Ka hana a me ka ikehu kinetic - nā pilikia a me nā hoʻonā

Work-Kinetic energy :

1. A 5000-kg car accelerated from rest to 20 m/s. Determine the net hana done on the car.

ʻIke ʻia:

Ka Mass (m) = 5000 kg

Initial speed (v_o) = 0 m/s (car rest)

Final speed (v_t) = 20 m/s

makemake : hana upena

Lōlā:

The work-kinetic energy principle :

W_upena = ΔEK

W_upena = ½ m (v_t² - v_o²)

W_upena = hana upena

ΔEK = the change in kinetik energy

m = mass (kg),

v_t = final speed (m/s),

v_o = initial speed (m/s).

Pūnaewele:

W_upena = ½ m (v_t² - v_o²)

W_upena = ½ (5000)(20² - 0²)

W_upena = (2500)(400 – 0)

W_upena = (2500)(400)

W_upena = 1000,000 Joule

2. A 10-kg object accelerated from 5 m/s to 10 m/s. Determine the net work done on the object!

ʻIke ʻia:

Ka Mass (m) = 10 kg

Initial speed (v_o) = 5 m/s

Final speed (v_t) = 10 m/s

makemake : hana upena

Lōlā:

Pūnaewele:

W_upena = ΔEK

W_upena = ½ m (v_t² - v_o²)

W_upena = ½ (10)(10² - 5²)

W_upena = (5)(100 – 25)

W_upena = (5)(75)

W_upena = 375 Joule

3. A 2000-kg car decelerated from 10 m/s to 5 m/s. What is the work done on the car ?

ʻIke ʻia:

Car’s mass (m) = 2000 kg

Initial speed (v_o) = 10 m/s

Final speed (v_t) = 5 m/s

Makemake: hana upena

Lōlā:

Pūnaewele:

W_upena = ΔEK

W_upena = ½ m (v_t² - v_o²)

W_upena = ½ (2000)(5² - 10²)

W_upena = (1000)(25 – 100)

W_upena = (1000)(-75)

W_upena = -75,000 Joule

The minus sign indicates that the direction of displacement is opposite with the direction of the net force.

4. A 60-N constant force exerted on a 10-kg object for 12 seconds. The initial velocity of an object is 6 m/s and the direction of the object is the same as the direction of the force.

(1) Work done on the object is 30,240 Joule

(2) The final kinetic energy is 30,240 joule

(3) mana is 2,520 Watt

(4) Th increase in the kinetic energy of the object is 180 Joule

The correct statements are…

ʻIke ʻia:

Ikaika (F) = 60 N

Ka manawa (t) = 12 kekona

Ka nuipa o ka mea (m) = 10 kg

Ka wikiwiki mua (v_o) = 6 m/s

Makemake ʻia: The correct statements

Lōlā:

Acceleration of object :

∑F = ma

60 = 10 a

a = 60 / 10 = 6 m/s²

The final velocity :

v_t = v_o + ma

v_t = 6 + (6)(12)

v_t = 6 + 72

v_t = 78m/s

ka mamao traveled in 12 seconds :

s = v_o t + 1/2 ma²

s = (6)(12) + 1/2 (6)(12)²

s = 72 + (3)(144)

s = 72 + 432

s = 504 meters

(1) Work done by force

W = F s = (60)(504) = 30,240 Joule

(2) The final kinetic energy

KE = 1/2 m v_t² = 1/2 (10)(78)² = (5)(6084) = 30,420 Joules

(3) Power

P = W / t = 30,240 / 12 = 2,520 Joule/second

(4) The increase in the kinetic energy

ΔKE = 1/2 m v_t² – 1/2 m v_o² = 1/2 m (v_t² - v_o²) = 1/2 (10)(78² - 6²) = 5 (6084 –36) = 5 (6048)

ΔKE = 30,240 Joule

5. The larger work is done by object number…

Lōlā:

Net work = change of th kinetic energy

W_upena = ½ m (v_t² - v_o²)

The larger work :

W₁ = ½ (8)(4² - 2²) = (4)(16 – 4) = (4)(12) = 48 Joules

W₂ = ½ (8)(5² -3²) = (4)(25 – 9) = (4)(16) = 64 Joules

W₃ = ½ (10)(6² - 5²) = (5)(36 – 25) = (5)(11) = 55 Joules

W₄ = ½ (10)(4² - 0²) = (5)(16 – 0) = (5)(16) = 80 Joules

W₅ = ½ (20)(3² - 3²) = (10)(9 – 9) = (10)(0) = 0 Joules

6. A 4000-kg car travels along straight line at 25 m/s. The car is decelerated so that the car’s final velocity is 15 m/s. What is the work done on the car.

