4



4.1 WORK

Work is defined as the product of magnitude of the displacement with the component of the force parallel to the displacement.

• Unit S.I for work – Nm @ Joule (J)

• Formula work:

Example 1

A student lift a 25 kg bag onto his shoulders through a vertical distance of 3 m. what is the work done by the student?

Solution:

F = ( 25 ) ( 9.81 )

= 245.25 N

W = ( 245.25 ) ( 3 )

= 735.75 Nm @ Joule

When F is not parallel to s, then we must take the component of F which is parallel to the s.

Formula:

Example 2:

A gardener pushes the lawn mower with a constant force of 90N at an angle of 40o to the horizontal, how much work does he do in pushing it a horizontal distance of 7.5m?

Solution:

W = F cos ( x s

= 90 cos40° x 7.5

= 517 J

Work done raising an object against the force of gravity

The force required to lift an object of mass, m, near the surface of the earth is mg, that is, the weight of the object, directed upwards. When it is raised to a height, h, the object is displaced by a distance h in the direction of the force.

Example 3:

A student lift a 25 kg bag onto his shoulders through a vertical distance of 3 m. what is the work done by the student?

Solution:

F = ( 25 ) ( 9.81 )

= 245.25 N

W = ( 245.25 ) ( 3 )

= 735.75 Nm @ Joule

Exercise 1

How much work is done when a force of 5 kN is used to move a box through 600mm in the direction of the force.

Exercise 2:

A box of 20 kg mass is dragged across a smooth floor using a rope which is makes an angle of 30° to the horizontal. How much work is done if the box is being dragged for 12 m.

Exercise 3:

Find the work done in raising 100 kg of a box through a vertical distance of 3m.

Exercise 4:

How much work done by a force of 40 N acting 37° above the horizontal in pulling a block 8 m along a horizontal surface?

Exercise 5:

In raising 200 kg box 10 000 J of work is performed. How high is it raised ?

Exercise 6:

A force did 1500 Joules of work on an object in 7.5 m. How big was the force?

Exercise 7:

A man with a mass of 55 kg ascends a flight of 10 steps. Each step has a height of 0.25 m. Calculate the work that has been done by the man.

4.2 ENERGY

Energy is defined as the capacity to do work.

It is a scalar quantity and has the units of joules, (J).

Type of energy

Generally energy can be divided into two groups:

Mechanical Energy:

Energy due to position in a field force or energy due to movement

Non-mechanical Energy:

Energy that does not fall into the above category

4.2.1 KINETIK ENERGY

Definition: kinetic energy is the energy of motion

The kinetic energy of a body may be defined as the amount of work it can do before being brought to rest.

Formula:

Example 4:

A car of mass 1000 kg travelling at 30m/s has its speed reduced to 10m/s by a constant breaking force over a distance of 75m.

Find:

i. The cars initial kinetic energy

ii. The final kinetic energy

iii. The breaking force

Solution:

i. Initial kinetic energy = ½ mv2

= 500 x 302

= 450000J

ii. Final Kinetik Energy = ½ mv2

= 500 x 102

= 500000J

iii. Change in kinetic energy = 400 kJ

Example 5:

The kinetic energy of a boat is calculated at 52,000 J. If the boat has a mass of 39,000 kg, with what velocity is it moving?

Solution:

Ek= ½ mv²

v² = 2Ek

m

v² = 2(52000)

39000

v = 1.63 m/s

Example 6:

A rocket of mass 1.5x104 kg accelerates at 220 m/s2 for 29s from an initial speed of 5200 m/s.

i. How fast will be rocket be travelling after the 29 s?

ii. How much Kinetic Energy has the rocket gained?

Solution:

i. v = u + at

= 5200 + (220 x 29)

= 5200 + 6380

= 11580m/s

ii. Initial Kinetic Energy

Ek= ½ mv²

=0.5 x (1.5x104) x (5200)2

=2.028 x 1011 J

Final Kinetic Energy

Ek= ½ mv²

=0.5 x (1.5x104) x (11580)2

=1.006 x 1012J

so Kinetic Energy gained = final E - initial E

= 1.006 x 1012J - 2.028 x 1011J

= 8.032 x 1011J

Exercise 8:

A pistol fires a 3g bullet with a velocity of 400 m/s. How much kinetic energy is given to the bullet.

