The rate of doing work is equal to the rate of using energy since the a force transfers one unit of energy when it does one unit of work.
Work:
If a person pushes a wall and the wall does not move, though the person may sweat and physically become tired, he would not have done any work. But if the person pushes a trolley and the trolley moves it is said work is done.
Therefore,
Work is the product of force and distance moved in the direction of the force.
Thus,
Work = Force (f) * distance (d) moved in the direction of the force.
SI unit of work is Joules.
Example:
A sack of maize, which weighs 800N, is lifted to a height of 2m. What is work done against gravity?
Data given:
Force (f) = 800N
Height distance = 2m
Work d>
Solution:
Work done (w.d) = force (f) x distance (d)
w.d = 800N x 2m
= 1600 Joules
Work done (w.d) = 1600Joules
Energy:
Energy can be defined as capacity of doing work.
Energy has the same SI unit like that of work, and that is Joules (J)
There are different forms of energy such as:
- Chemical energy
- Heat energy
- Light energy
- Sound energy
- Electrical energy
- Nuclear energy
- Solar energy
Types of energy;
There are two types of chemical energy, which are:
- Potential energy
- Kinetic energy
- Potential energy
Potential energy is the energy possessed by a body mass in its position or state. - Kinetic energy
Kinetic energy – is the energy possessed by a body due to its motion
Consider when the body is vertically thrown upwards with an initial velocity ‘u’ from the ground.
Consider when the body is vertically thrown upwards with an initial velocity ‘u’ from the ground.
At the ground
The height is zero and initial velocity is at maximum so as to attain maximum height.
Therefore K.E = ½ MV2 will be maximum
K.Emax = ½ mv2
Where K.E = Kinetic energy
M = Mass of the object/body
V=Velvety
P.E = Mgh
Where
P.E = Potential energy
P.E = Potential energy
M= Mass of the object
H = Height of the object
g = gravitation force
P.E = Mgh will be zero because P.E
= M*g*0 (body at the ground where k=0)
Neglecting the air resistance, as the body moves upwards its velocity decreases it also experiences gravitational force (g) pulling downwards towards the earth’s centre.
The maximum Height Attained
- The final velocity of the body will be zero (V=0)
Therefore K.E = ½ mv2
K.E = ½ m(0)2
K.E = 0
P.E = MgHmax
Note:
That the object drops from Hmax that is; it leaves with zero Kinetic Energy. At position A in fig. 8. The conservation of mechanical energy (M.E) is given as:
P.E + K.E = Constant
(The sum of P.E and K.E is constant throughout the motion of the object if the air resistance is neglected)
TRANSFORMATION OF ENERGY
The notion of energy is that energy is changed from one form into different forms using transducers.
Transducer is a device used to transform energy from one form to another.
Example:
Transducer is a device used to transform energy from one form to another.
Example:
- Battery converts chemical energy into electrical energy
- A generator converts mechanical energy into electrical energy.
- A motor converts electrical energy into mechanical energy.
TABLE SUMMARISING ENERGY TRANSFORMATIONS.
ORIGINAL ENERGY
|
TRANSDUCER
|
ENERGY TRANSFORMED
|
|
Battery
|
Electrical energy
|
|
Motor
|
Chemical energy
|
|
Generator
|
Electrical energy
|
|
Solar panel
|
Electrical energy
|
|
Motor car
|
Mechanical energy
|
|
Microphone
|
Sound energy
|
|
Heater
|
Heat energy
|
Experiment 01
Are:- To demonstrate pressure of potential energy.
Materials and Apparatus
- A heavy stone
- A bucket full of water
- A strong inelastic rope; and
- Smooth pulley
Procedures:
- Collect the heavy stone, using a strong rope tie it to a bucket of water
- Pass the rope over smooth pulley fixed to a support.
- Hold stationary the heavy stone at height “h” above the ground
- Release the stone
Results and observations:
- When the stone released the bucket of water will start to rise.
- The stone is said to have potentials energy because of its position above the ground.
- Lifting a body of mass “m” to a height “h” above the ground requires work to be done against gravity.
Work = Mgh
Example 01
A ball of mass 0.5 kg is kicked vertically upwards and rises to a height of 5m. Find the potential energy by the ball.
Data given:
Mass of the ball (Mb) = 0.5 kg
Height (h) = 5m
Gravitation force (g) 10N/kg
Potential energy (P.E) = ?
Height (h) = 5m
Gravitation force (g) 10N/kg
Potential energy (P.E) = ?
Solution:
Potential energy (P.E)= mgh
= 0.5kg x 10N/kg x 5m
= 25 NM
1NM= 1Joules
Potential energy (P.E) = 25 Joules.
= 0.5kg x 10N/kg x 5m
= 25 NM
1NM= 1Joules
Potential energy (P.E) = 25 Joules.
Experiment 02
Aim: to investigate the law of conservation of energy by a simple pendulum.
Materials and Apparatus:
A pendulum bob and light inelastic string
Materials and Apparatus:
A pendulum bob and light inelastic string
Producers:
- Pull the bob of a simple pendulum in position A so that it is at height “h” above position B.
- Release the bob so that it swings to position C via the lowest position B and back to A.
Consider the figure below:
Observation
When the bob is at position A, it possesses potential energy only due to the height “h” which is equal to “Mgh”.
As it swings downwards to position B, the height decreases, and as the result it loses potential energy.
When the bob is at position A, it possesses potential energy only due to the height “h” which is equal to “Mgh”.
As it swings downwards to position B, the height decreases, and as the result it loses potential energy.
- The bob has Vmax and hence K.Emax at B. The height at B is zero, thus the P.E is zero.
