Mechanical Energy:
Mechanical energy is the energy of a moving object. Any object that is in motion process mechanical energy.
Work:
Work wasd into in ss 1 topics. I wish to review work in this part of the note because we shall make use of he formula of work to solve some problems in this part of the notes.
Definition of Work:
Work is done when a force move an object through a distance in the direction of the force. Work also defined as the product of the force that act on an object and the distance through which the object moved. Symbol of work is W.
Unit of Work:
The s.I. unit of work is Joule. The symbol of joule is J.
Formula for calculating work:
The formula for calculating work is as stated below.
Work = force * distance moved in the direction of force .
Work = force * distance.
Worked Examples:
1. Calculate the work done when a force of 4.5 N moves 23 kg mass through a distance of 2.4 m.
Solution:
Data given in the question:
Force = 4.5 N, mass = 23 kg, distance = 2.4 m.
Formula: work = force * distance
Substitution: work = 4.5 * 2.4..
➡ work = 10.8 Joules
2. If a certain force F moved an object through a distance of 2340cm, calculate the magnitude of the force if he works done is 34.0 J.
Solution:
Data given in the question:
Force = 34 J, distance = 2340 cm = 2340 /100 = 23.4 m
Formula: work = force * distance
Substitution: 34 = force * 23.4
Make force the subject: 34 / 23.4.
➡ force = 1.50 Joule
Etc.
Definition of Joule:
Joule is defined as the unit of work done when a force of one Newton moves an object through a distance of one meter.
Work Done By Incline Force:
An incline force is a force that act at an angle to the horizontal. Such a force has two directions or components, the vertical direction and the horizontal direction. The vertical direction of the force is the force that will lift the object up while the horizontal direction of the force is the force that will move the object along the horizontal surface as shown in the figure below.
From the left beside, force F act at an angle @ to the horizontal. The force has two directions, the vertical force Fy and the horizontal force Fx . the horizontal force Fx is the force that will move the object in the horizontal direction along the ground while the force Fy is the force that will lift the object upward in the vertical direction.
Formula for calculating the vertical and horizontal directions of a force acting at an angle:
From the figure by the right, we will derive the formula for calculating the directions of force F that act at an angle @ using the SHO CAH TOA concept thus:
Vertical direction of the force F:
SOH : sin @ = opposite / hypotenuse ➡ sin @ = Fy / F
➡The vertical force Fy = F * Sin@
Horizontal direction of the force F:
CAH : Cos @ = adjacent / hypotenuse ➡ Cos @ = Fx / F
➡The horizontal force Fx = F * Cos @
Now, we can derive the formula for calculating work done by an inclined force as follows:
Work Done by a vertical force in lifting the object:
Work done in the vertical direction
= vertical force * vertical distance ( or height )
Recall that vertical force Fy = F * Sin @. Let us substitute it in the formula above. The we will get,
Work done in vertical direction by Fy
= Fy * vertical distance
= F * Sin @ * vertical distance ( or height )
Also,
Work done by horizontal force Fx:
= Fx * horizontal distance
Recall at Fx = F * Sin @. Let us substitute for Fx in the above formula. Then,
Work done by horizontal force Fx:
= Fx * horizontal distance
= F * Cos @ * horizontal distance
Worked Examples:
*From the figure, if the 240 joules of work is done in raising the tip of the block of mass 0.54 kg to a height of 0.5 m and the force F is applied at angle 45° to the vertical, calculate the :
I. Magnitude of the force F
II. The force that is acting in the horizontal direction
III. Work done in pulling the block along the ground through 2.5 m.
Solution:
Data given in the question:
Mass = 0.54 kg, work done in vertical direction = 240J, height = 0.5 m, angle o the vertical = 45°,
Since we are given the value of the work done in lifting the object vertically, we will use the formula for calculating work done in the vertical direction to calculate the magnitude of F. Therefore,
I. Magnitude of F:
Work done in vertical direction = vertical force * height
= Fy * height
Substitution: 245 = Fy * 0.5
Make Fy the subject: Fy = 245 / 0.5.
➡Vertical force Fy = 490 Newtons.
Now, we will use the formula for calculating the vertical force Fy to calculate the force F as follows:
Formulà: Vertical force Fy = F * Sin @
Substitution: 490 = F * Sin 45°. ➡ 490 = F * 0.7071
Make F the subject: F = 490 / 0.7071. ➡. F = 692.97 N
II. The force that act in the horizontal direction:
We can use Pythagoras formula to calculate the horizontal force Fx as follows:
From figure 2 above, and from our calculation, Fy = 490N, F = 692.97N
Formula: F² = (Fy)² + (Fx)²
Substitution: 692.97² = 490² + (Fx)²
480138.14 = 240100 + (Fx)²
Make Fx the subject of the formula: (Fx)² = 480138.14 – 240100
. (Fx)² = 240038.14. ➡ Fx = √240038.14 ➡Fx = 489.94 N
Or
We can also use the conventional method to calculate Fx as follows:
Formula : horizontal force Fx = F * Cos @
Substitution: horizontal force Fx = 692.79 * Cos 45°
. Fx = 692.79 * 0.7071➡. Fx = 490 N
III. Work done in pulling the block through 2.5 m
Formula: Work done = force * distance
Work done = Fx * distance
Substitution: work done = 490 * 2.5
➡ work done = 1225 Joules
Or
We can bring in the formula of Fx, which is Fx = F * Cos @ into the formula of work done in the horizontal direction.
Therefore, we will get:
Work done in horizontal direction = Fx * distance
Work done in horizontal direction = F * Cos @ * distance
Substitution: work done = 692.97 * Cos 45° * 2.5
. work done = 692.97 * 0.7071 * 2.5.
➡ work done = 1224.998 Joules
*A rope tensioned by 120 N is tied to a toy car at an angle of 30° to the horizontal. Calculate with the aid of diagram:
I. The effective force that act in the vertical direction.
II. The effective force that act in the horizontal direction.
III. The work one in lifting the toy head 0.25 m high.
IV. The work done in pulling the toy through 600 cm.
Solution:
Data given in the question:
Angle = 30°, force F = 120 N, height = 0.25m, distance = 600 cm = 600 / 100 = 6m.
From figure 2,
i. The effective force that act in the vertical direction:
Using SOH CAH TOA,
Sin @ = opposite / hypotenuse. ➡ Sin @ = Fy / F
Make Fy the subject: vertical force Fy = F * Sin @
Substitution: Fy = 120 * Sin 30° ➡ Fy = 120 * 0.5
Vertical force Fy = 60 N
ii. The effective force that in the horizontal direction Fx:
Also, Using SOH CAH TOA, we will get
Cos @ = adjacent /hypotenuse. ➡ Cos @ = Fx / F
Make Fx the subject: horizontal force Fx = F * Cos @
Substitution: Fx = 120 * Cos 30° ➡ Fx = 120 * 0.8660
Horizontal force Fx = 103.92 Newtons
iii. The work done in lofting the toy head:
Work done in vertical direction = vertical force * height
Work done in vertical direction = Fy * height
Substitution: work done = 60 * 6
➡ work done = 360 Joules
iv. The work done in pulling the toy car:
Formula:
Work done in horizontal direction = horizontal force * distance
Work done = Fx * distance
Substitution: work done = 103.92 * 6
➡ Work done = 623.52 Joules
Mechanical energy is the energy of a moving object. Any object that is in motion process mechanical energy.
Work:
Work wasd into in ss 1 topics. I wish to review work in this part of the note because we shall make use of he formula of work to solve some problems in this part of the notes.
Definition of Work:
Work is done when a force move an object through a distance in the direction of the force. Work also defined as the product of the force that act on an object and the distance through which the object moved. Symbol of work is W.
Unit of Work:
The s.I. unit of work is Joule. The symbol of joule is J.
Formula for calculating work:
The formula for calculating work is as stated below.
Work = force * distance moved in the direction of force .
Work = force * distance.
Worked Examples:
1. Calculate the work done when a force of 4.5 N moves 23 kg mass through a distance of 2.4 m.
Solution:
Data given in the question:
Force = 4.5 N, mass = 23 kg, distance = 2.4 m.
Formula: work = force * distance
Substitution: work = 4.5 * 2.4..
➡ work = 10.8 Joules
2. If a certain force F moved an object through a distance of 2340cm, calculate the magnitude of the force if he works done is 34.0 J.
Solution:
Data given in the question:
Force = 34 J, distance = 2340 cm = 2340 /100 = 23.4 m
Formula: work = force * distance
Substitution: 34 = force * 23.4
Make force the subject: 34 / 23.4.
➡ force = 1.50 Joule
Etc.
Definition of Joule:
Joule is defined as the unit of work done when a force of one Newton moves an object through a distance of one meter.
Work Done By Incline Force:
An incline force is a force that act at an angle to the horizontal. Such a force has two directions or components, the vertical direction and the horizontal direction. The vertical direction of the force is the force that will lift the object up while the horizontal direction of the force is the force that will move the object along the horizontal surface as shown in the figure below.
From the left beside, force F act at an angle @ to the horizontal. The force has two directions, the vertical force Fy and the horizontal force Fx . the horizontal force Fx is the force that will move the object in the horizontal direction along the ground while the force Fy is the force that will lift the object upward in the vertical direction.
