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Unformatted text preview: Chapter 8 jCHAPTER 8  Conservation of Energy 1. The potential energy of the spring is zero when the spring is not compressed ( x = 0). For the stored potential energy, we have U = ! kx f 2 – 0; 35.0 J = ! (82.0 N/m) x f 2 – 0, which gives x f = 0.924 m . 2. For the potential energy change we have ? U = mg ? y = (5.0 kg)(9.80 m/s 2 )(1.5 m) = 74 J . 3. For the potential energy change we have ? U = mg ? y = (58 kg)(9.80 m/s 2 )(3.8 m) = 2.2 × 10 3 J . 4. ( a ) With the reference level at the ground, for the potential energy change we have ? U = mg ? y = (66.5 kg)(9.80 m/s 2 )(2660 m – 1500 m) = 7.56 × 10 5 J . ( b ) The minimum work would be equal to the change in potential energy: W min = ? U = 7.56 × 10 5 J . ( c ) Yes , the actual work will be more than this. There will be additional work required for any kinetic energy change, and to overcome retarding forces, such as air resistance and ground deformation. 5. ( a ) With the reference level at the ground, for the potential energy we have U a = mgy a = (2.20 kg)(9.80 m/s 2 )(2.40 m) = 51.7 J . ( b ) With the reference level at the top of the head, for the potential energy we have U b = mg ( y b – h )= (2.20 kg)(9.80 m/s 2 )(2.40 m – 1.70 m) = 15.1 J . ( c ) Because the person lifted the book from the reference level in part ( a ), the potential energy is equal to the work done: 51.7 J . In part ( b ) the initial potential energy was negative, so the final potential energy is not the work done, which was still 51.7 J. 6. For the potential energy U = 3 x 2 + 2 xy + 4 y 2 z , we find the components of the force from F x = – ? U /? x = – 6 x – 2 y – 0 = – (6 x + 2 y ); F y = – ? U /? y = – 0 – 2 x – 8 yz = – (2 x + 8 yz ); F z = – ? U /? z = – 0 – 0 – 4 y 2 = – (4 y 2 ). Thus the force is F = – (6 x + 2 y ) i – (2 x + 8 yz ) j – (4 y 2 ) k . 7. ( a ) Because the force F = (– kx + ax 3 + bx 4 ) i is a function only of position, it is conservative . ( b ) We find the form of the potential energy function from U = – ? F ∙ d r = – ? (– kx + ax 3 + bx 4 ) i ∙ (d x i + d y j + d z k ) = – ? (– kx + ax 3 + bx 4 )d x = ! kx 2 – # ax 4 – $ bx 5 + constant ....
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This note was uploaded on 10/19/2009 for the course PHYS 1A taught by Professor Musumeci during the Spring '08 term at UCLA.
 Spring '08
 Musumeci
 Energy, Potential Energy

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