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Unformatted text preview: 10/6/09 Week 6: Work and Kinetic Energy The first midterm exam is THIS FRIDAY. It covers chapters 15 PDF slides will be posted 10/6/09 Why did we worry about the dot product? Definition: Work W F s ≡ ⋅ r r An object has a constant force acting on it. F r The object undergoes the displacement (no reference to any other movement details!) s r W F s ≡ ⋅ r r The work done is If : 10/6/09 W F s ≡ ⋅ r r Units: Newton meters 1 Joule = 1 Newton meter 10/6/09 The simplest case: θ = 10/6/09 Pick up something heavy mg r F r 1 s r The work done by is 1 W F s = ⋅ r r 1 W Fs = ( 29 1 cos 0 o W Fs = F r What’s the work done by ? mg r ( 29 1 cos 180 o W mgs = 1 W mgs =  Notice: if a = F mg = the n 10/6/09 Pick up something heavy mg r F r 1 s r What’s the work done by the net force ? net F r net F = r net W = 10/6/09 Now move it horizontally mg r F r 2 s r 2 W F s = ⋅ r r W = ( 29 2 cos 90 o W Fs = The work done by is F r What’s the work done by ? mg r ( 29 1 cos 90 o W mgs = W = 10/6/09 W F s ≡ ⋅ r r 10/6/09 The next standard example: something being dragged with a constant force m φ Example 6.2 10/6/09 Freebody diagram φ mg fr F N F x y F 20 m 14,700 N 5000 N 36.9 3500 N o fr s w mg F F φ = = = = = = s r Find work done by each force & total 10/6/09 Freebody diagram φ mg fr F N F x y F Consider each force in turn s r φ F r s r 1 W F s = ⋅ r r 1 cos W Fs φ = ( 29 ( 29 ( 29 1 5000 N 20 m cos 36.9 o W = 80,000 J ≈ 10/6/09 Freebody diagram φ mg fr F N F x y F Consider each force in turn s r s r 2 W mg s = ⋅ r r ( 29 2 cos 90 o W mgs = 2 0 J W = mg r 10/6/09 Freebody diagram φ mg fr F N F x y F Consider each force in turn s r s r 3 fr W F s = ⋅ r r ( 29 3 cos 180 o fr W F s = ( 29 ( 29 ( 29 3 3500 N 20 m 1 W = fr F r 70,000 J =  10/6/09 Freebody diagram φ mg fr F N F x y F Consider each force in turn s r s r 4 N W F s = ⋅ r r ( 29 4 cos 90 o N W F s = 4 0 J W = N F 10/6/09 So, the total work done is: 1 2 3 4 tot W W W W W = + + + 80,000 J + 0 J 70,000 J + 0 J tot W = 10,000 J tot W = Alternativel y, We can find the work done by the net force 10/6/09 Freebody diagram φ mg fr F N F x y F 20 m 14,700 N 5000 N 36.9 3500 N o fr s w mg F F φ = = = = = = x componen t , cos net x fr F F F φ =  ( 29 ( 29 , 5000 cos 36.9 3500 N o net x F N =  , 500 N net x F ≈ y componen t , 0 N net y F = 10/6/09 , 500 N net x F ≈ s r net F r ( 29 ( 29 ( 29 500 N 20 m cos 0 o tot net W F s = ⋅ = r r 10,000 J tot W = 10/6/09 So what?...
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This note was uploaded on 05/04/2009 for the course PHY 160 taught by Professor Bareto during the Spring '09 term at George Mason.
 Spring '09
 bareto
 Energy, Force, Kinetic Energy, Work

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