# Work and kinetic energy – problems and solutions

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

__Known :__

Mass (m) = 5000 kg

Initial speed (v_{o}) = 0 m/s (car rest)

Final speed (v_{t}) = 20 m/s

__Wanted__ : net work

__Solution :__

The work-kinetic energy principle :

W_{net} = ΔEK

W_{net} = ½ m (v_{t}^{2} – v_{o}^{2})

W_{net} = net work

ΔEK = the change in kinetik energy

m = mass (kg),

v_{t} = final speed (m/s),

v_{o} = initial speed (m/s).

Net work :

W_{net} = ½ m (v_{t}^{2} – v_{o}^{2})

W_{net} = ½ (5000)(20^{2} – 0^{2})

W_{net} = (2500)(400 – 0)

W_{net} = (2500)(400)

W_{net} = 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!

__Known :__

Mass (m) = 10 kg

Initial speed (v_{o}) = 5 m/s

Final speed (v_{t}) = 10 m/s

__Wanted__ : net work

__Solution :__

Net work :

W_{net} = ΔEK

W_{net} = ½ m (v_{t}^{2} – v_{o}^{2})

W_{net} = ½ (10)(10^{2} – 5^{2})

W_{net} = (5)(100 – 25)

W_{net} = (5)(75)

W_{net} = 375 Joule

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

__Known :__

Car’s mass (m) = 2000 kg

Initial speed (v_{o}) = 10 m/s

Final speed (v_{t}) = 5 m/s

Wanted:__ net work__

__Solution :__

Net work :

W_{net} = ΔEK

W_{net }= ½ m (v_{t}^{2} – v_{o}^{2})

W_{net} = ½ (2000)(5^{2} – 10^{2})

W_{net} = (1000)(25 – 100)

W_{net }= (1000)(-75)

W_{net} = -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) Power is 2,520 Watt

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

The correct statements are…

__Known :__

Force (F) = 60 N

Time interval (t) = 12 seconds

Mass of object (m) = 10 kg

Initial velocity (v_{o}) = 6 m/s

__Wanted :__ The correct statements

__Solution :__

Acceleration of object :

∑F = m a

60 = 10 a

a = 60 / 10 = 6 m/s^{2}

The final velocity :

v_{t} = v_{o} + a t

v_{t} = 6 + (6)(12)

v_{t} = 6 + 72

v_{t} = 78 m/s

The distance traveled in 12 seconds :

s = v_{o} t + 1/2 a t^{2 }

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_{t}^{2} = 1/2 (10)(78)^{2 }= (5)(6084) = 30,420 Joule

(3) Power

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

(4) The increase in the kinetic energy

ΔKE = 1/2 m v_{t}^{2 }– 1/2 m v_{o}^{2 }= 1/2 m (v_{t}^{2 }– v_{o}^{2}) = 1/2 (10)(78^{2 }– 6^{2}) = 5 (6084 –36) = 5 (6048)

ΔKE = 30,240 Joule

5. The larger work is done by object number…

__Solution :__

Net work = change of th kinetic energy

W_{net} = ½ m (v_{t}^{2} – v_{o}^{2})

The larger work :

W_{1} = ½ (8)(4^{2} – 2^{2}) = (4)(16 – 4) = (4)(12) = 48 Joule

W_{2} = ½ (8)(5^{2} –3^{2}) = (4)(25 – 9) = (4)(16) = 64 Joule

W_{3} = ½ (10)(6^{2} – 5^{2}) = (5)(36 – 25) = (5)(11) = 55 Joule

W_{4} = ½ (10)(4^{2} – 0^{2}) = (5)(16 – 0) = (5)(16) = 80 Joule

W_{5} = ½ (20)(3^{2} – 3^{2}) = (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.

Known :

Mass (m) = 4000 kg

The initial velocity (v_{o}) = 25 m/s

The final velocity (v_{t}) = 15 m/s

__Wanted ____:__ Work done on car

__Solution :__

W_{net} = ½ m (v_{t}^{2} – v_{o}^{2}) = ½(4000)(15^{2}-25^{2}) = (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 is 10 m/s^{2}, then what is the kinetic energy of ball at the height of 2 meters.

__Known :__

Mass (m) = 0.1 kg

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

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ The kinetic energy at the height of 2 meters.

__Solution :__

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.__

__The initial mechanical energy ____= the gravitational potential energy.__

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

__The ____final ____mechanical energy ____= 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?

__Known :__

Mass (m) = 1000 kg

Initial velocity (v_{o}) = 0

Final velocity (v_{t}) = 5 m/s

__Wanted :__ Work (W) done by car

__Solution :__

W_{net} = ½ m (v_{t}^{2} – v_{o}^{2})

Work done by car :

W_{net} = ½ (1000)(5^{2} – 0^{2}) = (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/s^{2}. Acceleration due to gravity is 10 ms^{-2}. What is th work done by the weight force when ball reaches the maximum height.

