236_PartUniversity Physics Solution

# 236_PartUniversity Physics Solution - Potential Energy and...

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Potential Energy and Energy Conservation 7-25 EVALUATE: 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) IDENTIFY and SET UP: Apply other AA BB K UW KU ++ =+ to the motion from A to B . EXECUTE: 0, A K = 2 1 2 B B K mv = 0, A U = 2 1 el, 2 , B UU k x == where 0.25 m B x = other F B WW F x Thus 22 11 . 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 (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) IDENTIFY: 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 . BC x x + SET UP: The motion from A to B to C is described in Figure 7.75. other B C K EXECUTE: other 0 W = 2 1 2 5.0 J 1.25 J 3.75 J Km v −= (from part (a)) 2 1 2 1.25 J Uk x 0 C K = (instantaneously at rest at point closest to wall) 2 1 2 CC x = Figure 7.75 Thus 2 1 2 3.75 J 1.25 J C kx += 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. = EVALUATE: 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. IDENTIFY: Apply Eq.(7.14) to the motion of the student. SET 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. EXECUTE: (a) The speed of the brother at a given height h above the point of maximum compression is then found from 1 (4 ) , w v mgh g ⎛⎞ ⎜⎟ ⎝⎠ or 2 1 1 0 (4 ) kg x vx g h g h wx =− = . Therefore, (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 , 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 1 01 1 ( 0.53m) 0.53m 1 xx x x + + will be different. Note that on a planet with lower g , 1 x will be smaller and h will be larger. EVALUATE: 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. IDENTIFY: 22 / x ad x d t = , / y y d t = . x x Fm a = , y y a = . xy UF d x F d y =+ . SET UP: 00 0 (cos ) sin d tt dt ω ωω =− . 000 (sin ) cos d dt = . 0 1 cos sin td t t = , 0 1 sin cos t t . / x vd x d t = , / y y d t = . EKU . EXECUTE: (a) 2 2 /, . xx x x d t x a mx == −= = 2 2 yy y y d t yFm a my = = (b) 2 2 1 () 2 U F dx F dy m xdx ydy m x y ⎡⎤ + = + = + ⎣⎦ ∫∫ (c) 0 0 /s i n ( / ) . x v d x d tx txy y = 0 0 /c o s ( / ) . y v d y d ty tyx x + = + (i) When 0 x x = and 0, 0 x yv and y vy = , 2 2 2 2 2 0 0 0 0 0 111 1 , a n d ( ) 222 2 K mv v my U mx E K U m x y = = = + = + (ii) When 0 x = and 0 , x yyv x and 0 y v = , 2 2 22 2 0 0 00 0 11 1 ,a n d ( ) 2 K mx U m y E K U m x y = + = + EVALUATE: The total energy is the same at the two points in part (c); the total energy of the system is constant.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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236_PartUniversity Physics Solution - Potential Energy and...

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