{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

201_PartUniversity Physics Solution

# 201_PartUniversity Physics Solution - 6-18 6.68 Chapter 6...

This preview shows pages 1–3. Sign up to view the full content.

6-18 Chapter 6 (c) tot 2 1 W K K = gives 2 k 2 1 1 0 2 μ mgx mv = . 2 2 1 2 2 k (4.50 m/s) 10.3 m 2 2(0.100)(9.80 m/s ) v x μ g = = = . E VALUATE : The box goes farther when the friction coefficient doesn°t increase. 6.68. I DENTIFY : Use Eq.(6.7) to calculate W . S ET U P : 1 0 x = . In part (a), 2 0.050 m x = . In part (b), 2 0.050 m x = − . E XECUTE : (a) 2 2 2 3 2 3 4 2 2 2 0 0 ( ) 2 3 4 x x k b c W Fdx kx bx cx dx x x x = = + = + . 2 2 3 3 4 2 2 2 (50.0 N/m) (233 N/m ) (3000 N/m ) W x x x = + . When 2 0.050 m x = , 0.12 J W = . (b) When 2 0.050 m, x = − 0.17 J W = . (c) It°s easier to stretch the spring; the quadratic 2 bx term is always in the x -direction, and so the needed force, and hence the needed work, will be less when 2 0 x > . E VALUATE : When 0.050 m x = , 4.75 N x F = . When 0.050 m x = − , 8.25 N x F = . 6.69. I DENTIFY and S ET U P : Use m = F a " " to find the tension force T . The block moves in uniform circular motion and rad . = a a " " (a) The free-body diagram for the block is given in Figure 6.69. E XECUTE : x x F ma = 2 v T m R = 2 (0.70 m/s) (0.120 kg) 0.15 N 0.40 m T = = Figure 6.69 (b) 2 2 (2.80 m/s) (0.120 kg) 9.4 N 0.10 m v T m R = = = (c) S ET U P : The tension changes as the distance of the block from the hole changes. We could use 2 1 x x x W F dx = to calculate the work. But a much simpler approach is to use tot 2 1 . W K K = E XECUTE : The only force doing work on the block is the tension in the cord, so tot . T W W = 2 2 1 1 1 1 2 2 (0.120 kg)(0.70 m/s) 0.0294 J K mv = = = 2 2 1 1 2 2 2 2 (0.120 kg)(2.80 m/s) 0.470 J K mv = = = tot 2 1 0.470 J 0.029 J 0.44 J W K K = = = This is the amount of work done by the person who pulled the cord. E VALUATE : The block moves inward, in the direction of the tension, so T does positive work and the kinetic energy increases. 6.70. I DENTIFY : Use Eq.(6.7) to find the work done by F . Then apply tot 2 1 W K K = . S ET U P : 2 1 dx x x = − . E XECUTE : 2 1 2 1 2 1 1 x x W dx x x x α α = = . 26 2 1 9 1 17 (2.12 10 N m )((0.200 m ) (1.25 10 m )) 2.65 10 J W = × × = − × . Note that 1 x is so large compared to 2 x that the term 1 1/ x is negligible. Then, using Eq. (6.13)) and solving for 2 v , 17 2 5 2 5 2 1 27 2 2( 2.65 10 J) (3.00 10 m/s) 2.41 10 m/s. (1.67 10 kg) W v v m × = + = × + = × × (b) With 2 1 0, K W K = = − . Using 2 W x α = − , 26 2 10 2 2 27 5 2 1 1 2 2(2.12 10 N m ) 2.82 10 m. (1.67 10 kg)(3.00 10 m/s) x K mv α α × = = = = × × ×

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

View Full Document
Work and Kinetic Energy 6-19 (c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their original values, and the speed is 5 3.00 10 m/s × . E VALUATE : As the proton moves toward the uranium nucleus the repulsive force does negative work and the kinetic energy of the proton decreases. As the proton moves away from the uranium nucleus the repulsive force does positive work and the kinetic energy of the proton increases. 6.71. I DENTIFY and S ET U P : Use / x v dx dt = and / . x x a dv dt = Use m = F a " " to calculate F " from . a " E XECUTE : (a) 2 ( ) , x t t t α β 3 = + 2 ( ) 2 3 x dx v t t t dt α β = = + 4.00 s: t = 2 3 2 2(0.200 m/s )(4.00 s) 3(0.0200 m/s )(4.00 s) 2.56 m/s.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}