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Work and Energy Chapter IV Work and Energy The objective of this chapter is to introduce the energy tools used in mechanics to solve problems.Kinetic Energy Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 9 Work and Energy Chapter IV The Kinetic Energy Theorem Ek B - Ek A = ?WAB F ?ext Demonstration Consider a particle moving under the action of a resultant force F -> between A and B. The work done by F -> for an elemental displacement dr -> is : dW = F -> ?Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 18 Work and Energy Chapter IV Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 19 Work and Energy Chapter IV Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 20 Work and Energy Chapter IV Solution Calculation of the total mechanical energy at equilibrium: We know that : ET = Ek + Ep At equilibrium:

• the velocity of the system is zero V = 0 => Ek = 0
• and the potential energy of the system Ep = Ep(elastic) + Ep(gravitational) So: Ep = 1 2 kx 2 + mgh NA: Ep = 1 2 200(0.02) 2 + 0.1 x 10 x 0.5 = 1.54 J Hence, the total mechanical energy at equilibrium: ET = Ep = 1.Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 5 Work and Energy Chapter IV B- Work of a variable force When force is variable, the calculation of work for this variable force involves first defining the increment of work, denoted as dW, done by a force F ?No Work Done Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 4 Work and Energy Chapter IV Exercise 1 A body weighing 4 kg ascends a ramp inclined at 30o over a distance of 15 m. The driving force is F = 30 N. Calculate the work done by each force acting on this body.Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 15 Work and Energy Chapter IV Examples of potential energy a) Gravitational potential energy Ep = mgh + Cst If we set Ep =0 for h=0, then the constant (Cst) is equal to 0 Gravitational potential energy m h It is the energy that a body possesses due to its position in a gravitational field Work and Energy Chapter IV b- Elastic Potential Energy It is potential energy stored as a result of deformation of an elastic object such as spring.Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 12 Work and Energy Chapter IV Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 13 Work and Energy Chapter IV Exercise 4 : A particle of mass m moving along a straight trajectory is subjected to a force F(x), the variations of which with respect to x are depicted in figure below.Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 1 Work and Energy Chapter IV A- Work done by a constant force work done by a constant force F -> during a rectilinear displacement AB, is defined as the scalar product of the force F -> and the displacement AB. WAB(F ?0 5 10 15 20 25 30 0,0 0,5 1,0 x(m) F(N) Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 14 Work and Energy Chapter IV 4.Potential energy The kinetic energy Ek of a particle is associated with its motion.dr -> Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 6 Work and Energy Chapter IV Exercise 2: A body m is subjected to a force F -> , moving along the trajectory OABCO, as shown in figure opposite.Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 8 Work and Energy Chapter IV Kinetic energy Ek , is the energy possessed by an object due to its motion.= m dV dt Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 10 Work and Energy Chapter IV We substitute the expression for F -> into equation (): () => dW = m dV dt ?< 0 negative work Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 3 Work and Energy Chapter IV ?acting through an infinitesimal displacement dr ?.Introduction 2.1.?0 <= ?< ?2 ?> 0 ??2 < ?<= ?< 0 ??= 90????3.

Work and Energy Chapter IV
Work and Energy
The objective of this chapter is to introduce the energy tools used in mechanics to solve
problems. Indeed, sometimes the fundamental principle of dynamics is not sufficient or
not appropriate to reach a solution.

1. Introduction


2. Work Done by a Force
   Work done by a force measures the energy transfer when the force causes an object to
   move.
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 1
   Work and Energy Chapter IV
   A- Work done by a constant force
   work done by a constant force 𝐹
   →
   during a rectilinear displacement 𝐴𝐵, is defined as
   the scalar product of the force 𝐹
   →
   and the displacement 𝐴𝐵.
   𝑊𝐴𝐵(𝐹
   Ԧ
   ) = 𝐹
   Ԧ. 𝐴𝐵
    𝑊𝐴𝐵 𝐹
   Ԧ
   = 𝐹 𝐴𝐵 𝑐𝑜𝑠 𝛼
   Where :
   𝑊𝐴𝐵 𝐹
   Ԧ
   : is the work done in joules (J).
   𝐴𝐵 : is the displacement of the object in meters (m).
    : Angle between 𝐹
   →
   and 𝐴𝐵.
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 2
   Work and Energy Chapter IV
   b. Nature of work
    = 0
   90o
   ✓ Positive work
   𝐹
   →
   motive force  0 ≤ 𝛼 <
   𝜋
   2
    𝑐𝑜𝑠 𝛼 > 0
    𝑊 𝐹
   Ԧ
   = 𝐹 𝐴𝐵 𝑐𝑜𝑠 𝛼 > 0 positive work
   ✓ Negative work
   𝐹
   →
   resistive force 
   𝜋
   2
   < 𝛼 ≤ 𝜋 𝑐𝑜𝑠 𝛼 < 0
    𝑊 𝐹
   Ԧ
   = 𝐹 𝐴𝐵 𝑐𝑜𝑠 𝛼 < 0 negative work
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 3
   Work and Energy Chapter IV
    = 90o
   𝐹
   →
   force with zero net work  𝛼 = 90°
    𝑊 𝐹
   Ԧ
   = 𝑜 No work done
   ✓ No Work Done
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 4
   Work and Energy Chapter IV
   Exercise 1
   A body weighing 4 kg ascends a ramp inclined at 30o over a distance of 15 m.
   The driving force is F = 30 N.
   Calculate the work done by each force acting on this body. Given static and
   dynamic friction coefficients, respectively μs = 0.4 and μd = 0.2.
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 5
   Work and Energy Chapter IV
   B- Work of a variable force
   When force is variable, the calculation of work for this variable force involves first
   defining the increment of work, denoted as dW, done by a force 𝐹
   Ԧ
   acting through an
   infinitesimal displacement 𝑑𝑟
   Ԧ.
   Vectors used to define
   work.
   B
   A
   
   Path
   𝑑𝑊 = 𝐹
   Ԧ ⋅ 𝑑𝑟
   Ԧ
   = 𝐹 𝑑𝑟 cos 𝛼
   The total work done by the force during the displacement
   from point A to point B (refer to Figure opposite) is
   expressed as:
   𝑊𝐴𝐵 𝐹
   →
   = න
   𝐴
   𝐵
   𝐹
   →
   ⋅ 𝑑𝑟
   →
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 6
   Work and Energy Chapter IV
   Exercise 2:
   A body m is subjected to a force 𝐹
   →
   ,
   moving along the trajectory OABCO,
   as shown in figure opposite.
   Given that: 𝑭
   →
   = −𝒚𝒊
   Ԧ + 𝒙𝒋
   Ԧ
   ,
   Calculate the work done by this force for the body to move
   from O → A→B→ C→O.
   0,5 1,0 1,5 2,0
   0,0
   0,5
   1,0
   1,5
   2,0
   A
   B
   O
   y=x
   y=x
   2
   /2
   x(m)
   y(m)
   C
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 7
   Work and Energy Chapter IV
   Concept of Power: The power p of a force 𝐹
   →
   ​ represents the work done by this
   force per unit of time. In simple terms, power measures how quickly work is
   done.
   Average power
   𝑝 =
   𝑊(𝐹
   Ԧ
   )
   𝛥𝑡
   Instantaneous power :
   𝑝 =
   𝑑𝑊(𝐹
   Ԧ)
   𝑑𝑡
   =
   𝐹
   Ԧ
   𝑑𝑟
   Ԧ
   𝑑𝑡
   = 𝐹
   Ԧ
   ณ
   𝑑𝑟
   Ԧ
   𝑑𝑡
   𝑉
   ⇒ 𝑃 = 𝐹
   Ԧ. 𝑉
   The SI unit for power is the watt (W), equivalent to one joule per second.
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 8
   Work and Energy Chapter IV
   Kinetic energy Ek
   , is the energy possessed by an object due to its motion.
   Mathematically, it is expressed as:
   𝑬𝑲 =
   𝟏
   𝟐
   𝒎𝑽
   𝟐
   Where:
   Ek
   is the kinetic energy,
   m is the mass of the object,
   and V is its velocity.


