 29 cm. If g= 9,8 ms-2. How much the done effort?

We have :

x0 = 25 cm       Δx = x = x1 – xo

x1= 29                   Δx = 29 – 25 = 4 cm

g = 9,8 ms-2

F = 10 N

Will find. :                   W = ?

Solution :

F = k . Δx

10 = k. 4

k =  = 2,5 N/cm = 250 N/m

W = ½ . k Δx2

W = ½ . 250 . (0,04)2

W = 125 . 0,0016

W = 0,2 joule

C. POWER ( P)

Power is level of effort which done per time unit or dissociation energy of diatomic every second.

P = W/t

P          = power ( watt)

P = F . v

W       = effort ( joule)

t          = time ( sekon)

F         = force ( Newton)

v         = velocity ( ms-1)

Other power unit:

  • watt kilo ( Kw) = 1000 watt
  • horse power ( HP) = horse power ( DK) = PK
  • 1 HP = 1 DK = 1 PK = 746 watt

Example of problem and solving:

Motorcycle which mass 80 kg run with speed 18 km/jam pushed with certain force during 10 second so that the speed become 72 km/jam. Determine level of effort and power at the motorcycle!

We have :

m  = 80 kg

v0 = 18 km/jam =5ms-1

v1 = 72 km/jam = 20ms-1

t    = 10 sekon

Will find :

a) W = ?

b) P = ?

Solution:

a) W    = AEk

W = ½ . m . (v12 – v02)

W = ½ . m . (202 – 52)

W = ½ . 80 . (400 – 25)

W = 40 . 375

W = 15000 joule

b) P      = W/t = 15000/10 =  1.500 watt

D. SIMPLE PLANE

Example of simple plane: helper, grub screw, inclined plane, lifted with pulley, and others.

Example of problem and solving:

Earth payload which mass 3 ton lifted mechanically until elevation 3 m during 15 second. If g= 10 ms-2; How much is the engine power? How much efficiency of engine if the strength 7500 watt.

We have :

m         = 3 ton = 3000 kg

g         = 10 m-2

t          = 15 sekon

h          = 3 m

Pinput = 7500 watt

Will find.

a) P = ?

b) η = ?

Solution :

a) P =  W/t

P      =     m.g.h /t

P =     3000 . 10 . 3    / 15

P = 6000 watt

b) η     =  (6000/7500) x 100%

η = 80%

III. EXCERCISE

A. Choose an accurate Solution.

  1. Dimension from dissociation energy of diatomic is……a. ML2T

    b. MLT2

    c. ML2T3

    d. MLT1

    e. ML-2T3

  2. Kinetic energy owned by earth substance of 4 kg is 18 joule, hence the speed is ……

    a. 2 ms-1

    b. 3 ms-1

    c. 9 ms-1

    d. 12 ms-1

    e. 15 ms-1

  3. An automobile pulled with force level off 20 N above slippery wall, so that beam makes a movement as far as 10 m. Effort done by force for removing the substance is

a. 200 J

b. 20 J

c. 2 J

d. 0,2 J

e. 0,02 J

4. A child is pushing wall with force of 60 N and wall does not move, The work done by child is ….

a. 6000 J                               d. 0,6 J
b. 600 J                                  e. 0
c. 60 J

5. Someone push car with force 400 N, his hand make angel 60° with direction of displacement of car shifting as far as 20 m. The work done is ….

a. 8000 J                               d. 400 J

b. 4000 J                                e. 20 J

c. 600 J

6. Level of effort to move car (mass of car and its contents is 1000 kg) from finite silent state reach speed 72 km/hour (friction is disregarded).

a. 1,25 x 104 J                        d. 6,25 x 105 J

b. 2,50 x 104J                         e. 4,00 x 105 J

c. 2,00 x l05 J

7. A ball with earth of 20 gram shot off with angle of elevation of 30° and speed of 40 m/s. If friction with atmosphere was disregarded hence ball potential energy (in joule at the highest position) :

a. 2                        d. 6

b. 4                         e. 8

c. 5

8.  A lamp is hung housing roof as high as 4 m and lamp earth of 250 gram. If g = 10 ms-2, hence lamp has potential energy equal:

a. 10 Joule                             d. 16 Joule

b. 12 Joule                              e. 20 Joule

c. 14 Joule

9.  Elastic potential energy owned by:

a. oil

b. food

c. drink

d. electricity

e. gum which pulled

10.  The stone which mass 0,5 kg vertical thrown upward to the above. Potential energy of the stone reaches at the highest position is 100 J. If g = 10 ms-2, the maximum elevation of the stone is ….

a. 15 m                                  d. 30 m

b. 20 m                                  e. 10 m

c. 25 m

11.  Something moves at circle orbit with radius R. Then the orbit radius turned into 2 R but the kinetic energy fixed. Transformation of centripetal force object is …..

a. 0,25 times                                     d. 2 times

b. 0,50 times                                     e. 4 times

c. 1 times

12.  Two substances A and B the earth are the same ( m) each makes a move with permanent speed 2 ms-1 and 1 ms-1 the ratio of kinetic energy A and B is …

a. 1 : 4                                              d. 2 : 1

b. 1 : 2                                               e. 1 : 1

c. 4 : 1

13.   A the earth substance 2 kg makes a move with speed of 2 ms-1. Some substances make a move with speeds of 5 ms-1. Total effort which done for several the moment is….

a. 4 Joule                                           d. 21 Joule

b. 9 Joule                                            e. 25 Joule

c. 15 Joule

14.  The earth substance 1 kg moves horizontally as far as 1 m. If g = 10 ms-2 effort done by gravity is ….……

a. 102 J                                              d. 104 J

b. 12 J                                               e. zero

c. 10 J

15.  The mass substance I kg which is initially static with force level of 10 N. Speed of substance after moving as far as 5 m is ….

a. 7,5 ms-1 d. 20 ms-1

b. 10 ms-1 e. 30 ms-1

c. 15 ms-1

16.  In order to move beam as far as 10 m required by force of 200 N with direction of 60° to flat, hence level of effort is……

a. 2000 J                                           d. 500 J

b. 1000J                                             e. 270 J

c. 620 J

17.  Earth pile beater of 200 kg got out of elevation of 5 m above the pile. If g = 9,8 ms-2, hence effort which done equal…

a. 12000 J                                         d. 5000 J

b. 9800 J                                            e. 1000 J

c. 6000 J

18.  Earth steel of 3 ton lifted mechanically crane, so that during 15 second, upwards as high as 4 m. If g = 10ms-2, hence the power is…..

a. 1500 watt                                     d. 8000 watt

b. 3000 watt                                      e.  12000 watt

c. 6000 watt

An automobile pulled with force level off

Tags: , ,

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: