Work Energy Theorem Practice Problems

Work-Kinetic energy :

1. A 5000-kg automobile accelerated from rest to 20 m/southward. Determine the net work done on the machine.

Known :

Mass (m) = 5000 kg

Initial speed (v _o ) = 0 g/s ( automobile residual )

Final speed (v _t ) = twenty k/s

Wanted : net piece of work

Solution :

The work-kinetic energy principle :

Due west _internet = Δ EK

W _net = ½ g (v _t ² – v _o ² )

W _net = internet work

Δ EK = the alter in kinetik energy

m = mass (kg),

v _t = final speed (m/s),

5 _o = initial speed (1000/s).

Net piece of work :

W _net = ½ m (v _t ² – five _o ² )

W _cyberspace = ½ (5000)(xx ² – 0 ^two )

Due west _net = (2500)(400 – 0)

Due west _net = (2500)(400)

W _net = 1000,000 Joule

two. A 10-kg object accelerated from 5 k/south to x m/s. Determine the net work washed on the object!

Known :

Mass (m) = 10 kg

Initial speed (five _o ) = five m/south

Final speed (five _t ) = 10 1000/s

Wanted : internet work

Solution :

Net piece of work :

Westward _net = Δ EK

Westward _net = ½ m (5 _t ² – 5 _o ² )

W _cyberspace = ½ (10)(10 ⁱⁱ – 5 ² )

Due west _cyberspace = (five)(100 – 25)

W _net = (v)(75)

West _internet = 375 Joule

3. A 2000-kg car decelerated from 10 chiliad/s to 5 1000/south. What is the work done on the car ?

Known :

Car's m donkey (m) = 2000 kg

Initial speed (v _o ) = x yard/s

Final speed (five _t ) = 5 m/south

Wanted: net piece of work

Solution :

Net work :

W _internet = Δ EK

W _net = ½ 1000 (v _t ^two – 5 _o ⁱⁱ )

West _internet = ½ (2000)(5 ^two – 10 ⁱⁱ )

Due west _cyberspace = (1000)(25 – 100)

W _net = (g)(-75)

W _net = -75,000 Joule

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

4. A sixty-N constant force exerted on a x-kg object for 12 seconds. The initial velocity of an object is 6 m/southward and the direction of the object is the same equally the direction of the strength.

(one) Piece of work done on the object is thirty,240 Joule

(ii) The terminal kinetic free energy is xxx,240 joule

(iii) Power is ii,520 Watt

(iv) Thursday increase in the kinetic energy of the object is 180 Joule

The correct statements are…

Known :

Strength (F) = 60 N

Time interval (t) = 12 seconds

Mass of object (m) = 10 kg

Initial velocity (v_o) = six m/s

Wanted : The correct statements

Solution :

Dispatch of object :

∑F = grand a

60 = 10 a

a = lx / 10 = half-dozen m/sⁱⁱ

The final velocity :

v_t = v_o + a t

v_t = 6 + (6)(12)

v_t = vi + 72

v_t = 78 chiliad/s

The distance traveled in 12 seconds :

s = five_o t + 1/2 a t²

south = (6)(12) + one/2 (half dozen)(12)²

s = 72 + (three)(144)

due south = 72 + 432

south = 504 meters

(1) Work washed past force

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

(2) The final kinetic energy

KE = 1/ii thousand five_t ² = 1/2 (10)(78)² = (five)(6084) = xxx,420 Joule

(iii) Power

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

(4) The increment in the kinetic free energy

ΔKE = 1/2 m v_t ² – 1/2 thousand 5_o ⁱⁱ = ane/2 m (v_t ² – v_o ²) = 1/2 (10)(78² – 6²) = 5 (6084 –36) = five (6048)

ΔKE = xxx,240 Joule

v. The larger piece of work is washed by object number…

Solution :

Net piece of work = alter of th kinetic energy

W_net = ½ grand (v_t ² – v_o ²)

The larger work :

Due west₁ = ½ (8)(4² – twoⁱⁱ) = (4)(16 – 4) = (4)(12) = 48 Joule

Due west₂ = ½ (viii)(5² –3ⁱⁱ) = (4)(25 – 9) = (4)(xvi) = 64 Joule

West₃ = ½ (10)(half dozen² – vⁱⁱ) = (5)(36 – 25) = (5)(11) = 55 Joule

Westward₄ = ½ (10)(4ⁱⁱ – 0²) = (5)(xvi – 0) = (5)(sixteen) = lxxx Joule

West₅ = ½ (xx)(3² – three²) = (x)(nine – 9) = (10)(0) = 0 Joule

half dozen. A 4000-kg car travels along straight line at 25 1000/s. The car is decelerated so that the car's final velocity is xv m/s. What is the piece of work washed on the machine.

