UNITS, PHYSICAL QUANTITIES AND VECTORS QUESTIONS

1.1 Starting with the definition find the number of (a) kilometers in 1.00 mile and (b) feet in 1.00 km.

Answer:

1.1. IDENTIFY: Convert units from mi to km and from km to ft.
SET UP: 1 in. = 2.54 cm, 1 km = 1000 m, 12 in. = 1 ft, 1 mi = 5280 ft.
EXECUTE:
(a) 1.00 mi = (1 00 mi) ((5280 ft)/(1 mi))((12 in)/(1 ft))((2.54 cm)/(1 in))(1m/(〖10〗^2 cm))(1km/(〖10〗^3 m)) = 1.61 km

(b) 1.00 km =  (1.00 km) ((〖10〗^3 m)/1km)((〖10〗^2 cm)/1m)((1 in)/(2.54 cm))((1 ft)/(12 in)) = 3.28 × 103 ft
EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km.

1.2 According to the label on a bottle of salad dressing, the volume of the contents is 0.473 liter (L). Using only the conversions and express this volume in cubic inches.

Answer:

 1.2. IDENTIFY: Convert volume units from L to in.3.
SET UP: 1 L =1000 cm3. 1 in. = 2.54 cm
EXECUTE:
0 473 L×((〖10〗^3 cm)/1L)×((1 in.)/(2.54 cm))^3 = 28.9 in.3

EVALUATE: 1 in.3 is greater than 1 cm3, so the volume in in.3 is a smaller number than the volume in cm3, which is 473 cm3.

1.3 How many nanoseconds does it take light to travel 1.00 ft in vacuum? (This result is a useful quantity to remember.)

Answer:

1.3. IDENTIFY: We know the speed of light in m/s. t = d/v. Convert 1.00 ft to m and  t from s to ns.
SET UP: The speed of light is v = 3.00 ×10^8 m/s. 1 ft = 0.3048 m. 1 s =10^9 ns.
EXECUTE:
t = (0.3048 m)/(3.00×〖10〗^8 m/s) = 1.02×〖10〗^(-9) = 1.02 ns
EVALUATE: In 1.00 s light travels 3.00×〖10〗^8 m = 3.00×〖10〗^5 km =1.86×〖10〗^5 mi.

1.4 The density of gold is What is this value in kilograms per cubic meter?

Answer:

1.4. IDENTIFY: Convert the units from g to kg and from cm3 to m3.
SET UP: 1 kg =1000 g. 1 m =1000 cm.
EXECUTE:
19.3 g/cm^3 ×((1 kg)/1000g)×(100cm/1m)^3 = 1.93×〖10〗^4 kg/m^3 

EVALUATE: The ratio that converts cm to m is cubed, because we need to convert 〖cm〗^3 to  m^3.

1.5 The most powerful engine available for the classic 1963 Chevrolet Corvette Sting Ray developed 360 horsepower and had a displacement of 327 cubic inches. Express this displacement in liters (L) by using only the conversions and 1 in. = 2.54 cm.
Answer:
1.5. IDENTIFY: Convert volume units from in.3 to L.
SET UP: 1 L =1000 cm3. 1 in. = 2.54 cm.
EXECUTE: (327 in.3) × (2.54 cm/in.)3 × (1 L/1000 cm3) = 5.36 L
EVALUATE: The volume is 5360 cm3. 1 cm3 is less than 1 in.3, so the volume in cm3 is a larger number than the volume in in.3.

1.6 A square field measuring 100.0 m by 100.0 m has an area of 1.00 hectare. An acre has an area of If a country lot has an area of 12.0 acres, what is the area in hectares?
Answer:
 IDENTIFY: Convert ft2 to m2 and then to hectares.
SET UP: 1.00 hectare =1.00×104 m2. 1 ft = 0.3048 m.
EXECUTE: The area is (12 0 acres)(〖43 600 ft 〗^2/(1 acres)) ((0.3048 m)/(1.00 ft))^2 ((1.00 hectare)/(1.00×〖10〗^4 m^2 ))= 4.86 hectares.
EVALUATE: Since 1 ft = 0.3048 m, 1 ft2 = (0.3048)2 m2.