ʻIke ʻia:

Nuipaʻa (m) = 4000 kg

ʻO ka wikiwiki mua (v_o) = 25 m/s

ʻO ka wikiwiki hope loa (v_t) = 15 m/s

makemake : Work done on car

Lōlā:

W_upena = ½ m (v_t² - v_o²) = ½(4000)(15²-25²) = (2000)(225-625) = (2000)(-400) = -800,000 Joule = -800 kJ

7. A 0.1-kg thrown horizontally at 6 m/s from the height of 5 meters. If the ka wikiwiki o ke koʻikoʻi he 10 m/s², then what is the kinetic energy of ball at the height of 2 meters.

ʻIke ʻia:

Nuipaʻa (m) = 0.1 kg

The change in height (h) = 5 m – 2 m = 3 meters

Ka wikiwiki ma muli o ke koʻikoʻi (g) = 10 m/s²

Makemake ʻia: The kinetic energy at the height of 2 meters.

Lōlā:

Projectile motion can be understood by analyzing the horizontal and vertical components of the motion separately. Motion in horizontal direction analyzed as the constant velocity motion and motion in vertical direction analyzed as free fall motion or vertical motion.

ka loiloi mua ka ikaika mechanical = ka ikehu hiki ke hoʻopaʻa ʻia.

PE = m g h = (0.1)(10)(3) = 3 Joule.

ka hope loa ka ikaika mechanical = the kinetic energy.

KE = 3 Joule.

8. A 1000-kg car accelerated from rest and travels at 5 m/s. What is the work done by car?

ʻIke ʻia:

Nuipaʻa (m) = 1000 kg

Ka wikiwiki mua (v_o) = 0

Ka wikiwiki hope loa (v_t) = 5 m/s

Makemake ʻia: Work (W) done by car

Lōlā:

W_upena = ½ m (v_t² - v_o²)

Work done by car :

W_upena = ½ (1000)(5² - 0²) = (500)(25 – 0) = (500)(25) = 12,500 Joules

9. A 500-gram ball thrown vertical upward from the surface of earth with the initial velocity 10 m/s². Acceleration due to gravity is 10 ms^-2. What is th work done by the weight force when ball reaches the maximum height.

Kaulana :

Ka nuipa o ka pōpō (m) = 500 grams = 0.5 kg

Ka wikiwiki mua (v_o) = 10 m/s²

Ka wikiwiki hope loa (v_t) = 0 (velocity at the highest point)

Ka wikiwiki ma muli o ke koʻikoʻi (g) = 10 m/s²

Makemake ʻia: Work (W) don by weight

Lōlā:

The net work done by net force on an object = the change in the kinetic energy.

W_upena = ΔEK = EK_t – EK_o

Wnet = ½ m v_t² – ½ mv_o² = ½ m (v_t² - v_o²)

KE_t = the final kinetic energy, KE_o = the initial kinetic energy, m = mass of object, v_t = the final velocity of object, v_o = initial velocity of object.

Pūnaewele:

W_upena = ½ m (v_t² - v_o²) = ½ (0.5)(0² - 10²)

W_upena = (0.25)(-100) = -25 Joule

Minus sign indicates that the direction of displacement is opposite to the weight of the ball. The direction of ball is upright and the direction of weight is downright.

10. A 1-kg object free fall with the height difference = 2.5 meters. Acceleration due to gravity is 10 m.s^-2. What is the work done on the object?