3. Kinetic Energy
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 9
   Work and Energy Chapter IV
   The Kinetic Energy Theorem
   𝐸𝑘 𝐵 − 𝐸𝑘 𝐴 = ෍𝑊𝐴𝐵 𝐹
   Ԧ𝑒𝑥𝑡
   Demonstration
   Consider a particle moving under the action of a resultant force 𝐹
   →
   between A and B. The work done by 𝐹
   →
   for an elemental displacement 𝑑𝑟
   →
   is :
   𝑑𝑊 = 𝐹
   →
   ⋅ 𝑑​𝑟
   →
   (∗)
   We apply the second law of Newton : 𝐹
   Ԧ
   = 𝑚𝑎
   Ԧ 𝑎𝑛𝑑 𝑎
   Ԧ
   =
   𝑑𝑉
   𝑑𝑡
   ⇒ 𝐹
   Ԧ
   = 𝑚
   𝑑𝑉
   𝑑𝑡
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 10
   Work and Energy Chapter IV
   We substitute the expression for 𝐹
   →
   into equation (*):
   (∗) ⇒ 𝑑𝑊 = 𝑚
   𝑑𝑉
   𝑑𝑡 ⋅ 𝑑𝑟
   →
   𝑎𝑛𝑑 𝑉 =
   𝑑𝑟
   →
   𝑑𝑡 ⇒ 𝑑𝑟
   →
   = 𝑉𝑑𝑡
   𝑑𝑊 = 𝑚
   𝑑𝑉
   𝑑𝑡 𝑉𝑑𝑡 ⇒ 𝑑𝑊 = 𝑚𝑉𝑑𝑉
   The work done by 𝐹
   →
   from A to B :
   ��׬
   𝐵
   𝑑𝑊(𝐹
   Ԧ
   �𝑉�׬ = (
   𝑉𝐵 𝑚 𝑉
   ถ
   𝑑𝑉
   𝑉𝑑𝑉 𝑐𝑜𝑠 0
   �𝑉�׬ =
   𝑉𝐵 𝑚𝑉𝑑𝑉 = ⇒
   𝑊(𝐹
   Ԧ
   ) =
   1
   2
   𝑚𝑉
   2
   𝑉𝐴
   𝑉𝐵
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 11
   Work and Energy Chapter IV
   so : 𝑊𝐴𝐵(𝐹
   Ԧ
   ) =
   1
   2
   𝑚𝑉𝐵
   2
   𝐸𝑘(𝐵)
   −
   1
   2
   𝑚𝑉𝐴
   2
   𝐸𝑘(𝐴)
   Hence:
   𝑊𝐴𝐵(𝐹
   Ԧ
   ) = 𝐸𝑘(𝐵) − 𝐸𝑘(𝐴)
   it is he Kinetic Energy Theorem.
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 12
   Work and Energy Chapter IV
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 13
   Work and Energy Chapter IV
   Exercise 4 :
   A particle of mass m moving along a straight trajectory is subjected to a
   force F(x), the variations of which
   with respect to x are depicted
   in figure below.
   Calculate the change in kinetic energy
   of this particle between
   the positions x=0 and x=30 (m).
   0 5 10 15 20 25 30
   0,0
   0,5
   1,0
   x(m)
   F(N)
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 14
   Work and Energy Chapter IV
   4.Potential energy
   The kinetic energy Ek of a particle is associated with its motion. There is another
   form of energy that is associated with its position; this energy is called potential
   energy. It is energy stored by a body and can later be transformed, for example, into
   kinetic energy when the body is set in motion.
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 15
   Work and Energy Chapter IV
   Examples of potential energy
   a) Gravitational potential energy
   𝐸𝑝 = 𝑚𝑔ℎ + 𝐶𝑠𝑡
   If we set Ep
   ​=0 for h=0,
   then the constant (𝐶𝑠𝑡) is equal to 0
   Gravitational potential energy
   m
   h
   It is the energy that a body possesses due to its
   position in a gravitational field
   Work and Energy Chapter IV
   b- Elastic Potential Energy
   It is potential energy stored as a result of deformation of an elastic object such as
   spring.
   When a spring is compressed or stretched,
   it stores elastic potential energy.
   This energy is released when
   the spring returns to its original shape.
   𝐸𝑝 =
   1
   2
   𝑘𝑥
   2 𝑥 = 𝐿 − 𝐿0
   x: The elongation of the spring.
   k: is the stiffness constant.
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 17
   Work and Energy Chapter IV


4. Mechanical Energy of a System
   The mechanical energy of a system, denoted as ET
   ​, is the sum of its kinetic
   energy and potential energy.
   𝐸𝑇 = 𝐸𝑘 + 𝐸𝑝
   If the mechanical energy of a system increases or decreases, it means that it has
   received or released energy from the external environment. If there is no exchange of
   energy between the system and the external environment, the system is said to be
   isolated. In this case, its energy 𝐸𝑇 remains constant.
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 18
   Work and Energy Chapter IV
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 19
   Work and Energy Chapter IV
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 20
   Work and Energy Chapter IV
   Solution
   Calculation of the total mechanical energy at equilibrium:
   We know that :
   𝐸𝑇 = 𝐸𝑘 + 𝐸𝑝
   At equilibrium:

• the velocity of the system is zero 𝑉 = 0 ⇒ 𝐸𝑘 = 0


• and the potential energy of the system 𝐸𝑝 = 𝐸𝑝(𝑒𝑙𝑎𝑠𝑡𝑖𝑐) + 𝐸𝑝(gravitational)
  So: 𝐸𝑝 =
  1
  2
  𝑘𝑥
  2 + 𝑚𝑔ℎ
  𝑁𝐴: 𝐸𝑝 =
  1
  2
  200(0.02)
  2 + 0.1 × 10 × 0.5 = 1.54 𝐽
  Hence, the total mechanical energy at equilibrium: 𝑬𝑻 = 𝑬𝒑 = 𝟏. 𝟓𝟒 𝑱.
  Dr IACHACHENE FARIDA FHC-UMBB-2024-2025
  Work and Energy Chapter IV

5. Conservative force
   Potential energy exists exclusively for a specific category of forces
   known as conservative forces, or forces that can be derived from a
   potential function.
   Properties of a Conservative Force
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 22
   Work and Energy Chapter IV
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   Work and Energy Chapter IV
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 24
   Work and Energy Chapter IV
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   Work and Energy Chapter IV
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   Work and Energy Chapter IV
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   Work and Energy Chapter IV
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   Work and Energy Chapter IV
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   Work and Energy Chapter IV
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   Work and Energy Chapter IV
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   Work and Energy Chapter IV
   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025
   Work and Energy Chapter IV
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   Dr IACHACHENE FARIDA FHC-UMBB-2024-2025 34
   Work and Energy Chapter IV
   Work and Energy Chapter IV