Ps7 - P207 Fall 2007 Solutions Assignment 7 1 Chapter 8 Problem 28(a The total mechanical energy of the box is the sum of the potential energy of

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P207 - Fall 2007 Solutions Assignment 7 1. Chapter 8, Problem 28 (a) The total mechanical energy of the box is the sum of the potential energy of the spring, the potential energy of gravity and the kinetic energy. Let x = 0 when the spring is in the unstretched position. The change in height is h = x sin θ . E mech = U spring + U gravity + K 0 = 1 2 kx 2 mgx sin θ + 1 2 mv 2 When x = 0, E mech = 0 and mechanical energy is conserved so we can solve for v as a function of x v = p 2 m ( mgx sin θ 1 2 kx 2 ) P 1 2 (1) When x = 0 . 1m, v = p 2 2 kg ((2 kg)(0 . 1 m)(9 . 8 m / s 2 ) sin(40 ) 1 2 (120N / m)(0 . 1 m) 2 ) P 1 2 = 0 . 81 m / s (b) We need to determine the values of x in Equation 1 such that v = 0. We have that mg sin θ = 1 2 kx x = 2 mg sin θ k = 2(2 kg)(9 . 8 m / s 2 ) sin 40 120N / m = 0 . 21 m (c) We have that when the box stops when x = x 0 = 2 mg sin θ/k . Then the force on the box is F = kx + mg sin θ and the acceleration is a = F/m 1
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= kx/m + g sin θ = k p 2 mg sin θ k P 1 m + g sin θ = g sin θ = 9 . 8 m / s 2 sin(40 ) = 6 . 3 m / s 2 Note that at x = 0, we see by inspection that a (0) = g sin θ . Then by symmetry the acceleration at the other point of zero velocity, x = x 0 , that the acceleration a ( x 0 ) = a (0) = g sin θ , which is exactly what we found above. (d) The acceleration is in the direction of the spring restoring force, up the ramp. 2. Chapter 8, Problem 40 (a) The total mechanical energy E mech = U + 1 2 mv 2 = 12 J+4 J = 16 J. The speed of the particle at x = 3 . 5 m is v = p 2( E mech U A ) m P 1 2 = p 2(16 J 9 J) 0 . 2 kg P 1 2 = 8 . 4 m / s (b) The speed of the particle at x = 6 . 5 m is v = p 2( E mech 0) m P 1 2 = p 2(16 J) 0 . 2 kg P 1 2 = 12 . 6 m / s (c) The turning point on the right side is where the potential energy is equal to the total mechanical energy. That would be at position x = 7 m+Δ x when E mech = ( U D / (1 m))Δ x or x = 7 m + ( E mech /U D )(1 m) = (16 J / 24 J) = 7 . 67 m (d) The turning point on the left side is where the potential energy is equal to the total mechanical energy. Since the total energy is 16 J, the turning point will be between x 1 = 1 m and x 3 = 3 m at some distance Δ x from x 1 . U
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This note was uploaded on 01/22/2009 for the course PHYS 207 taught by Professor Liepe during the Fall '08 term at Cornell University (Engineering School).

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Ps7 - P207 Fall 2007 Solutions Assignment 7 1 Chapter 8 Problem 28(a The total mechanical energy of the box is the sum of the potential energy of

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