ʻIke ʻia:

Mass of ball (m) = 1 kg

Ka wikiwiki mua (v_o) = 0 m/s

Kiʻekiʻe (h) = 2.5 mika

Ka wikiwiki ma muli o ke koʻikoʻi (g) = 10 m/s²

Makemake ʻia: Net work during displacement

Lōlā:

Final velocity of ball (v_t)

Calculated using the equation of free fall motion. ʻIke ʻia: Ka wikiwiki ma muli o ke koʻikoʻi (g) = 10 m/s², The change in height of ball (h) = 2.5 meters. Makemake ʻia: Final velocity.

v_t² = 2 g h = 2(10)(2.5) = 2(25)

v_t = √2(25)

v_t = 5√2

Net work = the change in kinetic energy

W_upena = ΔEK = ½ m (v_t² - v_o²) = ½ (1){(5√2)² - 0²}

W_upena = ½ (25)(2) = 25 Joule

11. A 2-kg object travels at 72 km/hour. After travels 400 meters, the final velocity of object is 144 km/hour. Acceleration due to gravity is 10 ms^-2. Find the net work.

ʻIke ʻia:

Ka nuipa o ka mea (m) = 2 kg

Ka wikiwiki mua (v_o) = 72 km/jam = 20 m/s

Ka wikiwiki hope loa (v_t) = 144 km/jam = 40 m/s

Ka mamao (s) = 400 mika

Ka wikiwiki ma muli o ke koʻikoʻi (g) = 10 m/s²

Makemake ʻia: The net work

Lōlā:

The net work = changes of the kinetic energy

W_upena = ΔEK = ½ m (v_t² - v_o²) = ½ (2)(40² - 20²}

W_upena = ½ (2)(1600 – 400) = 1200 Joule

12. A 2-kg object travels at 2 ms^-1. The work done ob the object is 21 Joule. What is the final velocity of object.

ʻIke ʻia:

Nuipaʻa (m) = 2 kg

Ka wikiwiki mua (v_o) = 2 m/s

Hana (W) = 21 Joule

Makemake ʻia: ka wikiwiki hope loa (v_t)

Lōlā:

W_upena= ΔEK

W_upena= 1/2 mv_t² -1/2 m v_o²

W_upena = 1/2 m (v_t² - v_o²)

21 = 1/2 (2) (v_t² - 2²)

21 = (ʻumi_t² - 2²)

21 = v_t² - 4

v_t² 21 + 4 = 25

v_t = √25

v_t = 5m/s

13. A 8 N constant force acts on an object with mass of 16 kg. If the object initially at rest, then determine the speed of the object after force acts on the object for 4 seconds.

ʻIke ʻia:

Constant force (F) = 8 Newton

Ka nuipa o ka mea (m) = 16 kg

Ka wikiwiki mua o ka mea (v)_o) = 0 m/s

Time interval force acts on object (t) = 4 seconds

Makemake ʻia: ʻO ka wikiwiki hope loa (v_t)

Lōlā:

Work = The change in the kinetic energy

W = KE final – KE initial

W = ½ mv_t² – ½ mv_o²

W = ½ mv_t² - 0

W = ½ mv_t² —— Equation 1

Work = Force x Displacement

W = F d

W = 8 d

Use the equation of nonuniform linear motion below to calculate displacement (d) :

d = v_o t + ½ ma²

d = displacement, v_o = initial velocity, t = time interval, a = acceleration

d = 0 + ½ a t² = ½ ma² —-> a = (v_t - v_o) / t = v_t /t

d = ½ (v_t /t)t²

^d = ½ (v_tne) t

Change displacement (d) on equation of Work with displacement (d) in this equation :

W = 8 d

W = 8(1/2)(v_t)(t)

W = (4)(v_t)(t) —— equation 2

Equation 1 = Equation 2

W = W

½ mv_t² = (4)(v_t)(t)

½ mv_t = (4)(t)

½ (16)(v_t) = 4(4)

8 v_t = 16

v_t = 16/8

v_t = 2 meters/second

14. To increase the speed of an object become 2 times of the initial speed, determine work required in the process…

ʻIke ʻia:

Ka nuipa o ka mea (m) = 1 kg

Ka wikiwiki mua (v_o) = 1 m/s

Ka wikiwiki hope loa (v_t) = 2 x initial speed = 2 x 1 = 2 m/s

Makemake ʻia: hana

Lōlā:

The initial kinetic energy :

KE initial = ½ m v_o² = ½ (1)(1)² = ½ (1)(1) = ½ (1) = 0.5

The final kinetic energy when the speed of object becomes 2 time of its initial speed :

KE final = ½ m v_t² = ½ (1)(2)² = ½ (4) = 2

Theorem of work-kinetic energy :

Work = The change in kinetic energy

Work = The final kinetic energy– the initial kinetic energy

Work = 2 – 0.5

Work = 1.5

The initial kinetic energy = 0.5

Work = 3 x 0.5 = 1.5

Required work 3 times of its initial kinetic energy.