Known :

Mass (chiliad) = 4000 kg

The initial velocity (v_o) = 25 thousand/s

The final velocity (five_t) = 15 m/s

Wanted : Piece of work done on car

Solution :

W_net = ½ m (v_t ^two – five_o ²) = ½(4000)(15^two-25²) = (2000)(225-625) = (2000)(-400) = -800,000 Joule = -800 kJ

7. A 0.1-kg thrown horizontally at six m/s from the pinnacle of 5 meters. If the dispatch of gravity is 10 m/due south², so what is the kinetic free energy of brawl at the height of 2 meters.

Known :

Mass (m) = 0.one kg

The alter in height (h) = 5 thou – 2 one thousand = three meters

Dispatch due to gravity (g) = 10 g/s²

Wanted : The kinetic energy at the pinnacle of 2 meters.

Solution :

Projectile motion can be understood past analyzing the horizontal and vertical components of the motion separately. Motion in horizontal direction analyzed as the abiding velocity motion and movement in vertical direction analyzed every bit free fall motion or vertical motion.

The initial mechanical free energy = the gravitational potential energy.

PE = k 1000 h = (0.1)(ten)(3) = three Joule.

The final mechanical energy = the kinetic free energy.

KE = 3 Joule.

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

Known :

Mass (m) = 1000 kg

Initial velocity (five_o) = 0

Final velocity (v_t) = 5 k/due south

Wanted : Work (W) washed by car

Solution :

West_net = ½ m (5_t ² – 5_o ²)

Work done by automobile :

W_net = 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 x m/s². Acceleration due to gravity is 10 ms^-2. What is thursday work washed by the weight force when ball reaches the maximum acme.

Known :

Mass of ball (yard) = 500 gram = 0.5 kg

Initial velocity (v_o) = x g/south²

Final velocity (v_t) = 0 (velocity at the highest indicate)

Acceleration due to gravity (m) = 10 yard/s²

Wanted : Work (Westward) don by weight

Solution :

The net work done by cyberspace forcefulness on an object = the change in the kinetic free energy.

W_cyberspace = ΔEK = EK_t – EK_o

Wnet = ½ m v_t ² – ½ k v_o ^two = ½ m (five_t ² – v_o ^two)

KE_t = the concluding kinetic energy, KE_o = the initial kinetic energy, thousand = mass of object, v_t = the final velocity of object, 5_o = initial velocity of object.

Cyberspace piece of work :

W_net = ½ m (v_t ⁱⁱ – five_o ⁱⁱ) = ½ (0.5)(0² – 10²)

Due west_net = (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 ane-kg object free fall with the peak difference = 2.5 meters. Acceleration due to gravity is x thou.s^-2. What is the work done on the object?

Known :

Mass of ball (g) = 1 kg

Initial velocity (v_o) = 0 m/s

Elevation (h) = ii.v meters

Acceleration due to gravity (g) = 10 m/s^two

Wanted : Internet work during displacement

Solution :

Concluding velocity of ball (v _t )

Calculated using the equation of costless fall move. Known : Acceleration due to gravity (thousand) = x m/southⁱⁱ, The change in height of brawl (h) = 2.5 meters. Wanted : Concluding velocity.

v_t ² = two grand h = two(x)(ii.5) = 2(25)

5_t = √2(25)

v_t = 5√2

Net work = the modify in kinetic energy

W_net = ΔEK = ½ one thousand (5_t ² – five_o ²) = ½ (one){(v√2)² – 0²}

West_net = ½ (25)(ii) = 25 Joule

11. A 2-kg object travels at 72 km/hour. Afterward travels 400 meters, the final velocity of object is 144 km/60 minutes. Acceleration due to gravity is 10 ms^-2. Find the net work.

Known :

Mass of object (one thousand) = 2 kg

Initial velocity (five_o) = 72 km/jam = 20 m/south

Final velocity (five_t) = 144 km/jam = 40 thou/s

Altitude (s) = 400 meters

Dispatch due to gravity (m) = 10 m/s²

Wanted : The net piece of work

Solution :

The net work = changes of the kinetic energy

W_net = ΔEK = ½ m (v_t ² – v_o ²) = ½ (ii)(40^two – 20²}

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

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

Known :

Mass (g) = two kg

Initial velocity (v_o) = 2 m/south

Piece of work (Westward) = 21 Joule

Wanted : final velocity (v_t)

Solution :

W_internet= ΔEK

W_net= ane/2 k v_t ² -1/two m v_o ^two

Due west_internet = 1/2 chiliad (v_t ² – v_o ²)

21 = 1/2 (2) (v_t ⁱⁱ – 2ⁱⁱ)

21 = (v_t ² – 2^two)

21 = five_t ^two – iv

v_t ^two = 21 + 4 = 25

v_t = √25

5_t = v m/s

one 3 . A 8 Due north constant force acts on an object with mass of 16 kg. If the object initially at residuum, then determine the speed of the object after strength acts on the object for 4 seconds.

