{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern