Physics Solution Manual for 1100 and 2101

Thus the pressure increment 1gh1 for liquid 1 must be

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e, KE has a maximum value as the object passes through its equilibrium position, which is position 2. EPE has a maximum value when the spring is maximally stretched at position 3. GPE has a maximum value when the object is at its highest point above the ground, that is, at position 1. 13. (a) At the instant the top block is removed, the total mechanical energy of the remaining system is all elastic potential energy and is 1 kA2 (see Equation 10.13), where A is the 2 amplitude of the previous simple harmonic motion. This total mechanical energy is conserved, because friction is absent. Therefore, the total mechanical energy of the ensuing simple harmonic motion is also 1 kA2 , and the amplitude remains the same as it was 2 k . Thus, when m the mass m attached to the spring decreases, the angular frequency increases. previously. The angular frequency ω is given by Equation 10.11 as ω = 14. (b) The angular frequency ω of oscillation of a simple pendulum is given by Equation 10.16 g ω = . It depends only on the magnitude g of the acceleration due to gravity and the L length L of the pendulum. It does not depend on the mass. Therefore, the pendulum with the greatest length has the smallest frequency. 15. 1.7 s 16. (c) When the energy of the system is dissipated, the amplitude of the motion decreases. The motion is called damped harmonic motion. 17. (a) Resonance occurs when the frequency of the external force equals the frequency of oscillation of the object on the spring. The angular frequency of such a system is given by k Equation 10.11 ω = . Since the frequency of the force is doubled, the new frequency m must be 2ω = 2 k 8k k . The frequency of system A is, in fact, ω = =2 . m 2m m Chapter 10 Answers to Focus on Concepts Questions 497 ∆L 18. (c) According to Equation 10.17 F = Y A , the force F required to stretch a piece of L 0 material is proportional to Young’s modulus Y, the amount of stretch ∆L, and the crosssectional area A of the material, but is inversely proportional to the initial length L0 of the FL material. Solving this equation for the amount of stretch gives ∆ L = 0 . Thus, the greater YA the cross-sectional area, the smaller is the amount of stretch, for given values of Young’s modulus, the initial length, and the stretching force. Thus, B stretches more than A, because B has the smaller cross-sectional area of solid material. 19. 0.50 × 10−6 m 20. 0.0017 498 SIMPLE HARMONIC MOTION AND ELASTICITY CHAPTER 10 SIMPLE HARMONIC MOTION AND ELASTICITY PROBLEMS ______________________________________________________________________________ 1. SSM REASONING AND SOLUTION Using Equation 10.1, we first determine the spring constant: FxApplied 89.0 N k= = = 4660 N/m x 0.0191 m Again using Equation 10.1, we find that the force needed to compress the spring by 0.0508 m is FxApplied = kx = (4660 N/m)(0.0508 m) = 237 N ______________________________________________________________________________ 2. REASONING AND SOLUTION According to Equation 10.1 and the data from the graph, the effective spring constant is Applied Fx 160 N 2 = 6.7 × 10 N/m x 0.24 m ______________________________________________________________________________ k= 3. = REASONING The force required to stretch the spring is given by Equation 10.1 as FxApplied = kx, where k is the spring constant and x is the displacement of the stretched spring from its unstrained length. Solving for the spring constant gives k = FxApplied / x . The force applied to the spring has a magnitude equal to the weight W of the board, so FxApplied = W . Since the board’s lower end just extends to the floor, the unstrained length x0 of the spring, plus the length L0 of the board, plus the displacement x of the stretched spring must equal the height h0 of the room, or x0 + L0 + x = h0. Thus, x = h0 − x0 − L0. SOLUTION Substituting FxApplied = W and x = h0 − x0 − L0 into Equation 10.1, we find k= FxApplied x = W 104 N = = 650 N/m h0 − x0 − L0 2.44 m − 1.98 m − 0.30 m ____________________________________________________________________________________________ Chapter 10 Problems 4. 499 REASONING The weight of the person causes the spring in the scale to compress. The amount x of compression, according to Equation 10.1, depends on the magnitude FxApplied of the applied force and the spring constant k. SOLUTION a. Since the applied force is equal to the person’s weight, the spring constant is Applied k= Fx = x 670 N 0.79 × 10 −2 m = 8.5 × 104 N / m (10.1) b. When another person steps on the scale, it compresses by 0.34 cm. The weight (or applied force) that this person exerts on the scale is ( )( ) FxApplied = k x = 8.5 × 104 N / m 0.34 × 10−2 m = 290 N (10.1) ______________________________________________________________________________ 5. SSM REASONING AND SOLUTION According to Newton's second law, the force required to accelerate the trailer is Fx = ma x (Equation 4.2a). The displacement x of the spring is given by Fx = −kx (Equation...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online