236_PartUniversity Physics Solution

236_PartUniversity Physics Solution - Potential Energy and...

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Potential Energy and Energy Conservation 7-25 E VALUATE : Our expression for y gives the reasonable results that 0 y = when k 0 μ = ; in the absence of friction the package returns to its starting point. 7.75. (a) I DENTIFY and S ET UP : Apply other A A B B K U W K U + + = + to the motion from A to B . E XECUTE : 0, A K = 2 1 2 B B K mv = 0, A U = 2 1 el, 2 , B B B U U kx = = where 0.25 m B x = other F B W W Fx = = Thus 2 2 1 1 2 2 . B B B Fx mv kx = + (The work done by F goes partly to the potential energy of the stretched spring and partly to the kinetic energy of the block.) (20.0 N)(0.25 m) 5.0 J B Fx = = and 2 2 1 1 2 2 (40.0 N/m)(0.25 m) 1.25 J B kx = = Thus 2 1 2 5.0 J 1.25 J B mv = + and 2(3.75 J) 3.87 m/s 0.500 kg B v = = (b) I DENTIFY : Apply Eq.(7.15) to the motion of the block. Let point C be where the block is closest to the wall. When the block is at point C the spring is compressed an amount , C x so the block is 0.60 m C x from the wall, and the distance between B and C is . B C x x + S ET UP : The motion from A to B to C is described in Figure 7.75. other B B C C K U W K U + + = + E XECUTE : other 0 W = 2 1 2 5.0 J 1.25 J 3.75 J B B K mv = = = (from part (a)) 2 1 2 1.25 J B B U kx = = 0 C K = (instantaneously at rest at point closest to wall) 2 1 2 C C U k x = Figure 7.75 Thus 2 1 2 3.75 J 1.25 J C k x + = 2(5.0 J) 0.50 m 40.0 N/m C x = = The distance of the block from the wall is 0.60 m 0.50 m 0.10 m. = E VALUATE : The work (20.0 N)(0.25 m) 5.0 J = done by F puts 5.0 J of mechanical energy into the system. No mechanical energy is taken away by friction, so the total energy at points B and C is 5.0 J. 7.76. I DENTIFY : Apply Eq.(7.14) to the motion of the student. S ET UP : Let 0 0.18 m x = , 1 0.71 m x = . The spring constants (assumed identical) are then known in terms of the unknown weight w , 0 4 kx w = . Let 0 y = at the initial position of the student. E XECUTE : (a) The speed of the brother at a given height h above the point of maximum compression is then found from 2 2 1 1 1 (4 ) , 2 2 w k x v mgh g = + or 2 2 2 1 1 0 (4 ) 2 2 k g x v x gh g h w x = = . Therefore, 2 2 (9.80 m/s )((0.71 m) (0.18 m) 2(0.90 m)) 3.13 m/s v = = , or 3.1 m/s to two figures. (b) Setting 0 v = and solving for h , 2 2 1 1 0 2 1.40 m, 2 kx x h mg x = = = or 1.4 m to two figures. (c) No; the distance 0 x will be different, and the ratio 2 2 2 1 1 1 0 1 1 ( 0.53 m) 0.53 m 1 x x x x x x + = = + will be different. Note that on a planet with lower g , 1 x will be smaller and h will be larger. E VALUATE : We are able to solve the problem without knowing either the mass of the student or the force constant of the spring.
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7-26 Chapter 7 7.77. I DENTIFY : 2 2 / x a d x dt = , 2 2 / y a d y dt = . x x F ma = , y y F ma = . x y U F dx F dy = + . S ET UP : 0 0 0 (cos ) sin d t t dt ω ω ω = − . 0 0 0 (sin ) cos d t t dt ω ω ω = . 0 0 0 1 cos sin t dt t ω ω ω = , 0 0 0 1 sin cos t dt t ω ω ω = − . / x v dx dt = , / y v dy dt = . E K U = + . E XECUTE : (a) 2 2 2 2 0 0 / , . x x x a d x dt x F ma m x ω ω = = − = = − 2 2 2 2 0 0 / , y y y a d y dt y F ma m y ω ω = = − = = − (b) 2 2 2 2 0 0 1 ( ) 2 x y U F dx F dy m xdx ydy m x y ω ω = − + = + = + (c) 0 0 0 0 0 0 / sin ( / ).
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