1.7 How many years older will you be 1.00 gigasecond from now? (Assume a 365-day year.)
Answer:
IDENTIFY: Convert seconds to years.
SET UP: 1 billion seconds =1×109 s. 1 day = 24 h. 1 h = 3600 s.
EXECUTE: 1.00 billion seconds = (1.00 109 s) (1h/3600s)(1d/(24 h))(1y/(365 d)) = 31.7 y.
EVALUATE: The conversion 1 y = 3.156 ×107 s assumes 1 y = 365.24 d,which is the average for one extra day every four years, in leap years. The problem says instead to assume a 365-day year.
 
1.8 While driving in an exotic foreign land you see a speed limit sign on a highway that reads 180,000 furlongs per fortnight. How many miles per hour is this? (One furlong is and a fortnight is 14 days. A furlong originally referred to the length of a plowed furrow.)
Answer:
IDENTIFY: Apply the given conversion factors.
SET UP: 1 furlong = 0.1250 mi and 1 fortnight = 14 days. 1 day = 24 h.
EXECUTE: (180 000 furlongs/fortnight) ((0.125 mi)/(1 furlong ))((1 fortnight)/(14 d))((1 d)/(24 h)) = 67 mi/h
EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number.
 
1.9 A certain fuel-efficient hybrid car gets gasoline mileage of 55.0 mpg (miles per gallon). (a) If you are driving this car in Europe and want to compare its mileage with that of other European cars, express this mileage in Use the conversion factors in Appendix E. (b) If this car’s gas tank holds 45 L, how many tanks of gas will you use to drive 1500 km?
Answer:
IDENTIFY: Convert miles/gallon to km/L.
SET UP: 1 mi =1.609 km. 1 gallon = 3.788 L.
EXECUTE: (a) 55.0 miles/gallon = (55.0 miles/gallon) ((1.609 km)/1mi)((1 gallon)/(3.788 L)) = 23.4 km/L.
(b) The volume of gas required is (1500 km)/(23 4 km/L) = 64 1 L.
(64.1 L)/(45 L/tank) = 1.4 tanks.
EVALUATE: 1 mi/gal = 0.425 km/L. A km is very roughly half a mile and there are roughly 4 liters in a
gallon, so 1 mi/gal ~2/4 km/L, which is roughly our result.
 
 
1.10 The following conversions occur frequently in physics and are very useful. (a) Use 1 mi = 5280 ft and 1 h = 3600 s to convert 60 mph to units of ft/s (b) The acceleration of a freely falling object is 32ft/s2.  Use 1 ft = 30.48 cm to express this acceleration in units of m/s2. (c) The density of water is 1.0 g/cm3. Convert this density to units of kg/m3.
IDENTIFY: Convert units.
SET UP: Use the unit conversions given in the problem. Also, 100 cm = 1 m and 1000 g =1 kg.
EXECUTE: (a) ((60 mi)/h)((1 h)/3600s)((5280 ft)/1mi)=88 ft/s

(b) ((32 ft)/s^2 )((30.48 cm)/1ft)(1m/100cm)=9.8  m/s^2
(c) ((1.0 g)/〖cm〗^3 ) (100cm/1m)^3 (1kg/1000g)=〖10〗^3  kg/m^3
EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 =103 kg/m3 are exact. The relation
32 ft/s2 = 9.8 m/s2 is accurate to only two significant figures.

Newton's Law Question Grade X (Soal Hukum Newton Kelas X)

1. A 10 kg box is placed on a table. A horizontal force of 32 N is applied to the box. A frictional force of 7N is present between the surface and the box.

a. Draw a force diagram indicating all the horizontal forces acting on the box.

b. Calculate the acceleration of the box.

1. Sebuah kotak dengan massa 10 kg diletakkan di atas meja. Gaya horisontal sebesar 32N di kenakan pada kotak. Gaya gesek sebesar 7N terjadi diantara permukaan meja dan kotak.

a. Gambar diagram gaya yang mengindikasikan semua gaya horisontal yang bekerja pada balok.

b. Hitung percepatan balok

Answer:
Given that: m (box) = 10 kg
                   F = 32 N
                   f = 7 N
Asked: a. Draw a force diagram indicating all the horizontal forces acting on the box.
            b. Calculate the acceleration of the box.
Answer:
a.

b. To calculate the acceleration of the box we will be using the equation F_R = ma.
Therefore:
F_R = ma
F₁ + F_f = (10)(a)
32 − 7 = 10 a
25 = 10 a
a = 2,5m· s^-1towards the left

Jawab:
Diketahui; m (balok) = 10 kg
                   F = 32 N
                   f = 7 N
a. Gambar diagram gaya yang terjadi pada balok,

b. Untuk menghitung percepatan balok, kami akan menggunakan hukum II Newton;
F_R = ma
F₁ + F_f = (10)(a)
32 − 7 = 10 a
25 = 10 a
a = 2,5m· s^-1bergerak ke arah kiri.

