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Unformatted text preview: θ F θ F d Chapter 4 Work and Energy 4.1 Work and Kinetic Energy 4.2 Workenergy Theorem 4.3 Potential Energy 4.4 Total Mechanical Energy 4.5 Energy in a Block Attached to a Spring 4.6 Potential Energy Curve 4.7 Law of Conservation of Energy 4.1 Work and kinetic energy (1) Work Work done by the force F is d F d F d F W ⋅ = = = ) cos ( // θ where F // is the force along the displacement d . In other words, the work done is the dot product of the force and the displacement which is a scalar quantity. SI unit of work: Joule (J) = N ⋅ m Remarks: 1. The dot product of two vectors A and B is: z z y y x x B A B A B A AB B A + + = = ⋅ θ cos Note that 1 ˆ ˆ ˆ ˆ ˆ ˆ = ⋅ = ⋅ = ⋅ k k j j i i and ˆ ˆ ˆ ˆ ˆ ˆ = ⋅ = ⋅ = ⋅ i k k j j i . 2. Positive work is done on an object when the point of application of the force moves along the direction of the force. In contrast, negative work is done on an object when the point of application of the force moves opposite to the direction of the force. 3. If F or d is not constant, then the work done by the force is given by: ∫ ∑ ∑ ⋅ ≡ ∆ ⋅ = ∆ = C i i i i i s d F s F W W where i s ∆ are some extremely small segments that made up the path C taken by the object upon the application of the force. 1 F d y x C (0, 0) mg h ( h , h ) For example, the work done by the gravity when the ball drops in the hemisphere (see the figure) is given by: mgh dy mg j dy i dx j g m W h C = = + ⋅ = ∫ ∫ ) ˆ ˆ ( ) ˆ ( (2) Power Power v F dt s d F dt dW t W P t ⋅ = ⋅ = = ∆ ∆ = → ∆ lim It is the rate at which work is done. If the force and the velocity are in the same direction, P = Fv. SI unit of power: Watt (W) = J/s (3) Kinetic energy Kinetic energy 2 2 1 mv K = It is the energy of motion. SI unit of kinetic energy: Joule (J) = N ⋅ m Example 41 A 75.0kg person slides a distance of 5.00 m on a straight water slide, dropping through a vertical height of 2.50 m. How much work does gravity do on the person? Answer: The gravity along the displacement has a magnitude mg cos θ . By the definition of work done, the work done of the gravity on the person is given by: W = ( mg cos θ ) d = mg ( h / d ) d = mgh = (75.0)(9.8)(2.50) = 1837.5 J h = 2.5m d = 5.00m mg θ θ 2 f k N F 0.902 m 1.62 m θ mg Example 42 A 47.2kg block of ice slides down an incline 1.62 m long and 0.902 m high. When the ice slides down the incline, a worker pushes up on the ice parallel to the incline so that it slides down at a constant speed. Given that the coefficient of kinetic friction between the ice and the incline is 0.110. Find (a) the force exerted by the worker, (b) the work done by gravity on the ice, and (c) the work done by the worker on the ice block....
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This note was uploaded on 02/22/2012 for the course CHEM yscn0027 taught by Professor Drtong during the Fall '10 term at HKU.
 Fall '10
 drtong

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