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**Unformatted text preview: **Week 9 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution is the same, but you will need to repeat part of the calculation to find out what your answer should have been. WebAssign Problem 1: A 10.1-kg uniform board is wedged into a corner and held by a spring at a 50.0° angle, as the drawing shows. The spring has a spring constant of 176 N/m and is parallel to the floor. Find the amount by which the spring is stretched from its unstrained length. REASONING AND SOLUTION In order to find the amount the spring stretches we need to calculate the force that acts on the spring. The magnitude of this force is F . Since the board is in equilibrium, the net torque acting on it is zero. Taking the axis of rotation to be at the corner and assuming the board has a length L , the net torque is Σ τ = FL sin 50.0° -mg ( L /2) cos 50.0° = 0 Solving for F gives cos 50.0 2 sin 50.0 2 tan 50.0 mg mg F ° = = ° ° The amount x by which the spring stretches is equal to the magnitude F of the force applied to it divided by the spring constant k (see Equation 10.1). Thus, 2 (10.1 kg)(9.80 m/s ) 0.236 m 2 tan 50.0 2(176 N/m) tan 50.0 F mg x k k = = = = ° ° WebAssign Problem 2: Concept Simulation 10.3 at www.wiley.com/college/cutnell illustrates the concepts pertinent to this problem. An 0.80-kg object is attached to one end of a spring, as in Figure 10.6 , and the system is set into simple harmonic motion. The displacement x of the object as a function of time is shown in the drawing. With the aid of these data, determine (a) the amplitude A of the motion, (b) the angular frequency ω (c) the spring constant k , (d) the speed of the object at t = 1.0 s, and (e) the magnitude of the objects acceleration at t = 1.0 s. REASONING AND SOLUTION a. Since the object oscillates between ± 0.080 m, the amplitude of the motion is 0.080 m . b. From the graph, the period is T = 4.0 s. Therefore, according to Equation 10.4, ϖ = 2 π T = 2 4.0 s = 1.6 rad/s c. Equation 10.11 relates the angular frequency to the spring constant: = k / m . Solving for k we find k = 2 m = (1.6 rad/s) 2 (0.80 kg) = 2.0 N/m d. At t = 1.0 s, the graph shows that the spring has its maximum displacement. At this location, the object is momentarily at rest, so that its speed is 0 m/s v = . e. The acceleration of the object at t = 1.0 s is a maximum, and its magnitude is a max = A 2 = (0.080 m)(1.6 rad/s) 2 = 0.20 m/s 2 WebAssign Problem 3: A 3.0-kg block is between two horizontal springs. Neither spring is strained when the block is at the position labeled x = 0 m in the drawing. ...

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