2. Passengers with an unknown total mass, m, climb into the lift. A lift, mass 250 kg, is initially at rest on the ground floor of a tall building. The lift accelerates upwards at 1.6 m•s^-2. The cable supporting the lift exerts a constant upward force of 4100 N. Use g = 10 m•s^-2.

a. Draw a labeled force diagram indicating all the forces acting on the lift while it accelerates upwards.

b. What is the maximum mass, m, of the passengers the lift can carry in order to achieve a constant upward acceleration of 1,6 m•s^-2.

2. Penumpang tanpa diketahui massa total, m, naik sebuah lift. Lift dengan massa 250 kg mula mula diam. Lift mengalami percepatan keatas dengan 1.6 m•s^-2 . kabel yang menarik lift menerima gaya keatas sebesar 4100 N. Gunakan g = 10 m•s^-2.
a. Gambar diagram gaya yang mengindikasikan semua gaya yang terjadi pada lift ketika lift mengalami percepatan keatas.
b. Berapa massa maksimum, m, penumpang sehingga lift dapat mencapai percepatan konstan keatas sebesar 1,6 m•s^-2.

Answer:
a. Step 1 : Draw a force diagram.

b. Step 2 : Find the mass, m.
Let us look at the lift with its passengers as a unit. The mass of this unit will be (250 + m) kg and the force of the Earth pulling downwards (F_g) will be (250 + m) x 10.
If we apply Newton’s Second Law to the situation we get:
F_net = ma
F_C − F_g = ma
7700 − (250 + m)(10) = (250 + m)(1,6)
7700 − 2500 − 10 m = 400 + 1,6 m
4800 = 11,6 m
m = 413,79 kg

Jawab:
a. Langkah 1: Gambar diagram gaya

b. Langkah 2: Menghitung massa, m

Kita anggap lift beserta penumpang sebagai satu sistem. massa penumpang dan lift menjadi (250 + m) kg dan gaya gravitasi bumi yang menarik kebawah (F_g) menjadi (250 + m) x 10.
Jika kita menggunakan hukum Newton II ke sistem, dapat ditulis sbb:
F_net = ma
F_C − F_g = ma
7700 − (250 + m)(10) = (250 + m)(1,6)
7700 − 2500 − 10 m = 400 + 1,6 m
4800 = 11,6 m
m = 413,79 kg

3. A 50 kg crate is placed on a slope that makes an angle of 30◦ with the horizontal. The box does not slide down the slope. Calculate the magnitude and direction of the frictional force and the normal force present in this situation.

3.Sebuah kotak 50kg ditempatkan pada sebuah bidang miring dengan sudut 30◦ terhadap garis horisontal. Kotak tidk meluncur kebawah. Hitung nilai dan arah gaya gesek dan gaya normal yang terjadi pada situasi diatas.

Answer:
Step 1 : Draw a force diagram
Draw a force diagram and fill in all the details on the diagram. This makes it easier to understand the problem.

Figure 1.1: Friction and the normal forces on a slope

Step 2 : Calculate the normal force

The normal force acts perpendicular to the surface (and not vertically upwards).
It’s magnitude is equal to the component of the weight perpendicular to the slope.

Therefore:

N = F_g cos 30◦
N = 490 cos 30◦
N = 224N perpendicular to the surface

Step 3 : Calculate the frictional forceThe frictional force acts parallel to the surface and up the slope. It’s magnitude is
equal to the component of the weight parallel to the slope. Therefore:
F_f = F_g sin 30◦
F_f = 490 sin 30◦
F_f = 245N up the slope

Jawab:
Langkah 1: Lukislah diagram gaya
Lukislah diagram gaya dan tulis semua detail pada diagram. Hal ini memudahkan untuk mengerti permaslahan dalam soal.

Figure 1.1: Gaya gesek dan gaya normal pada bidang miring.

Langkah 2: Perhitungan gaya normal

Gaya normal merupakan gaya yang tegak lurus dengan permukaan ( dan tidak keatas secara vertikal).
Besar gaya normal sama dengan komponen berat yang tegak lurus dengan kemiringan.