Known :

Constant force (F) = eight Newton

Mass of object (m) = xvi kg

Initial speed of object (v _o ) = 0 m/due south

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

Wanted : The final speed (5 _t )

Solution :

Work = The change in the kinetic energy

W = KE final – KE initial

Westward = ½ m v _t ² – ½ m v _o ²

W = ½ grand v _t ² – 0

Westward = ½ thousand five _t ² —— Equation 1

Work = Force x Displacement

West = F d

West = 8 d

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

d = v _o t + ½ a t ²

d = displacement, v _o = initial velocity, t = time interval, a = acceleration

d = 0 + ½ a t ² = ½ a t ² —-> a = (v _t – five _o ) / t = v _t / t

d = ½ (5 _t / t) t ²

^d = ½ (5 _t ) t

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

W = 8 d

Westward = 8(1/2)(five _t )(t)

West = (four)(v _t )(t) —— equation 2

Equation 1 = Equation 2

Due west = W

½ grand v _t ² = (four)(v _t )(t)

½ g 5 _t = (iv)(t)

½ (sixteen)(five _t ) = 4(iv)

8 five _t = 16

5 _t = 16 / 8

v _t = 2 meters/2d

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

Known :

Mass of object (grand) = i kg

Initial speed (v _o ) = 1 m/s

Final speed (v _t ) = 2 x initial speed = two 10 1 = 2 m/s

Wanted : Work

Solution :

The initial kinetic free energy :

KE initial = ½ m five _o ^two = ½ (1)(1) ² = ½ (ane)(i) = ½ (1) = 0.five

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

KE terminal = ½ grand v _t ^two = ½ (1)(two) ² = ½ (4) = 2

Theorem of work-kinetic energy :

Piece of work = The alter in kinetic energy

Work = The final kinetic free energy– the initial kinetic free energy

Work = 2 – 0.v

Piece of work = one.five

The initial kinetic free energy = 0.5

Work = 3 x 0.5 = 1.v

Required piece of work 3 times of its initial kinetic energy.

15 . A car with mass of 1500 kg moves with speed of 36 km/60 minutes on a linear and smooth horizontal route. The machine accelerated to 72 km/hour. Determine the work required to acceleration the car.

Known :

Mass of car (yard) = 1500 kg

Initial speed of car (v _o ) = 36 km/hour = 36,000 meters / 3600 second = x meters/2nd

Concluding speed of car (5 _t ) = 72 km/hr = 72,000 meters / 3600 2nd = twenty meters/second

Wanted : Work required to accelerates the car

Solution :

Theorem of piece of work-kinetic energy :

W = EK final – EK initial

Westward = ½ k v _t ^two – ½ m five _o ² = ½ chiliad (five _t ⁱⁱ –v _o ² )

W = ½ (1500)(twenty ⁱⁱ – 10 ⁱⁱ )

W = ½ (1500)(400 – 100)

W = ½ (1500)(300)

West = (1500)(150)

W =225,000 Joule

sixteen . An object with mass of ii kg initially moves at speed of 72 km.hour ^-1 . After motility in horizontal directly road equally far as 400 k, the speed of the object is 144 km.60 minutes ^-1 . Decide the total piece of work on the object.

Known :

Mass of object (k) = ii kg

Initial speed (five _o ) = 72 km/hour = 72,000 meters / 3600 second = xx m/s

Final speed (v _t ) = 144 km/hour = 144,000 meters / 3600 second = 40 m/s

Displacement of object = 400 meters

Wanted : Net piece of work on the object

Solution :

Theorem of work-kinetic energy states that the net work acts on an object same every bit the modify of the kinetic energy of the object.

Westward net = KE final – KE initial

W internet = ½ m v _t ⁱⁱ – ½ m five _o ²

West net = ½ g (v _t ² – v _o ² )

W net = ½ (2)(40 ² – 20 ⁱⁱ )

W net = 1600 – 400

W net = 1200 Joule

Description :

W = Work, KE = kinetic free energy

Kinetic energy

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

Known :

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

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

Wanted: Kinetic free energy

Solution :

KE = ane/two m v ^two

KE = ane/2 (0,01 kg)(100 m/due south) ²

KE = 1/ii (0,01 kg)(10.000 m ² /s ² )

KE = (0,01 kg)(5000 grand ² /s ^two )

KE = 50 kg thousand ² /due south ²

KE = 50 Joule

[wpdm_package id='1191′]

1. Work done by force issues and solutions
2. Work-kinetic energy bug and solutions
3. Piece of work-mechanical energy principle bug and solutions
4. Gravitational potential energy issues and solutions
5. Potential energy of elastic spring issues and solutions
6. Power problems and solutions
7. Application of conservation of mechanical energy for free autumn motion
8. Application of conservation of mechanical energy for upwards and down motion in free fall motility
9. Awarding of conservation of mechanical energy for motion on a curve surface
10. Application of conservation of mechanical energy for motility on an inclined aeroplane
11. Awarding of conservation of mechanical energy for projectile move

Work Energy Theorem Practice Problems,

Source: https://physics.gurumuda.net/work-and-kinetic-energy-problems-and-solutions.htm

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