Pembahasan Pekerjaan Rumah : Usaha dan Energi

1. Sebuah benda meluncur di atas papan kasar sejauh 5 m, mendapat perlawanan gesekan dengan papan sebesar 180 newton. Berapa besarnya usaha dilakukan oleh benda tersebut.

Diket :   s = 5 m

F = 180 N

Ditanya : W = …?

Jawab : W = F . s

W = 180 . 5

W = 900 joule

1. Gaya besarnya 60 newton bekerja pada sebuah gaya. Arah gaya membentuk sudut 30˚ dengan bidang horizontal. Jika benda berpindah sejauh 50 m. Berapa besarnya usaha ?

Diket :   F = 60 N

α = 30˚

s = 50 m

Ditanya : W = …?

Jawab : W = F .cos α. s

W = 60 .cos 30˚  50

W = 60 . . 50

W = 1500  J

Soal Esay Usaha dan Energi

1. Hitunglah besar usaha yang diperlukan untuk mempercepat mobil yang bermassa 1500 kg dari 25 m/s menjadi 30 m/s! Jawab : ……………………………………………………………………………………………………………………………………. 2. Dua buah benda memiliki massa masing – masing m1 = 20 kg, dan m2 = 5 kg memiliki energy kinetik yang sama besar. Jika benda m1bergerak dengan kelajuan 8 m/s, tentukan kelajuan benda m2! Jawab : …………………………………………………………………………………………………………………………………….

3. Perhatikan gambar di atas ! Sebuah benda bermassa 0,5 kg dijatuhkan dari Ketinggian 10 m. Hitunglah besar energy kinetik Benda pada saat berada 4 m diatas tanah! Jawab : …………………………………………………………………………………………………………………………………….

4. sebuah bola dilemparkan dari sebuah gedung dengan ketinggian 20 m. kecepatan awal bola ketika dilempar sama dengan 12 m/s dengan arah 30’ di atas garis horizontal. Berapakah kelajuan bola sesaat sebelum menumbuk tanah ? ( g = 10 m/s2) Jawab : ……………………………………………………………………………………………………………………………………. 5. Hitunglah usaha yang diperlukan uintuk menghentikan sebuah electron ( m = 9,11 x 10 -31 kg) yang bergerak dengan kecepatan 1,50 x 106 m/s! Jawab : ……………………………………………………………………………………………………………………………………. Soal Remidial

1. Apakah yang anda ketahui tentang : a. Usaha b. Energi c. Besaran skalar Jawab : ……………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………. ……………………………………………………………………………………………………………………………………. 2. Sebuah benda mempunyai massa 15 kg bergerak lurus dengan kecepatan 5 m/s2. Jika benda mula- mula diam, hitunglah besar usaha yang di ubah menjadi energy kinetic setelah 5 detik ! Jawab : ……………………………………………………………………………………………………………………………………. 3. Sebuah pompa dengan debit 20 m3 tiap menit dari sumber kedalaman 12 m. hitunglah daya yang diperlukan pompa itu! Jawab : ……………………………………………………………………………………………………………………………………. 4. Seorang anak bermassa 45 kg berjalan menaiki tangga yang tingginya 20 m dalam selang waktu 5 menit. Bila g = 10 m/s2 , hitunglah daya yang dikeluarkan oleh orang tersebut! Jawab : ……………………………………………………………………………………………………………………………………. 5. Gaya sebesar 50 N bekerja pada benda sehingga berpindah sejauh 100 m. hitunglah usaha yang dikerjakan oleh gaya bila sudut antara gaya dan perpindahan 45’ ? Jawab : …………………………………………………………………………………………………………………………………….

Soal Pengayaan

1. Perhatikan gambar di atas ! Sebuah benda bermassa 2 kg meluncur sejauh 2 m. jika gesekan antara benda diabaikan, hitunglah usaha yang dilakukan gaya tersebut ! ( 10 m/s2 ) Jawab : ……………………………………………………………………………………………………………………………………. 2. Sebuah benda mempunyai massa 2 kg mengalami jatuh benda dari ketinggian 100m di atas permukaan tanah. Apabila g = 10m/s2, tentukan besar energy mekanik benda pada saat berada pada : a. Titik maximum b. 30 m di atas tanah c. Titik minimum Jawab : a……………………………………………………………………………………………………………………………………. b……………………………………………………………………………………………………………………………………. c……………………………………………………………………………………………………………………………………. 3. Hitunglah usaha yang dilakukan pada sebuah benda bermassa 30 kg yang mula – mula diam kemudian padanya bekerja sebuah gaya konsatan 90N selama 8 sekon ! ( arah gaya searah denagn arak gerak ) Jawab : …………………………………………………………………………………………………………………………………….

4. Perhatikan gambar dia atas ! Sepeda motor bermassa 800 kg bergerak menanjak pada daerah perbukitan. Jarak lintasan yang ditempuh sejauh 20 m. apabila diketahui kelajuan awal di P adalah 72 km/jam dan di Q 36 km/jam, maka hitungla besar gaya gesekan yang dikerjakan permukaan jalan pada ban sepeda motor! ( g = 10 m/s2 ) Jawab : …………………………………………………………………………………………………………………………………….

5. Sebuah bandul bola mainan berayun dari keadaan diam hingga ketinggian berkurang 45 cm. jika percepatan gravitasi 10 m/s2. Hitunglah kelajuan bandul! Jawab : …………………………………………………………………………………………………………………………………….

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:

E_k = 1/2 mv²

m = mass (kg)

v = velocity (ms^-1)

E_k = 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:         E_k = ?

Solution:          E_k = ½ mv²

E_k = ½ . 2000 . 10²

E_k = 1000.100   = 100.000 joule

E_k = 100 Kilo joule

2. Potential Energy (E_p)

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

E_P = 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 ….

E_p = mgh

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

3. Mechanic Energy (Em)

Mechanic energy is total of kinetic energy and potential energy.

E_m = E_k + E_p

E_m = 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:         E_m = ?

Solution:

sin a    = h/s

½         = h/6

h          = 6 . ½

h = 3 m

E_m = E_k + E_p

E_k = ½ mv² + mgh

E_k = ½ . 40 . 2² + 40 . 9,8 . 2

E_k = 80 + 1176

E_k = 1256 joule

4. Law of Energy Constancy

The law of energy constancy is:

Large of mechanic energy is constant.

E_m1 = E_m2

E_k1 + E_p1 = E_k2 + E_p2

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

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 (F_g), 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 = ΔE_k

W = E_{k end} – E_{k start}

W = ½ mv_t² – 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          v_t   = ?

Solution

F . s                 = ½ . m . (v_t²– v₀²)

150.50             = ½ .100.(v_t² – 5²)

7.500               = 50 . (v_t² – 25)

7.500               = 50 . v_t² – 1250

7.500 + 1250   = 50 v_t²

v_t² = 175

v =  5√7 ms^-1 = 5 √7 m/s

• The Relation of Work and Potential Energy

W   = F . s

W = m . g (h_A – h_B)

W = m . g h_A – m.g.h_B

W  = E_pA – E_pB

W  = Δ E_p

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.h_A – m.g.h_B

h_A = 10m                                             W = m.g.(h_A – h_B)

h_B = 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 – x_o

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    = AE_k

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

P_input = 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. ML²T² 

   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