Sehingga:

N = F_g cos 30◦
N = 490 cos 30◦
N = 224N tegak lurus terhadap permukaan

Langkah 3: Hitung gaya gesek
Gaya gesek terjadi sejajar dengan permukaan dan keatas sejajar dengan permukaan. Nilai gaya gesek sama dengan komponen berat yang sejajar dengan kemiringan.
Sehingga:
F_f = F_g sin 30◦
F_f = 490 sin 30◦
F_f = 245N berarah keatas

4. Find the tension in each cord in Fig.1.2. if the weight of the suspended object is w.

Fig 1.2. Chord Tension

4. Tentukan besar tegangan tali masing-masing pada gambar 1.3. Jika berat barang yang tergantung adalah w.

Gambar 1.3. Tegangan tali

Answer:
IDENTIFY: Apply ΣF = ma to the object and to the knot where the cords are joined.
SET UP: Let + y be upward and +x be to the right.
EXECUTE: (a) T_C = w, T_A sin30° + T_B sin 45° = T_C = w, and T_A cos30° − T_B cos45° = 0. Since
sin 45° = cos 45°, adding the last two equations gives T_A(cos30° + sin30°) = w, and so
T_A = w/1.366 = 0 732w. then
T_B = T_A cos 30°/ sin 30° = 0 897w

Jawab:
(a) T_C = w,
     T_A sin30° + T_B sin 45° = T_C = w, persamaan 1
     T_A cos30° − T_B cos45° = 0. persamaan 2
karena sin 45° = cos 45°, penambahan persmaan 1 dan 2 menghasilkan persamaan 3;
T_A(cos30° + sin30°) = w, sehingga
T_A = w/1.366 = 0 732w. maka
T_B = T_A cos 30°/ sin 30° = 0 897w
 

5. A 15.0-kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A28.0-kg counterweight is suspended from the other end of the rope, as shown in Fig. 1.4. The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the tension in the rope while the load is moving? How does the tension compare to the weight of the load of bricks? To the weight of the counterweight?

Fig 1.4. tension on pulley

5. Sebuah muatan berisi batu bata dengan massa 15 kg bergantung disalah satu ujung tali, dimana gesekan katrol dapat diabaikan. Beban dengan massa 28 kg tergantung di ujung tali yang lain, seperti ditunjukkan pada gamber 1.4. Sistem mula mula diam a). Lukislah digram gaya yang bekerja pada sistem. b). Berapa besar percepatan keatas muatan batu bata? c). Berapa besar tegangan tali ketika beban bergerak? Berapa perbandingan tegangan tali dan muatan batu bata? dan perbandingan tegangan tali dan beban?

Fig 1.4. Tegangan tali pada katrol

Answer:

IDENTIFY: Apply ΣF = ma to the load of bricks and to the counterweight. The tension is the same at
each end of the rope. The rope pulls up with the same force (T ) on the bricks and on the counterweight.
The counterweight accelerates downward and the bricks accelerate upward; these accelerations have the same magnitude.
(a) SET UP: The free-body diagrams for the bricks and counterweight are given in Figure 1.5.

Figure 1.15

(b) EXECUTE: Apply ΣF_y = ma_y to each object. The acceleration magnitude is the same for the two objects. For the bricks take +y to be upward since a for the bricks is upward. For the counterweight take +y to be downward since a is downward.

bricks: ΣF_y = ma_y
T − m₁g = m₁a
counterweight: ΣF_y = ma_y
m₂g − T = m₂a
Add these two equations to eliminate T:
(m₂ − m₁)g = (m₁ + m₂)a

(c) T − m₁g = m₁a gives
T = m₁(a + g) = (15.0 kg)(2.96 m/s² + 9.80 m/s²) =191 N
As a check, calculate T using the other equation.
m₂g − T = m₂a gives
T = m₂(g − a) = 28.0 kg(9.80 m/s² − 2.96 m/s²) =191 N, which checks.

EVALUATE: The tension is 1.30 times the weight of the bricks; this causes the bricks to accelerate upward. The tension is 0.696 times the weight of the counterweight; this causes the counterweight to accelerate downward. If m₁ = m₂, a = 0 and T = m₁g = m₂g. In this special case the objects don’t move. If m₁ = 0, a = g and T = 0; in this special case the counterweight is in free fall. Our general result is correct in these two special cases.

