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بسم الله الرحمن الرحيم Dr- Sonia Reda.

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Presentation on theme: "بسم الله الرحمن الرحيم Dr- Sonia Reda."— Presentation transcript:

1 بسم الله الرحمن الرحيم Dr- Sonia Reda

2 Kinetic energy and Work
chapter 7 Kinetic energy and Work By Dr\ Sonia Reda Dr- Sonia Reda

3 chapter 7 Kinetic energy and Work
What is energy 7.3 Kinetic energy 7.4 Work 7.5 Work and kinetic Energy 7.6 Work done by the gravitational force 7.7 Work done by a Spring force Power

4 Outline Chapter 7 Work and Kinetic energy
Work done by a net force results in kinetic energy Some examples: gravity, spring, friction

5 However, this definition is too vague to be of help to us now.
What is Energy?   The term energy is so broad that a clear definition is difficult to write. Technically, Energy is a scalar quantity associated with the state (or condition) of one or more objects. However, this definition is too vague to be of help to us now. Dr- Sonia Reda

6 Kinetic Energy Kinetic energy K is energy associated with the state of motion of an object. For an object of mass m whose speed v is well below the speed of light, Kinetic energy K is: Unit for Kinetic energy is: Kinetic energy is a scalar quantity. Dr- Sonia Reda

7 Work Work W is energy transferred to or from an object by means of a force acting on the object. Energy transferred to the object is positive work, Energy transferred from the object is negative work. Dr- Sonia Reda

8 Properties of Work Only the force component along the object’s displacement will contribute to work. The force component perpendicular to the displacement does zero work. A force does positive work when it has a vector component in the same direction displacement, A force does negative work when it has a vector component in the opposite direction. Work is a scalar quantity. Dr- Sonia Reda

9 Finding an Expression for Work
                                                                                                                                     we can use Eq to write, for components along the x axis, v2 =vo2 + 2axd By multiplying the above Eq with ½ m Dr- Sonia Reda

10 Finding an Expression for Work
                                                                                                                                     Dr- Sonia Reda

11 Work-Kinetic Energy Theorem
Change in KE work done by all forces Chap 7.3 DK  Dw Dr- Sonia Reda

12 Vector sum of all forces acting on the body
Work-Kinetic Energy Theorem SF x Vector sum of all forces acting on the body xf xi Work and Energy 7.3 Work and Kinetic energy Note that in the slide F is the net force acting on the body. = 1/2mvf2 – 1/2mvi2 = Kf - Ki = DK Work done by net force = change in KE Dr- Sonia Reda

13 Checkpoint 1 A particle moves along an x axis. Does the kinetic energy of the particle increase, decrease, or remain the same if the particle’s velocity changes (a) from −3 m/s to −2 m/s and (b) from −2 m/s to 2 m/s? (c) In each situation, is the work done on the particle positive, negative, or zero? Dr- Sonia Reda

14 Example 7-3 During a storm, a crate of crepe is sliding across a slick, oily parking lot through a displacement while a steady wind pushes against the crate with a force The situation and coordinate axes are shown in Fig. 7-5. (a) How much work does this force do on the crate during the displacement?                            . Dr- Sonia Reda

15 Work done by the wind force on crate :
(a)  How much work does this force from the wind do on the crate during the displacement? SOLUTION:  Work done by the wind force on crate : The wind force does negative work, i.e. kinetic energy is taken out of the crate. Dr- Sonia Reda

16 (b)  If the crate has a kinetic energy of 10 J at the beginning of displacement , what is its kinetic energy at the end of ? SOLUTION:  Dr- Sonia Reda

17 Work done by ALL forces = change in KE
Gravitation and work Work done by me (take down as +ve) = F.(-h) = -mg(-h) = mgh h Work done by gravity = mg.(-h) = -mgh F ________ Total work by ALL forces (W) = mg =DK Lift mass m with constant velocity This example resolves the situation when there are 2 forces, which are equal and opposite. The change in KE is zero. Work done by ALL forces = change in KE DW = DK What happens if I let go? Dr- Sonia Reda

18 Work Done by a Spring Force
The spring force given by Hooke’s Law: The work done by spring force: Dr- Sonia Reda

19 Compressing a spring =DK What happens if I let go?
Compress a spring by an amount x F -kx x Work done by me Fdx = kxdx = 1/2kx2 Work done by spring -kxdx =-1/2kx2 Here also the two forces are equal and opposite Total work done (DW) = =DK What happens if I let go? Dr- Sonia Reda

20 Moving a block against friction at constant velocity
d Work done by me = F.d Work done by friction = -f.d = -F.d Total work done = Here again the two forces are equal and opposite The force here (friction) is different from the gravity force or the spring force. Friction is NOT a conservative force What happens if I let go? NOTHING!! Gravity and spring forces are Conservative Friction is NOT!! Dr- Sonia Reda

21 Sample Problem 7-8 In Fig. 7-11, a cumin canister of mass m = 0.40 kg slides across a horizontal frictionless counter with speed v = 0.50 m/s. It then runs into and compresses a spring of spring constant k = 750 N/m. When the canister is momentarily stopped by the spring, by what distance d is the spring compressed? Dr- Sonia Reda

22 SOLUTION:  We assume the spring is massless. Work done by the spring on the canister is negative. This work is : Kinetic energy change of the canister is : Therefore, Dr- Sonia Reda

23 Power The rate at which work is done by a force is called the power.
The average power due to the work done by a force during that time interval as We define the instantaneous power P as the instantaneous rate of doing work, so that W = F . Δx Dr- Sonia Reda

24 The units of power Dr- Sonia Reda

25 Sample Problem 7-10 Figure 7-14 shows constant forces and acting on a box as the box slides rightward across a frictionless floor. Force is horizontal, with magnitude 2.0 N; force is angled upward by 60° to the floor and has magnitude 4.0 N. The speed v of the box at a certain instant is 3.0 m/s. Dr- Sonia Reda

26 (a)  What is the power due to each force acting on the box at that instant, and what is the net power? Is the net power changing at that instant? SOLUTION:  The kinetic energy of the box is not changing. The speed of the box remains at 3 m/s. The net power does not change. Dr- Sonia Reda

27 (b) If the magnitude of is, instead, 6
(b)  If the magnitude of is, instead, 6.0 N, what now is the net power, and is it changing? SOLUTION:  There is a net rate of transfer of energy to the box. The kinetic energy of the box increases. The net power also increases. Dr- Sonia Reda

28 THANK شكـراً YOU Dr- Sonia Reda


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