Chapter 11 Work. Chapter Goal: To develop a deeper understanding of energy and its conservation Pearson Education, Inc.

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1 Chapter 11 Work Chapter Goal: To develop a deeper understanding of energy and its conservation.

2 Motivation * * There are also ways to gain or lose energy that are thermal, but we will not study these in Ch. 11. We will talk about Eth, but not its transfer 2013 Pearson Education, into or Inc. out of system.

3 Motivation

4 Basic energy model now includes Work * * I would frame this as "The lifter does work W ext in opposition to the conservative gravitational force, and this adds ΔU 2013 to Pearson the Education, system." Inc. The system is the weights.

5 A math technique we need: the Dot Product

6 Last but not least... the rate energy changes

7 Work and Kinetic Energy The word work has a very specific meaning in physics. Work is energy transferred to or from a body or system by the application of force. E.g. A pitcher is increasing the ball s kinetic energy by doing work on it.

8 The Basic Energy Model The energy of a system is a sum of its kinetic energy K, its potential energy U, and its thermal energy E th. The change in system energy is: Notation alert: Some places you will see W ext and some just W. Sigh. When you see this equ, always think to yourself "W comes from sources and forces external to the system whose energy is E sys ". 1. Energy can be transferred to or from a system by doing work W on the system. This process changes the energy of the system: ΔE sys = W. 2. Energy can be transformed within the system among K, U, and E th. These processes don t change the energy of the system: ΔE sys = 0.

9 Work and Kinetic Energy in One dimension Consider a force acting on a particle which moves along the s-axis. The force component F s causes the particle to speed up or slow down, transferring energy to or from the particle. The force does work on the particle equal to: The units of work are N m, where 1 N m = 1 kg m 2 /s 2 = 1 J.

10 What's wrong with this picture? (Nothing...) Consider a force acting on a particle which moves along the s-axis. Question: How can particle stay on the s-axis if the force vector, is not parallel to that axis? Answer: This particular force does work on the particle equal to:

11 The Work-Kinetic Energy Theorem (WET) Recall that the net force is the vector sum of all the forces acting on a particle. The net work is the sum W net = ΣW i, where W i is the work done by each force. IDEA: The net work done on a particle causes the particle s kinetic energy to change.

12 Analogy: WET and Impulse-Momentum Theorem The impulse-momentum theorem is: The work-kinetic energy theorem is: Impulse and work are both the area under a force graph. However, the horizontal axes are different! (Also, impulse is really a vector while work is a scalar.)

13

14 How much work is done by a constant force? The force does "positive work" The force does no work The force does "negative work"

15 QuickCheck 11.1 A constant force pushes a particle through a displacement. In which of these three cases does the force do negative work? A B C D Both A and B. E Both A and C.

16 QuickCheck 11.1 A constant force pushes a particle through a displacement. In which of these three cases does the force do negative work? A B C D Both A and B. E Both A and C.

17 Force Perpendicular to the Direction of Motion The figure shows a particle moving in uniform circular motion. At every point in the motion, F s, the component of the force parallel to the instantaneous displacement, is zero. The particle s speed, and hence its kinetic energy, doesn t change, so W = ΔK = 0. A force everywhere perpendicular to the motion does no work.

18 An easier case: Work Done by a Constant Force A force acts with a constant strength and in a constant direction as a particle moves along a straight line through a displacement. The work done by this force is: Here θ is the angle makes relative to.

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20 Work done by a variable force The force does work on the particle that moves along a path characterized by s is equal to: If the force F(s) is dependent on s, we evaluate the integral. We do this either geometrically, by finding the area under the curve, or by doing the integration symbolically.

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