Exercise 9:

The velocity of a 3 kg trolley is increased from 10 m/s to 16 m/s over a distance of 10 m. Find the charge in the kinetic energy of the trolley.

Exercise 10:

The kinetic energy of a car is 20 kJ. If the car has a mass of 1,500 kg, with what velocity is it moving?

4.2.2 POTENTIAL ENERGY

Definition: Potential energy, or stored energy, is the ability of a system to do work due to its position or internal structure.

A body is at rest on the earth's surface. It is then raised a vertical distance h above the surface. The work required to do this is the force required times the distance h.

Since the force required is it's weight, and weight, w = mg, then the work required is mgh.

The body now possesses this amount of energy - stored as potential energy - it has the capacity to do this amount of work, and would do so if allowed to fall to earth.

Formula:

.

Example 7:

What is the potential energy of a 10kg mass located 100m above the surface of the earth.

Solution:

Potential energy Ep = mgh

= 10 x 9.81 x 100

= 9810 J

= 9.81 kJ

Example 8:

A box has a mass of 5.8kg. The box is lifted from the garage floor and placed on a shelf. If the box gains 145J of Potential Energy (Ep), how high is the shelf?

Solution:

Potential energy Ep = mgh

h = Ep

mg

= 145

(5.8 x 9.81)

= 2.55 m

Example 9:

A man climbs on to a wall that is 3.6m high and gains 2268J of potential energy. What is the mass of the man?

Solution:

Potential energy Ep = mgh

m = Ep

gh

= 2268

( 9.81x3.6 )

= 64.22 kg

Exercise 11:

A 4 kg rock is sitting on a 3 m ledge. How much potential energy does it have?

Exercise 12:

A 30 kg child climbs 15 meters up a tree. When he stops to have a look around, what is the

child’s potential energy?

Exercise 13:

A 2 kg box rests on a table top 0.6 m above the floor. What is the potential energy of the box with respect to the floor? What is its potential energy with respect to a counter top 1.1 m above the floor?

4.2.4 CONSERVATION OF ENERGY

The principle of conservation of energy state that the total energy of a system remains constant. Energy cannot be created or destroyed but may be converted from one form to another.

Without considering the air resistance, for the free fall object, at the top, from where the object is being dropped, it has only potential energy and no kinetic energy as velocity is zero. When object is dropped it starts falling down under effect of gravity and as it falls down, its potential energy is converted into kinetic energy. As a result the kinetic energy of the object increases and potential energy decreases. At any point the decrease in potential energy is equal to the increase in kinetic energy. The moment when object hits the ground, the entire potential energy is converted into kinetic energy and this kinetic energy is equal to the potential energy of the object at the top.

[pic]

Example 10:

A 0.5 kg ball is dropped from height of 10 m from floor level, calculate the velocity of the ball when it reach the floor.

Solution:

Initial Potential energy, Ep = mgh1

E p = (0.5 x 9.81 x 10)

E p = 49.05 J

When reach the floor, Kinetic energy is equal to the initial potential energy

Ek = Ep = 49.05

Ek= ½ mv² = 49.05

v² = 49.05 x 2

0.5

V = 14 m/s

Example 11:

A boy drops a 1.2 kg of vase from height of 6.4 m.

(a) What is the kinetic energy of the vase when it is at a height of 4.2 m?

(b) With what speed will the vase hit the ground? (Neglect air resistance)

Solution:

Given: m = 1.2 kg

ho = 6.4 m

h = 4.2 m

(vo = 0)

First, find the total mechanical energy since this quantity is conserved while it is falling. Initially, with vo = 0, the vase’s total mechanical energy is all potential. Taking the ground as the zero reference point.

At h = 6.4 meter

i. Ep = mgh1

E p = (1.2 x 9.81 x 6.4)

E p = 75.34 J

At h = 4.2 meter

Ep2 = mgh2

Ep2 = (1.2 x 9.81 x 4.2)

Ep2 = 49.44 J

The kinetic energy is the change in potential energy, so

Ek = Ep1 - Ep2

= 75.34 – 49.44

= 25.9 J

ii. Just before the vase strikes the ground (h = 0, EK = 0), the total mechanical energy is all kinetic.