Formula for calculating the vertical and horizontal directions of a force acting at an angle:
From the figure by the right, we will derive the formula for calculating the directions of force F that act at an angle @ using the SHO CAH TOA concept thus:
Vertical direction of the force F:
SOH : sin @ = opposite / hypotenuse ➡ sin @ = Fy / F
➡The vertical force Fy = F * Sin@
Horizontal direction of the force F:
CAH : Cos @ = adjacent / hypotenuse ➡ Cos @ = Fx / F
➡The horizontal force Fx = F * Cos @
Now, we can derive the formula for calculating work done by an inclined force as follows:
Work Done by a vertical force in lifting the object:
Work done in the vertical direction
= vertical force * vertical distance ( or height )
Recall that vertical force Fy = F * Sin @. Let us substitute it in the formula above. The we will get,
Work done in vertical direction by Fy
= Fy * vertical distance
= F * Sin @ * vertical distance ( or height )
Also,
Work done by horizontal force Fx:
= Fx * horizontal distance
Recall at Fx = F * Sin @. Let us substitute for Fx in the above formula. Then,
Work done by horizontal force Fx:
= Fx * horizontal distance
= F * Cos @ * horizontal distance
Worked Examples:
*From the figure, if the 240 joules of work is done in raising the tip of the block of mass 0.54 kg to a height of 0.5 m and the force F is applied at angle 45° to the vertical, calculate the :
I. Magnitude of the force F
II. The force that is acting in the horizontal direction
III. Work done in pulling the block along the ground through 2.5 m.
Solution:
Data given in the question:
Mass = 0.54 kg, work done in vertical direction = 240J, height = 0.5 m, angle o the vertical = 45°,
Since we are given the value of the work done in lifting the object vertically, we will use the formula for calculating work done in the vertical direction to calculate the magnitude of F. Therefore,
I. Magnitude of F:
Work done in vertical direction = vertical force * height
= Fy * height
Substitution: 245 = Fy * 0.5
Make Fy the subject: Fy = 245 / 0.5.
➡Vertical force Fy = 490 Newtons.
Now, we will use the formula for calculating the vertical force Fy to calculate the force F as follows:
Formulà: Vertical force Fy = F * Sin @
Substitution: 490 = F * Sin 45°. ➡ 490 = F * 0.7071
Make F the subject: F = 490 / 0.7071. ➡. F = 692.97 N
II. The force that act in the horizontal direction:
We can use Pythagoras formula to calculate the horizontal force Fx as follows:
From figure 2 above, and from our calculation, Fy = 490N, F = 692.97N
Formula: F² = (Fy)² + (Fx)²
Substitution: 692.97² = 490² + (Fx)²
480138.14 = 240100 + (Fx)²
Make Fx the subject of the formula: (Fx)² = 480138.14 – 240100
. (Fx)² = 240038.14. ➡ Fx = √240038.14 ➡Fx = 489.94 N
Or
We can also use the conventional method to calculate Fx as follows:
Formula : horizontal force Fx = F * Cos @
Substitution: horizontal force Fx = 692.79 * Cos 45°
. Fx = 692.79 * 0.7071➡. Fx = 490 N
III. Work done in pulling the block through 2.5 m
Formula: Work done = force * distance
Work done = Fx * distance
Substitution: work done = 490 * 2.5
➡ work done = 1225 Joules
Or
We can bring in the formula of Fx, which is Fx = F * Cos @ into the formula of work done in the horizontal direction.
Therefore, we will get:
Work done in horizontal direction = Fx * distance
Work done in horizontal direction = F * Cos @ * distance
Substitution: work done = 692.97 * Cos 45° * 2.5
. work done = 692.97 * 0.7071 * 2.5.
➡ work done = 1224.998 Joules
*A rope tensioned by 120 N is tied to a toy car at an angle of 30° to the horizontal. Calculate with the aid of diagram:
I. The effective force that act in the vertical direction.
II. The effective force that act in the horizontal direction.
III. The work one in lifting the toy head 0.25 m high.
IV. The work done in pulling the toy through 600 cm.
Solution:
Data given in the question:
Angle = 30°, force F = 120 N, height = 0.25m, distance = 600 cm = 600 / 100 = 6m.
From figure 2,
i. The effective force that act in the vertical direction:
Using SOH CAH TOA,
Sin @ = opposite / hypotenuse. ➡ Sin @ = Fy / F
Make Fy the subject: vertical force Fy = F * Sin @
Substitution: Fy = 120 * Sin 30° ➡ Fy = 120 * 0.5
Vertical force Fy = 60 N
ii. The effective force that in the horizontal direction Fx:
Also, Using SOH CAH TOA, we will get
Cos @ = adjacent /hypotenuse. ➡ Cos @ = Fx / F
Make Fx the subject: horizontal force Fx = F * Cos @
Substitution: Fx = 120 * Cos 30° ➡ Fx = 120 * 0.8660
Horizontal force Fx = 103.92 Newtons
iii. The work done in lofting the toy head:
Work done in vertical direction = vertical force * height
Work done in vertical direction = Fy * height
Substitution: work done = 60 * 6
➡ work done = 360 Joules
iv. The work done in pulling the toy car:
Formula:
Work done in horizontal direction = horizontal force * distance
Work done = Fx * distance
Substitution: work done = 103.92 * 6
➡ Work done = 623.52 Joules
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