Darbs un kinētiskā enerģija – problēmas un risinājumi

Work-Kinetic energy :

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

Zināms:

Masa (m) = 5000 kg

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

Final speed (v_t) = 20 m/s

Wanted : tīkla darbs

šķīdums:

The work-kinetic energy principle :

W_neto = ΔEK

W_neto = ½ m (v_t² - v_o²)

W_neto = tīkla darbs

ΔEK = the change in kinetik energy

m = mass (kg),

v_t = final speed (m/s),

v_o = initial speed (jaunkundze).

Net work :

W_neto = ½ m (v_t² - v_o²)

W_neto = ½ (5000)(20² - 0²)

W_neto = (2500)(400 – 0)

W_neto = (2500)(400)

W_neto = 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!

Zināms:

Masa (m) = 10 kg

Initial speed (v_o) = 5 m/s

Final speed (v_t) = 10 m/s

Wanted : tīkla darbs

šķīdums:

Net work :

W_neto = ΔEK

W_neto = ½ m (v_t² - v_o²)

W_neto = ½ (10)(10² - 5²)

W_neto = (5)(100 – 25)

W_neto = (5)(75)

W_neto = 375 Joule

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

Zināms:

Car’s mass (m) = 2000 kg

Initial speed (v_o) = 10 m/s

Final speed (v_t) = 5 m/s

Vēlējās: tīkla darbs

šķīdums:

Net work :

W_neto = ΔEK

W_neto = ½ m (v_t² - v_o²)

W_neto = ½ (2000)(5² - 10²)

W_neto = (1000)(25 – 100)

W_neto = (1000)(-75)

W_neto = -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) jauda is 2,520 Watt

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

The correct statements are…

Zināms:

Spēks (F) = 60 N

Laika intervāls (t) = 12 sekundes

Mass of object (m) = 10 kg

Sākotnējais ātrums (v_o) = 6 m/s

Meklē: The correct statements

šķīdums:

Acceleration of object :

∑F = mA

60 = 10 a

a = 60 / 10 = 6 m/s²

The final velocity :

v_t = v_o + plkst.

v_t = 6 + (6)(12)

v_t = 6 + 72

v_t = 78m/s

The attālums traveled in 12 seconds :

s = v_o t + 1/2 a t²

s = (6)(12) + 1/2 (6)(12)²

s = 72 + (3)(144)

s = 72 + 432

s = 504 metri

(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)(336 000) = 336 000 džoulu

(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…

šķīdums:

Net work = change of th kinetic energy

W_neto = ½ m (v_t² - v_o²)

The larger work :

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

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

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

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

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

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.

Zināms:

Masa (m) = 4000 kg

Sākotnējais ātrums (v_o) = 25 m/s

Galīgais ātrums (v_t) = 15 m/s

Wanted : Work done on car

šķīdums:

W_neto = 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 acceleration of gravity ir 10 m/s², then what is the kinetic energy of ball at the height of 2 meters.

Zināms:

Masa (m) = 0.1 kg

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

Brīvās krišanas paātrinājums (g) = 10 m/s²

Meklē: The kinetic energy at the height of 2 meters.

šķīdums:

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.

Sākotnējais mehāniskā enerģija = gravitācijas potenciālā enerģija.

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

The galīgs mehāniskā enerģija = 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?

Zināms:

Masa (m) = 1000 kg

Sākotnējais ātrums (v_o) = 0

Galīgais ātrums (v_t) = 5 m/s

Meklē: Work (W) done by car

šķīdums:

W_neto = ½ m (v_t² - v_o²)

Work done by car :

W_neto = ½ (1000)(5² - 0²) = (500)(25 – 0) = (500)(25) = 12,500 Joule

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.

Zināms :

Bumbas masa (m) = 500 grami = 0.5 kg

Sākotnējais ātrums (v_o) = 10 m/s²

Galīgais ātrums (v_t) = 0 (velocity at the highest point)

Brīvās krišanas paātrinājums (g) = 10 m/s²

Meklē: Work (W) don by weight

šķīdums:

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

W_neto = ΔEK = EK_t – EK_o

Wnet = ½ m v_t² – ½ m v_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.

Net work :

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

W_neto = (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?

Zināms:

Bumbas masa (m) = 1 kg

Sākotnējais ātrums (v_o) = 0 m/s

Augstums (h) = 2.5 metri

Brīvās krišanas paātrinājums (g) = 10 m/s²

Meklē: Net work during displacement

šķīdums:

Final velocity of ball (v_t)

Calculated using the equation of free fall motion. Zināms: Brīvās krišanas paātrinājums (g) = 10 m/s², The change in height of ball (h) = 2.5 meters. Meklē: 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_neto = ΔEK = ½ m (v_t² - v_o²) = ½ (1){(5√2)² - 0²}

W_neto = 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.

Zināms:

Mass of object (m) = 2 kg

Sākotnējais ātrums (v_o) = 72 km/jam = 20 m/s

Galīgais ātrums (v_t) = 144 km/jam = 40 m/s

Attālums (s) = 400 metri

Brīvās krišanas paātrinājums (g) = 10 m/s²

Meklē: The net work

šķīdums:

The net work = changes of the kinetic energy

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

W_neto = 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.

Zināms:

Masa (m) = 2 kg

Sākotnējais ātrums (v_o) = 2 m/s

Darbs (W) = 21 džouli

Meklē: gala ātrums (v_t)

šķīdums:

W_neto= ΔEK

W_neto= 1/2 m v_t² -1/2 m v_o²

W_neto = 1/2 m (v_t² - v_o²)

21 = 1/2 (2) (v_t² - 2²)

21 = (v_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.

Zināms:

Constant force (F) = 8 Newton

Mass of object (m) = 16 kg

Initial speed of object (v_o) = 0 m/s

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

Meklē: Galīgais ātrums (v_t)

šķīdums:

Work = The change in the kinetic energy

W = KE final – KE initial

W = ½ m v_t² – ½ m v_o²

W = ½ m v_t² - 0

W = ½ m v_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 pie²

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

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

d = ½ (v_t / t) t²

^d = ½ (v_t) 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 m v_t² = (4)(v_t)(t)

½ m v_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…

Zināms:

Mass of object (m) = 1 kg

Sākotnējais ātrums (v_o) = 1 m/s

Beigu ātrums (v_t) = 2 x initial speed = 2 x 1 = 2 m/s

Meklē: Darbs ar

šķīdums:

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.

Zināms:

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

Meklē: Work required to accelerates the car

šķīdums:

Theorem of work-kinetic energy :

W = EK final – EK initial

W = ½ m v_t² – ½ m v_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.

Zināms:

Mass of object (m) = 2 kg

Sākotnējais ātrums (v_o) = 72 km/hour = 72,000 meters / 3600 second = 20 m/s

Beigu ātrums (v_t) = 144 km/hour = 144,000 meters / 3600 second = 40 m/s

Displacement of object = 400 meters

Meklē: Net work on the object

šķīdums:

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 m v_o²

W net = ½ m (v_t² - v_o²)

W net = ½ (2)(40² - 20²)

W net = 1600 – 400

W net = 1200 Joule

Apraksts:

W = Work, KE = kinetic energy

Kinētiskā enerģija

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

Zināms:

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

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

Vēlējās: Kinētiskā enerģija

šķīdums:

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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