[pic]

Thus,

[pic]

Example 12:

A 2.0-kilogram ball rolling along a flat surface starts up a hill. If the ball reaches a height of 0.6 meters, what was its initial speed?

Solution:

Ep = mgh1

E p = (2 x 9.81 x 0.6)

E p = 11.77 J

When the ball reaches 0.6 m of height, the initial kinetic energy is equal to the potential energy

Ek = Ep = 11.77 J

Ek= ½ mv² = 11.77

v² = 11.77 x 2

2

V = 3.43 m/s

Exercise 14:

A 600g ball is freely fall from a height of 8 meter, calculate the velocity of the ball when it reach the floor.

Exercise 15:

A ball of 2 kg of mass is dropped from a height of 5 m.

i. What is the kinetic energy of the vase when it is at a height of 3 m?

ii. With what speed will the vase hit the ground? (Neglect air resistance)

Exercise 16:

The floor of the basement of a house is 3.0 m below ground level, and the floor of the attic is 4.5 m above ground level.

i. What are the respective potential energies of 1.5-kg objects in the basement and attic, relative to ground level.

ii. What is the change in potential energy if the object in the attic is brought to the basement?

4.3 POWER

Power is defined as the rate of work done.

Units of power are watts (W).

• 1 watt = 1 joule per second.

• Also horsepower (hp). (1hp = 745.6 W.)

4.3.1 Power and velocity

Power can also be expressed in terms of force and velocity.

Example 13:

It takes 20 seconds to push the box in example 4.4 across the distance. What is the power?

Solution:

Work done = 3000 J

Power = 3000 (J)                   

20

= 150 W

Example 14:

An electric locomotive is pulling a train at a constant speed of 30 m/s. The train has a rolling resistance of 100 kN. Calculate the power of the electric locomotive.

Solution:

Power = force x speed = 100 000 x 30

= 3 000 000 W

= 3 MW

Example 15:

A constant force of 2kN pulls a crate along a level floor a distance of 10 m in 50s.

What is the power used?

Solution:

Work done = force x distance

= 2000 x 10

= 20000 J

Power = work done

time taken

= 20000

50

= 400 W

Example 16:

A hoist operated by an electric motor has a mass of 500 kg. It raises a load of 300 kg vertically at a steady speed of 0.2 m/s. Frictional resistance can be taken to be constant at 1200 N. What is the power required?

Solution:

Total mass m = 800 kg

Weight , w = 800 x 9.81

= 7849 N

Total force = 7848 + 1200

= 9048 N

Power = force F x speed v

= 9048 x 0.2

=1810 W

Exercise 17:

A woman runs a kilometer using a force of 210N and a power output of 500W. How long in minutes does it take this woman to complete 1 kilometer?

Exercise 18:

How much power is used if a force of 35N is used to push a box a distance of 10 meters in 5 seconds?

Exercise 19:

What is the power of a kitchen blender if it can perform 3,750 joules of work in 15 seconds?

4.3.2 EFFICIENCY

Definition – Efficiency is defined as the ratio of the useful output power or work to the total power or work put into the machine.

This ratio is normally expressed as a percentage.

Example 17:

Calculate the power output of a pump if its efficiency is 65 % and its the power input is 900 kW.

Solution:

( = Pout x 100%

Pin

Pout = Pin (

100%

Pout = 9000 x 65%

100%

= 5850 KW

Example 18:

If a machine has a 100W power input and 50W power output, calculate the efficiency

Solution:

[pic]

[pic]

[pic]

[pic]

Example 19:

An electric motor with a 2 hp output drives a machine with an efficiency of 75%. What is the energy output of the machine per second? (1hp = 745.6 Watt)

Solution:

Pin = 2 hp = 1491.2 W

( = Pout x 100%

Pin

Pout = 75% x 1491.2

100%

= 1118.4 W

Exercise 20:

Calculate the power output of a pump if its efficiency is 65 % and its the power input is 900 kW.