- As it swings towards C, the P.E increases and reaches its maximum again in position C, where the Kinetic Energy is zero. At position D, the energy of the bob is party potential and party Kinetic.
Principal of conservation of energy.
The law of conservation of energy state that “ Energy can neither be created nor destroyed but can only be converted from one form to another.”
This means the amount of energy is constant all the time.
This means the amount of energy is constant all the time.
Example 01
A stone of mass 2kg is released form a height of 2m above the ground. Find the potential energy of the stone when it is at the height of 0.5m above the ground.
Data given;
Mass of the stone (Ms) = 2kg
Height released (h) = 2m
Gravity (g) = 10N/kg
Potential energy = (P.E) ?
Data given;
Mass of the stone (Ms) = 2kg
Height released (h) = 2m
Gravity (g) = 10N/kg
Potential energy = (P.E) ?
Solution:
P.E at height g 2m
P.E = Mgh
= 21g x 10N/kg x 2m
P.E = 40 Joules
P.E at 2m = 40Joules
P.E at height g 2m
P.E = Mgh
= 21g x 10N/kg x 2m
P.E = 40 Joules
P.E at 2m = 40Joules
Than P.E at height of 0.5m
= 21g x 10N/kg x 0.5m
= 21g x 10N/kg x 0.5m
P.E at 0.5m 10Joules
100s of P.E = 40 Joules - 10Joules
= 30 Joules
= 30 Joules
According to conservation of energy the loss of P.E should be equal to the gain in K.E, when the air resistance is neglected.
K.E of the stone at 0.5 above the ground = 30 Joules
Example 02
A ball of mass 0.21kg is dropped from a height of 20m. on impact with the ground it loses 30J of energy. Calculate the height which it reaches on the rebound.
Data given;
Mass of ball (Mb) = 0.2kg
Height dropped (h) = 20m
Loos in energy (E) 30J
Height which reaches=?
Data given;
Mass of ball (Mb) = 0.2kg
Height dropped (h) = 20m
Loos in energy (E) 30J
Height which reaches=?
Solution;
Consider the figure below;
Consider the figure below;
At 20m above ground the initial energy of the ball = Mgh
= 0.2kg x 10N/kg x 20m
= 40 Joules.
So after the impact the ball loose 30J and the energy remaining is 40 J-10J
= 10Joules
At the top of rebound the energy of the ball = potential energy (P.E)
= 40 Joules.
So after the impact the ball loose 30J and the energy remaining is 40 J-10J
= 10Joules
At the top of rebound the energy of the ball = potential energy (P.E)
10J = Mgh
10J = 0.2kg x 10N/kgh
10J = 2No h
h = 10 Joules
2N
10J = 0.2kg x 10N/kgh
10J = 2No h
h = 10 Joules
2N
h=10 NM
2N
h = 5m
The height reaches (h) is 5m.
2N
h = 5m
The height reaches (h) is 5m.
POWER
Power is the rate of which work is done.
Power is the rate of which work is done.
- It is a measure of the rate at which energy changes.
- This means that whenever work is done energy changes into a different form.
Power = wok done (w)
Time (t)
Time (t)
P = w
t
The SI unit of power is Jules per second J/S or watts, W.
1 Joules per second = 1 watt
When 1 Joules of work is done per second the power produced is a watt. Watt is the unit for measuring electrical power.
t
The SI unit of power is Jules per second J/S or watts, W.
1 Joules per second = 1 watt
When 1 Joules of work is done per second the power produced is a watt. Watt is the unit for measuring electrical power.
RATE OF DOING WORK:
Suppose that two cranes each lift objects having masses of 200kg to a height of 12m. Crane A lifts its object in 10sec while crane B requires 15sec to lift its object. Assume they lift the objects at a constant velocity they do the same amount of work.
Work done = GPE
= Mgh
= (200kg) (9.8m/s2)(12m)
= 23520J
Each did a work that was equivalent to 23520J.
What is different for the two cranes is the rate at which they did the work or their generation of power.
The power of crane A can be calculated by;
= Mgh
= (200kg) (9.8m/s2)(12m)
= 23520J
Each did a work that was equivalent to 23520J.
What is different for the two cranes is the rate at which they did the work or their generation of power.
The power of crane A can be calculated by;
PA = w
t
PA = 23520J
10
PA = 2352watts
The power of crane B was;
t
PA = 23520J
10
PA = 2352watts
The power of crane B was;
PB = w
t
PB = 23520J
15sec
PB = 1568 watts.
Example:
How much power is required to accelerate a 1000kg car from rest to 26.7m/s in 8sec?
t
PB = 23520J
15sec
PB = 1568 watts.
Example:
How much power is required to accelerate a 1000kg car from rest to 26.7m/s in 8sec?
Solution:
The work done on the car increases its Kinetic energy.
Work done = AKE
½ MV2 – ½ MV2
= (1/2 X 1000kg) 26.7m/s) 2 – 0
= 3.56 * 105J
The work done on the car increases its Kinetic energy.
Work done = AKE
½ MV2 – ½ MV2
= (1/2 X 1000kg) 26.7m/s) 2 – 0
= 3.56 * 105J
the power required is given by;
P = w
T
P = 3.56 * 105J
8 sec
P = 4.45 x 104 watts
P = w
T
P = 3.56 * 105J
8 sec
P = 4.45 x 104 watts
Car engine is rated in horsepower (hp) where 1hp = 746watts. What is the required power measured in horsepower
P = 4.45 x 104watts * 1hp
746 watts
60 hp
Since work causes a change in energy. DE power can be considered as the rate of change of energy.
746 watts
60 hp
Since work causes a change in energy. DE power can be considered as the rate of change of energy.
P = DE
t
t
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