Chapter 7: Work & Energy

1. Chapter 7: Work & Energy

2. THE COURSE THEME is NEWTON’S LAWS OF MOTION! • Chs. 4, 5 & 6: Motion analysis with forces. • NOW (Chs. 7 & 8): An alternative analysis using the concepts of Work & Energy. • Easier? My opinion is yes! • Conservation of Energy:NOT a new law! • We’ll see that this is just Newton’s Laws of Motion re-formulated or re-expressed (translated) from Force Language to Energy Language.

3. We’ve expressed Newton’s Laws of Motion using concepts of position, displacement, velocity, acceleration, force. • Newton’s Laws with Forces: Quite general (macroscopic objects).In principle, could be used to solve any dynamics problem, But, often, they are very difficult to apply, especially to very complicated systems. So, alternate formulations have been developed. Often easier to apply. • One of these is an approach that uses Energy instead of Forceas the most basic physical quantity. • Newton’s Laws in a different language (Energy). Before we discuss these, we need to learn vocabulary inEnergy Language. • Energy: A very common term in everyday usage. Everyday meanings might not coincide with the PHYSICSmeaning! Every physical process involves energy or energy transfer or transformations. Energy in physics can be somewhat abstract.

4. So far, we’ve expressedNewton’s Laws of Motionin terms of forces & we’ve considered the dynamical properties of a particle by talking about various particle properties. • Now, we’ll take a different approach & talk about Systems & System Properties • System: A small portion of the universe which we focus on in a given problem. What the system is depends on the problem. • A System may be, for example: • A single particle. • A collection of particles. • A region of space. • May vary in size & shape, depending on the problem • In addition to aSystem, we also talk about theSystem Environment. The System interacts with environment at it’s boundaries.

5. Sect. 7-1: Work Done by Constant Force • Work:Precisely defined in physics. Describes what is accomplished by a force in moving an object through a distance. For an object moving under the influence of a Constant Force F, the work done (W) the product of the magnitude of the displacement (d)  the component of force parallel to the displacement (F||). W  F||d = Fd cosθ d

6. Work Done by a Constant Force Work W  F||d = Fd cosθ NOTE!Valid for aCONSTANTforceONLY!

7. W = F||d = Fd cosθ • Consider a simple special case when F & d are parallel: θ = 0, cosθ = 1  W = Fd • Example: d = 50 m, F = 30 N W = (30N)(50m) = 1500 N m • Work Units:Newton - meter = Joule 1 N m = 1 Joule = 1 J

8. W = F||d = Fd cosθ • A force can be exerted & do no work! Could have d = 0  W = 0 Could have F  d  θ = 90º, cosθ = 0  W = 0 Example, walking at constant velocity v with a grocery bag:

9. A Typical Problem An object displaced by force F on a frictionless, horizontal surface. Free body diagram is shown  Normal force FN & weight mg do no work in the process, since both are perpendicular to the displacement. Angles for forces: Normal force: θ = 90°,cosθ = 0 Weight: θ = 270 (or - 90°), cosθ = 0 FN W = F||d = Fd cosθ d

10. W = F||d = Fd cosθ Note: • W is a scalar(in contrast to forces, which are vectors). • However, W can have either a positive or a negative sign, since cosθcan be positive or negative. • IMPORTANT: Work (as we’ll see) is a Transfer of Energy The System either gains energy (if W > 0) or loses energy (if W < 0)

11. Example 7-1 W = F||d = Fd cosθ m = 50 kg, FP = 100 N, Ffr = 50 N, θ = 37º Note: The old text had this friction force!! Your text has taken the friction away. Note:d is the same as x in this figure!

12. Example (from another text) A man cleaning a floor pulls a vacuum cleaner with a forceF = 50 Nat an angle θ = 30ºwith the horizontal. Calculate the work done by the forceFas the vacuum cleaner is moved a distanced = 3.0 mto the right. F = 50 N W = Fd cosθ = (50)(3)cos(30º) = 130 J FN θ = 30º d = 3.0 m

13. Solving Work Problems • Sketch a free-body diagram. • Choosea coordinate system. • ApplyNewton’s Laws to determine any unknown forces. • Find the work done by a specific force. • Findthe net work by either • a. Find the net force & then find the work it does, or • b. Find the work done by each force & add.

14. Example 7-2: Work on a backpack (a)Calculate the work a hiker must do on a backpack of mass m = 15 kg in order to carry it up a hill of height h = 10 m, as shown. (b)Calculate the work done by gravity on the backpack. (c) Calculate the net work done on the backpack. For simplicity, assume that the motion is smooth & at constant velocity (zero acceleration). For the hiker, ∑Fy = 0 = FH - mg  FH = mg WH = FHdcosθ = FHh

15. Conceptual Example 7-3: Does the Earth do work on the Moon? The Moon revolves around the Earth in a nearly circular orbit, with approximately constant tangential speed, kept there by the gravitational force exerted by the Earth. Does gravity do (a) positive work (b) negative work, or (c) no work at all on the Moon?

16. Sect. 7-2: Scalar Product of Two Vectors • Pure math discussion for a short time. • We now know that the work done by a constant force is: W = F||d = Fd cosθ • Note that both F & d are vectors, while W is a scalar. • W has the mathematical form that mathematicians call the Scalar Product. Scalar Product of Two Vectors If A & B are vectors, their Scalar Product (or dot product) is defined as: So, we can write:

17. Scalar Product of two vectors A & B: • So, the work done by a constant force, W = Fd cosθ, can be written as • Basic math properties of the Scalar Product: • Commutative: • Distributive: A(B + C) = AB + AC

18. Scalar Product of 2 Vectors Using Components Scalar Product: • From the discussion of vector components & unit vectors: A 3 dimensional vector V can be written as V = Vxi + Vyj + Vzk i,j,k are unit vectors along the x,y,zaxes,  Vx,Vy,Vzare the x,y,zcomponents of V Unit vectors i,j,k are dimensionless & have length 1 • From scalar product definition & because the angles between i,j,k are all 90[cos(90°) = 0, cos(0°) = 1] we can easily show: ii = jj = kk = 1 ij= ik = jk = 0

19. For two 3 d vectors A & B using this notation: A = Axi + Ayj + Azk, B = Bxi + Byj + Bzk • Consider the scalar product of A & B in this form: AB = (Axi + Ayj + Azk)(Bxi + Byj + Bzk) • Rewriting this (long form, 9 terms!): AB = (AxBx)(ii) + (AxBy)(ij) + (AxBz)(ik) + (AyBx)(ji) + (AyBy)(jj) + (AyBz)(jk) + (AzBx)(ki) + (AzBy)(kj) + (AzBz)(kk) • Using ii = jj = kk = 1, ij= ik = jk = 0 gives AB = AxBx+ AyBy+ AzBz(1) This form will be convenient sometimes! • Consider the case where A = B. Then, (1) becomes: AA = AxAx+ AyAy+ AzAzor AA = (Ax)2+ (Ay)2+ (Az)2(2) or AA = |A|2 = square of the length ofA  The scalar product of a vector with itself is the square of the length of that vector.

20. Example 7-4: Using the dot product The force shown has magnitude FP = 20 N & makes an angle θ = 30° to the ground. Calculate the work done by this force, using the dot product, when the wagon is dragged a displacement d = 100 m along the ground.