Exercise 21:

A pulley system operates with 80% efficiency. If the work input is 500 joules, how much useful works produced?

Exercise 22:

The motor of an electric drill with efficiency of 90% has a power input of 3 hp. Calculate:

i. Power output, if given 1 hp = 745.6 Watt.

ii. Work is done by the drill in a time of 1 minute.

END OF CHAPTER (Questions)

Question 1

Give the definition, formula and SI unit for:

i. Work

ii. Energy

iii. Power

iv. Kinetic Energy

v. Potential Energy

Question 2

A 70 kg athlete can run 100m within 10 seconds. Calculate his Kinetic Energy:

Question 3

A 200 grams bullet move through the air with a velocity of 150 m/s at 250 m height. Calculate the total Potential energy and Kinetic Energy of the bullet.

Calculate in Joule the work done in each of the following cases:

i. A mass of 4 kg is lifted to a height of 20 m.

ii. A trolley is pulled with a force of 30 N for a distance of 9m

Question 4

A man pulls a 135 kg trolley with a force of 675 N for 8 m at an angle 35° from horizontal. Calculate the work done of the trolley (Neglect friction).

Question 5

A 10000J of work is raised up a box vertically. If the mass of the box is 20kg, calculate the height that can be reached by the box.

Question 6

A car is moving with a velocity of 20 m/s. If the mass of the car is 3000kg, calculate the kinetic energy possessed by the car. (Given g = 9.81 m/s2)

Question 7

A student with a weight of 650N is going up a staircase of 60 m height within 1.5 minutes. How much power is needed by the student to go up the stairs?

Question 8

Calculate the work done by 5 kW powered motor for 7 minutes.

Question 9

A pump can lift 150 kg water of height in 40 seconds from it source. Calculate the work done and the power of the pump.

Question 10

A ball of mass 10 kg falls from a height of 2.5m. Calculate the potential energy and kinetic energy possessed by the ball.

Question 11

Calculate the height of box of mass of 200kg can be lifted if the work done is 10000 J.

Question 12

Calculate the work done by the force; refer to the diagram as shown as figure Q13 below:

Question 13

A steel ball of mass 2 kg is released from a height of 8 m from the ground. On hitting the ground, the ball rebounds to a height of 3.2 m as shown in Figure Q14.

If air resistance can be neglected and the acceleration due to gravity g = 9.81 m/s2, find

i. The kinetic energy of the ball as it reaches the ground,

ii. The velocity of the ball on reaching the ground,

iii. The kinetic energy of the ball as it leaves the ground on rebound.

Question 14

The motor of an electric drill with efficiency of 80% has a power input of 1.5 hp. Calculate:

iii. Power output, if given 1 hp = 746 Watt.

iv. Work is done by the drill in a time of 30 sec.

[pic]

-----------------------

F

s

s

F cos (

W = F x s

20°

F

3.2 m

(

Magnetic

Electric

Elastic

Gravitational

Rotational

Linear

Potential

Kinetic

Heat

Sound

Light

Non-mechanical

Mechanical

Energy

F = w

m

h

Figure Q14

H"

w = mg

m

8.0≈

w = mg

m

8.0 m

m

P = W  

t

( = Pout x 100%

Pin

Where :

Ep = Potential Energy (J)

m = mass (kg)

g = gravitational acceleration (9.81 m/s2)

h = height (m)

Where:

Ek = Kinetic energy (J)

m = mass (kg)

v = velocity (m/s2)

h

Where :

W = work done (J)

F = Force (N)

S = distance moved in the direction of the force (s)

W = F cos( x s

Where :

W = work done (J)

F = Force (N)

S = distance moved in the direction of the force (s)

( = angle

Where:

( = efficiency (%)

Pout = power output (W)

Pin = power input (W)

(

F = 30 kN

S = 50 m

Ek = [pic] m v2

Ep = mg h

Where :

P = Power (Watt)

W = work done ( J) 

t = time taken (s)

Where :

P = Power (Watt)

F = Force (N)

V = velocity (m/s2) 

P = Fv  

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