ch06 - Chapter 6 Work and Energy 6.1 Work Done by a...

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Chapter 6 Work and Energy
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6.1 Work Done by a Constant Force Fs W = ( 29 J joule 1 m N 1 =
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6.1 Work Done by a Constant Force
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6.1 Work Done by a Constant Force ( 29 s F W θ cos = 1 180 cos 0 90 cos 1 0 cos - = = =
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6.1 Work Done by a Constant Force Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m. ( 29 ( 29 [ ] ( 29 J 2170 m 0 . 75 0 . 50 cos N 0 . 45 cos = = = s F W θ
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6.1 Work Done by a Constant Force ( 29 Fs s F W = = 0 cos ( 29 Fs s F W - = = 180 cos
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6.1 Work Done by a Constant Force Example 3 Accelerating a Crate The truck is accelerating at a rate of +1.50 m/s 2 . The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it?
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6.1 Work Done by a Constant Force The angle between the displacement and the normal force is 90 degrees. The angle between the displacement and the weight is also 90 degrees. ( 29 0 90 cos = = s F W
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6.1 Work Done by a Constant Force The angle between the displacement and the friction force is 0 degrees. ( 29 [ ] ( 29 J 10 2 . 1 m 65 0 cos N 180 4 × = = W ( 29 ( 29 N 180 s m 5 . 1 kg 120 2 = = = ma f s
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6.2 The Work-Energy Theorem and Kinetic Energy Consider a constant net external force acting on an object. The object is displaced a distance s , in the same direction as the net force. The work is simply s F ( 29 ( 29 s ma s F W = =
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6.2 The Work-Energy Theorem and Kinetic Energy ( 29 ( 29 2 2 1 2 2 1 2 2 2 1 o f o f mv mv v v m as m W - = - = = ( 29 ax v v o f 2 2 2 + = ( 29 ( 29 2 2 2 1 o f v v ax - = DEFINITION OF KINETIC ENERGY The kinetic energy KE of and object with mass m and speed v is given by 2 2 1 KE mv =
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6.2 The Work-Energy Theorem and Kinetic Energy THE WORK-ENERGY THEOREM When a net external force does work on and object, the kinetic energy of the object changes according to 2 2 1 2 f 2 1 o f KE KE o mv mv W - = - =
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6.2 The Work-Energy Theorem and Kinetic Energy Example 4 Deep Space 1 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If the 56.0-mN force acts on the probe through a displacement of 2.42×10 9 m, what is its final speed?
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6.2 The Work-Energy Theorem and Kinetic Energy 2 2 1 2 f 2 1 W o mv mv - = ( 29 [ ] s F θ cos W =
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6.2
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ch06 - Chapter 6 Work and Energy 6.1 Work Done by a...

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