Jawab:

Gunakan ΣF = ma pada sistem muatan batu bata di atas. Tegangan tali besarnya sama pada tiap-tiap ujungnya. Tali menerima gaya T pada ujung muatan batu bata dan beban penarik.
beban penarik mengalami percepatan kebawah dan muatan batu bata mengalami percepatan ke atas; percepatan pada sistem besarnya sama.
(a) diagram gaya yang bekerja pada muatan batu bata dan beban penarik diberikan pada gambar 1.5..

Figure 1.15

(b) Gunakan ΣF_y = ma_y untuk masing-masing obyek. Besar percepatan adalah sama untuk kedua sistem. untuk batu bata ambil +y untuk arah pergerakan ke atas. Untuk beban penarik ambil +y untuk arah percepatan ke bawah.

batu bata: ΣF_y = ma_y
T − m₁g = m₁a ------------------ (1)
beban penarik: ΣF_y = ma_y
m₂g − T = m₂a ------------------ (2)
Tambahkan kedua persamaan untuk mengeliminasi T:
(m₂ − m₁)g = (m₁ + m₂)a

(c) T − m₁g = m₁a ;  maka
T = m₁(a + g) = (15.0 kg)(2.96 m/s² + 9.80 m/s²) =191 N
Untuk pembuktian, hitung nilai T dengan menggunakan persamaan yang lain.
m₂g − T = m₂a ; maka
T = m₂(g − a) = 28.0 kg(9.80 m/s² − 2.96 m/s²) =191 N, terbukti.

Jadi, Besar tegangan tali 1.30 kali berat muatan batu bata; Hal ini menyebabkan batu bata mengalami percepatan ke atas. Besar tegangan 0.696 kali berat penarik; Hal ini menyebabkan batu bata mengalami percepatan ke bawah. Jika m₁ = m₂, a = 0 and T = m₁g = m₂g. Pada kasus khusus ini, benda tidak bergerak. Jika m₁ = 0, a = g and T = 0; pada kasus ini beban penarik akan mengalami gerak jatuh bebas. Hasil perhitungan diatas sesuai untuk kedua kasus ini.

6. Suatu benda bermassa m₁ = 60 kg dilewatkan melalui katrol seperti nampak pada gambar.
a. Dengan gaya berapa orang harus menarik tali agar benda m₁ terangkat dengan percepatan 2 m/s?
b. Seperti soal a, tetapi benda terangkat dengan kecepatan tetap?

Jawab:
Skema gaya-gaya yang bekerja pada gambar diatas:

a.  2T - W = m . a
     2T - m.g = m . a
     2T - 60. 10 = 60 . 2
     2T = 120 + 600
     2T = 720
     T = 360 N
b.  Jika benda terangkat dengan kecepatan tetap, maka percepatan benda a = 0 m/s2
     2T - W = 0
     2T - m.g = 0
     2T = 60 .  10
     2T = 600
     T = 300 N
 
7. Dua benda bermassa m1 dan m2 dihubungkan dengan sebuah katrol. Katrol ini kemudian ditarik oleh gaya F keatas. Jika m1 = 4kg, m2 = 2kg dan F = 60N, apa yang terjadi dengan kedua massa tersebut? Anggap massa tali dan massa katrol diabaikan.

Jawab:

Skema gaya-gaya yang bekerja pada gambar diatas:

Besar gaya yang diterima oleh T1 = F/2 dan T2 = F/2
Jadi, pada benda 1 bekerja gaya Newton sebagai berikut:
ΣF = m₁ . a
T1 - w₁ =m₁ . a
F/2 - m₁ . g = m₁ . a
60/2 - 4 . 10 = 4. a
-10 = 4a
a = -10/4
a = -2.5 m/s²
Tanda negatif menunjukkan bahwa benda bergerak kebawah.

pada benda 2 bekerja gaya Newton sebagai berikut:
ΣF = m₂ . a
T₂ - w₂ =m₂ . a
F/2 - m₂ . g = m₂ . a
60/2 - 2 . 10 = 2. a
10 = 2a
a = 10/2
a = 5 m/s²

8. The masses of blocks A and B in Fig. below are 20.0 kg and 10.0 kg, respectively. The blocks are initially at rest on the floor and are connected by a massless string passing over a massless and frictionless pulley. An upward force F is applied to the pulley. Find the accelerations a_a of block A and a_b of block B when F is (a) 124 N; (b) 294 N; (c) 424 N.

