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ps6 - P207 Fall 2007 Solutions Assignment 6 1 Chapter 7...

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P207 - Fall 2007 Solutions Assignment 6 1. Chapter 7, Problem 13 (a) The work done by the forces is Δ x F x + Δ y F y and we know that Δ y = 0 so the forces in the y-direction make no contribution. In the x-direction summationdisplay F x = | vector F 2 | cos θ − | vector F 1 | = (9 N) cos 60 5 N = 0 . 5 N Then W = Δ x summationdisplay F x = ( 3 m)( 0 . 5 N) = 1 . 5 J (b) The work on the trunk is translated into kinetic energy. Δ KE = W = 1 . 5J 2. (a) 0 5 10 15 20 25 30 velocity [m/s] 0 1 2 3 4 5 6 force [kN] 0 50 100 150 200 8 7 6 5 4 3 2 1 0 power [kW] time [seconds] (b) The car is accelerating due to the force of static friction that the road is exerting on the tires. F = ma = (1500 kg)(30 m / s) / (8 s) = 5625 N. (c) The acceleration a = v/t = (30 m / s) / (8 s) = 3 . 75 m / s 2 . Since the acceleration is constant, the average velocity over an interval Δ t is given by the average of initial and final velocities for the interval. Initial and final velocities are v i = at i and v f = at f . Then ( v ) = a t f + t i 2 . During the period 1
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(i) 0-2 s, ( v ) = a 1 2 ( t f + t i ) = (3 . 75 m / s 2 )(1 s) = 3 . 75 m / s ( P ) = F ( v ) = (5625 N)(3 . 75 m / s) = 21kW (ii) 2-4 s, ( v ) = a 1 2 ( t f + t i ) = (3 . 75 m / s 2 )(3 s) = 11 . 25 m / s ( P ) = F ( v ) = (5625 N)(11 . 25 m / s) = 63kW (iii) 4-6 s, ( v ) = a 1 2 ( t f + t i ) = (3 . 75 m / s 2 )(5 s) = 18 . 75 m / s ( P ) = F ( v ) = (5625 N)(3 . 75 m / s) = 105kW (iv) 6-8 s, ( v ) = a 1 2 ( t f + t i ) = (3 . 75 m / s 2 )(7 s) = 26 . 25 m / s ( P ) = F ( v ) = (5625 N)(3 . 75 m / s) = 147kW (d) The work done by the engine is W = Pt = 240kJ. If all of that work appears as the kinetic energy of the car, then W = 1 2 mv 2 v = radicalbigg 2 W m = parenleftbigg 2(240 , 000 J) 1500 kg parenrightbigg 1 2 = 18 m / s 3. (a) The flea stores enough energy to do the work necessary to achieve a height of 25 cm. W = Fd = mgd = (10 6 kg)(10 m / s 2 )(0 . 25 m) = 2 . 5 × 10 6 J (b) The energy stored in the spring is U = 1 2 kx 2 . Then k = 2 U x 2 = 2(2 . 5 × 10 6 J) (10 4 m) 2 = 500 N / m 4. Chapter 7, Problem 28 Let the applied force F = 80 N, that yields the initial displacement x i = 2cm. The spring constant k = | F/x i | = (80 N) / (0 . 02 m) = 4000N / m. The work done on a spring when changing its displacement from equilibrium from x i to x f = x i + Δ x is W = 1 2 k ( x 2 f x 2 i ) = 1 2 k (( x i + Δ x ) 2 x 2 i ) = 1 2 k x 2 + 2 x i Δ x ) (1) There are two solutions for Δ x . We can do the work on the spring by continuing to compress it beyond the original x i of -2cm. Or we can stretch it in the other 2
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direction, allowing it to relax to zero and then compress it in the positive direction. The solutions to Equation 1 are Δ x = 1 2 parenleftbigg 2 x i ± radicalBig 4 x 2 i + 8 W/k parenrightbigg = 1 2 parenleftBig 2( 0 . 02 m) ± radicalbig 4( 0 . 02 m) 2 + 8(4 J) / (4000N / m) parenrightBig = (0 . 02 ± 0 . 05) m The final position is x f = x i + Δ x = 0 . 02 m + (0 . 02 m ± 0 . 05 m) = ± 0 . 05 m 5. Chapter 7, Problem 54 (a) The work done on the block by the gravitational force is W g = F g x = mgx = ( 0 . 25 kg)(10 m / s 2 )( 0 . 12 m) = 0 . 3 J (b) The work done on the block by the spring is W s = integraldisplay x 0 kxdx = 1 2 kx 2 = 1 2 (2 . 5N / cm)(100cm / m)(0 . 12 m) 2 = 1 . 8 J (c) The final kinetic energy is the sum of the initial kinetic energy and the work done on the block. The final kinetic energy is zero. Therefore
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