15. A car with mass of 1500 kg moves with speed of 36 km/hour on a linear and smooth horizontal road. The car accelerated to 72 km/hour. Determine the work required to acceleration the car.

ʻIke ʻia:

Mass of car (m) = 1500 kg

Initial speed of car (v_o) = 36 km/hour = 36,000 meters / 3600 second = 10 meters/second

Final speed of car (v_t) = 72 km/hour = 72,000 meters / 3600 second = 20 meters/second

Makemake ʻia: Work required to accelerates the car

Lōlā:

Theorem of work-kinetic energy :

W = EK final – EK initial

W = ½ mv_t² – ½ mv_o² = ½ m (v_t² –v_o²)

W = ½ (1500)(20² - 10²)

W = ½ (1500)(400 – 100)

W = ½ (1500)(300)

W = (1500)(150)

W =225,000 Joule

16. An object with mass of 2 kg initially moves at speed of 72 km.hour^-1. After move in horizontal straight road as far as 400 m, the speed of the object is 144 km.hour^-1. Determine the total work on the object.

ʻIke ʻia:

Ka nuipa o ka mea (m) = 2 kg

Ka wikiwiki mua (v_o) = 72 km/hour = 72,000 meters / 3600 second = 20 m/s

Ka wikiwiki hope loa (v_t) = 144 km/hour = 144,000 meters / 3600 second = 40 m/s

Displacement of object = 400 meters

Makemake ʻia: Net work on the object

Lōlā:

Theorem of work-kinetic energy states that the net work acts on an object same as the change of the kinetic energy of the object.

W net = KE final – KE initial

W net = ½ m v_t² – ½ mv_o²

W net = ½ m (v_t² - v_o²)

W net = ½ (2)(40² - 20²)

W net = 1600 – 400

ʻupena W = 1200 Joule

Hō'ike:

W = Work, KE = kinetic energy

ʻO ka ikehu kinetic

17. A 10-gram bullet moving at a constant 100 m/s. What is the kinetic energy of the bullet.

ʻIke ʻia:

Mass of bullet (m) = 10 gram = 10/1000 kilogram = 1/100 kilogram = 0.01 kilogram

Bullet’s speed (v) = 100 meters/second

Makemake: ʻO ka ikehu kinetic

Lōlā:

KE = 1/2 m v²

KE = 1/2 (0,01 kg)(100 m/s)²

KE = 1/2 (0,01 kg)(10.000 m²/s²)

KE = (0,01 kg)(5000 m²/s²)

KE = 50 kg m²/s²

KE = 50 Joule

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1. Nā pilikia hana i hana ʻia e ka ikaika a me nā hoʻonā
2. Nā pilikia a me nā hoʻonā o ka ikehu hana-kinetic
3. Nā pilikia a me nā hoʻonā i nā kumumanaʻo ikehu hana-mekanika
4. Nā pilikia ikehu hiki ke umekaumaha a me nā hoʻonā
5. Potential energy of elastic spring problems and solutions
6. Nā pilikia mana a me nā hoʻonā
7. Ka hoʻohana ʻana i ka mālama ʻana i ka ikehu mechanical no ka neʻe hāʻule manuahi
8. Ka hoʻohana ʻana i ka mālama ʻana i ka ikehu mechanical no ka neʻe ʻana i luna a i lalo i ka neʻe hāʻule manuahi
9. Ka hoʻohana ʻana i ka mālama ʻana i ka ikehu mechanical no ka neʻe ʻana ma kahi ʻili piʻo
10. Ka hoʻohana ʻana i ka mālama ʻana i ka ikehu mechanical no ka neʻe ʻana ma kahi mokulele inclined
11. Ka hoʻohana ʻana i ka mālama ʻana i ka ikehu mechanical no ka neʻe ʻana o ka projectile