Answer:

Each cord receive F/2
a). F = 124 N
     Ta = Tb = F/2 = 124/2 = 62 N
     Wa > Ta and Wb > Tb, so the blocks are still at rest.
     a_a = a_b = 0 m/s2

b). F = 294 N
     Ta = Tb = F/2 = 294/2 = 147 N
     Wa > Ta, so the block A is still at rest.
     a_a= 0 m/s2
     Wb < Tb, so the block B is moving.
     ΣF = m_b . a_b
     T_b - w_b =m_b . a_b
    F/2 - m_b . g = m_b . a_b
  294/2 - 10 . 10 = 10. a_b
   147 - 100 = 10a_b
      10a_b = 47
           a_b = 4.7 m/s²
c). F = 424 N
     Ta = Tb = F/2 = 424/2 = 212 N
     Wa < Ta, so the block A is moving.
     ΣF = m_a . a_a
     T_a - w_a =m_a . a_b
    F/2 - m_a . g = m_a . a_a
  212 - 20 . 10 = 20. a_a
   212 - 200 = 20a_a
      20a_a = 12
           a_a = 0.6 m/s²
     Wb < Tb, so the block B is moving.
     ΣF = m_b . a_b
     T_b - w_b =m_b . a_b
    F/2 - m_b . g = m_b . a_b
  212 - 10 . 10 = 10. a_b
   212 - 100 = 10a_b
      10a_b =112
           a_b = 11.2 m/s²

9.One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig below.
a. If the buckets are at rest, what is the tension in each cord?

b. If the two buckets are pulled upward with an acceleration of 1.6 m/s2 by upper cord, calculate the tension in each cord.

Answer:

a. upper bucket = Object2
   lower bucket = Object1
Look at object 1  
ΣF = 0 (because object 1 is at rest)
T1 - w = 0
T1 = m . g
T1 = 3.2 kg . 10 m/s2
T1 = 32 N

Look at object 2
ΣF = 0 (because object 2 is at rest)
T2 - w - T1 = 0
T2 = m . g + T1
T2 = 32 N + 32 N
T2 = 64 N

b. Look at object 1  
ΣF = m . a (because the system is going up with a = 1.6 m/s2)
T1 - w =m . a
T1 = m . g + m . a
T1 = 3.2 kg . 10 m/s2 + 3.2kg . 1.6m/s2
T1 = 32 N + 5.12 N
T1 = 37.12 N

Look at object 2
ΣF = m . a (because the system is going up with a = 1.6 m/s2)
T2 - w1 - w2 = m . a
T2 = m1 . g + m2 . g + (m1+m2) . a
T2 = 3.2 kg . 10 m/s2 + 3.2kg . 10 m/s2 + (3.2+3.2) 1.6 m/s2
T2 = 32 N + 32 N + 10.24 N
T2 = 74.24 N

10. Untuk sistem yang ditunjukkan pada gambar disamping. Hitung tegangan tali T1, T2, dan T3 pada saat:
a. Sistem bergerak ke atas dengan percepatan a.
b. Sistem bergerak ke atas dengan kecepatan tetap.

Jawab:

a. Look at object 1  
ΣF = m . a (bergerak ke atas dengan percepatan a)
T1 - w =m . a
T1 = m1 . g + m1 . a
T1 = m1 (g + a)

Look at object 2
ΣF = m . a (bergerak ke atas dengan percepatan a)
T2 - w1 - w2 = m . a
T2 = m1 . g + m2 . g + (m1+m2) . a
T2 = (m1+m2). g + (m1+m2) . a
T2 = (m1+m2) . (a + g)

Look at object 3
ΣF = m . a (bergerak ke atas dengan percepatan a)
T3 - w1 - w2 - w3 = m . a
T3 = m1 . g + m2 . g + m3 . g + (m1+m2+m3) . a
T2 = (m1+m2+m3). g + (m1+m2+m3) . a
T2 = (m1+m2+m3) . (a + g)

b. Look at object 1  
ΣF = 0 (bergerak ke atas dengan kecepatan tetap)
T1 - w = 0
T1 = m1 . g

Look at object 2
ΣF = 0 (bergerak ke atas dengan kecepatan tetap)
T2 - w1 - w2 = 0
T2 = m1 . g + m2 . g
T2 = (m1+m2). g

 Look at object 3
ΣF = 0 (bergerak ke atas dengan kecepatan tetap)
T3 - w1 - w2 - w3 = 0
T3 = m1 . g + m2 . g + m3 . g
T2 = (m1+m2+m3). g