__Known __:

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

Initial velocity (v_{o}) = 10 m/s^{2}

Final velocity (v_{t}) = 0 (velocity at the highest point)

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ Work (W) don by weight

__Solution :__

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

W_{net} = ΔEK = EK_{t }– EK_{o }

Wnet = ½ m v_{t}^{2} – ½ m v_{o}^{2 }= ½ m (v_{t}^{2} – v_{o}^{2})

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_{net }= ½ m (v_{t}^{2} – v_{o}^{2}) = ½ (0.5)(0^{2} – 10^{2})

W_{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 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?

__Known :__

Mass of ball (m) = 1 kg

Initial velocity (v_{o}) = 0 m/s

Height (h) = 2.5 meters

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ Net work during displacement

__Solution :__

__Final velocity of ball ____(v___{t}__)__

Calculated using the equation of free fall motion. __Known :__ Acceleration due to gravity (g) = 10 m/s^{2}, The change in height of ball (h) = 2.5 meters. __Wanted :__ Final velocity.

v_{t}^{2} = 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_{net} = ΔEK = ½ m (v_{t}^{2} – v_{o}^{2}) = ½ (1){(5√2)^{2} – 0^{2}}

W_{net} = 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.

__Known :__

Mass of object (m) = 2 kg

Initial velocity (v_{o}) = 72 km/jam = 20 m/s

Final velocity (v_{t}) = 144 km/jam = 40 m/s

Distance (s) = 400 meters

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ The net work

__Solution :__

__The net work = changes of the kinetic energy__

W_{net }= ΔEK = ½ m (v_{t}^{2} – v_{o}^{2}) = ½ (2)(40^{2} – 20^{2}}

W_{net} = 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.

__Known :__

Mass (m) = 2 kg

Initial velocity (v_{o}) = 2 m/s

Work (W) = 21 Joule

__Wanted :__ final velocity (v_{t})

__Solution :__

W_{net}= ΔEK

W_{net}= 1/2 m v_{t}^{2} -1/2 m v_{o}^{2}

W_{net} = 1/2 m (v_{t}^{2} – v_{o}^{2})

21 = 1/2 (2) (v_{t}^{2} – 2^{2})

21 = (v_{t}^{2} – 2^{2})

21 = v_{t}^{2} – 4

v_{t}^{2 }= 21 + 4 = 25

v_{t} = √25

v_{t} = 5 m/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.

__Known :__

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

__Wanted :__ The final speed (v_{t})

__Solution :__

__Work = The change in the kinetic energy__

W = KE final – KE initial

W = ½ m v_{t}^{2} – ½ m v_{o}^{2 }

W = ½ m v_{t}^{2} – 0

W = ½ m v_{t}^{2} —— 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 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}* – v*_{o}*) / t = v*_{t}* / t *

*d = ½ (v*_{t}* / t) t*^{2 }

^{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_{t}^{2 }= (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…

__Known :__

Mass of object (m) = 1 kg

Initial speed (v_{o}) = 1 m/s

Final speed (v_{t}) = 2 x initial speed = 2 x 1 = 2 m/s

__Wanted :__ Work

__Solution :__

The initial kinetic energy :

KE initial = ½ m v_{o}^{2} = 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_{t}^{2} = 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.

__Known :__

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

__Wanted :__ Work required to accelerates the car

__Solution :__

Theorem of work-kinetic energy :

W = EK final – EK initial

W = ½ m v_{t}^{2} – ½ m v_{o}^{2} = ½ m (v_{t}^{2} –v_{o}^{2})

W = ½ (1500)(20^{2} – 10^{2})

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.

__Known :__

Mass of object (m) = 2 kg

Initial speed (v_{o}) = 72 km/hour = 72,000 meters / 3600 second = 20 m/s

Final speed (v_{t}) = 144 km/hour = 144,000 meters / 3600 second = 40 m/s

Displacement of object = 400 meters

__Wanted :__ Net work on the object

__Solution :__

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_{t}^{2} – ½ m v_{o}^{2}

W net = ½ m (v_{t}^{2} – v_{o}^{2})

W net = ½ (2)(40^{2} – 20^{2})

W net = 1600 – 400

W net = 1200 Joule

__Description :__

W = Work, KE = kinetic 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 (m) = 10 gram = 10/1000 kilogram = 1/100 kilogram = 0.01 kilogram

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

__Wanted:__ Kinetic energy

__Solution :__

KE = 1/2 m v^{2 }

KE = 1/2 (0,01 kg)(100 m/s)^{2}

KE = 1/2 (0,01 kg)(10.000 m^{2}/s^{2})

KE = (0,01 kg)(5000 m^{2}/s^{2})

KE = 50 kg m^{2}/s^{2}

KE = 50 Joule

### Ebook PDF Work - kinetic energy problems and solutions

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- Work-mechanical energy principle problems and solutions
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- Application of conservation of mechanical energy for up and down motion in free fall motion
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- Application of conservation of mechanical energy for projectile motion