# v1chap6 - Chapter 6 Work Kinetic Energy and Potential...

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Chapter 6 Work, Kinetic Energy and Potential Energy 6.1 The Important Stuf 6.1.1 Kinetic Energy For an object with mass m and speed v , the kinetic energy is defned as K = 1 2 mv 2 (6.1) Kinetic energy is a scalar (it has magnitude but no direction); it is always a positive number; and it has SI units o± kg · m 2 / s 2 . This new combination o± the basic SI units is known as the joule : 1joule = 1J = 1 kg · m 2 s 2 (6.2) As we will see, the joule is also the unit o± work W and potential energy U . Other energy units o±ten seen are: 1erg = 1 g · cm 2 s 2 = 10 - 7 J 1eV = 1 . 60 × 10 - 19 J 6.1.2 Work When an object moves while a ±orce is being exerted on it, then work is being done on the object by the ±orce. I± an object moves through a displacement d while a constant ±orce F is acting on it, the ±orce does an amount o± work equal to W = F · d = Fd cos φ (6.3) where φ is the angle between d and F . Work is also a scalar and has units o± 1N · m. But we can see that this is the same as the joule, defned in Eq. 6.2. 127

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128 CHAPTER 6. WORK, KINETIC ENERGY AND POTENTIAL ENERGY Work can be negative; this happens when the angle between force and displacement is larger than 90 . It can also be zero ; this happens if φ = 90 . To do work, the force must have a component along (or opposite to) the direction of the motion. If several diFerent (constant) forces act on a mass while it moves though a displacement d , then we can talk about the net work done by the forces, W net = F 1 · d + F 1 · d + F 1 · d + ... (6.4) = p s F P · d (6.5) = F net · d (6.6) If the force which acts on the object is not constant while the object moves then we must perform an integral (a sum) to ±nd the work done. Suppose the object moves along a straight line (say, along the x axis, from x i to x f ) while a force whose x component is F x ( x ) acts on it. (That is, we know the force F x as a function of x .) Then the work done is W = i x f x i F x ( x ) dx (6.7) ²inally, we can give the most general expression for the work done by a force. If an object moves from r i = x i i + y i j + z i k to r f = x f i + y f j + z f k while a force F ( r ) acts on it the work done is: W = i x f x i F x ( r ) dx + i y f y i F y ( r ) dy + i z f z i F z ( r ) dz (6.8) where the integrals are calculated along the path of the object’s motion. This expression can be abbreviated as W = i r f r i F · d r . (6.9) This is rather abstract! But most of the problems where we need to calculate the work done by a force will just involve Eqs. 6.3 or 6.7 We’re familiar with the force of gravity; gravity does work on objects which move ver- tically. One can show that if the height of an object has changed by an amount Δ y then gravity has done an amount of work equal to W grav = - mg Δ y (6.10) regardless of the horizontal displacement. Note the minus sign here; if the object increases in height it has moved oppositely to the force of gravity. 6.1.3 Spring Force
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v1chap6 - Chapter 6 Work Kinetic Energy and Potential...

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