65. A 1.50-kg mass is placed on the end of a spring that has a spring constant of The mass-spring system rests on a frictionless incline that is at an angle of 30° from the horizontal (Fig. 5.28). The system is eased into its equilibrium position, where it stays. (a) Determine the change in elastic potential energy of the system. (b) Determine the system’s change in gravitational potential energy. (a) 0.154 J (b) 65. (a) The component of the weight, of the object along the incline (parallel to mg, the spring) is equal to sin . (See Exercise 4-69.) This is the force that stretches the spring. mg = = = = 0.0420 m. x = = (175 N/m)(0.0420 m)2 0 = . Us k x2 k xo2 (b) The vertical distance the mass moves down is equal to sin = (0.0420 m) sin x 30 = 0.0210 m. = = (1.50 kg)(9.80 m/s2)(0.0210 m 0) = . Ug mgy
65. A 1.50-kg mass is placed on the end of a spring that has a spring constant of The mass-spring system rests on a frictionless incline that is at an angle of 30° from the horizontal (Fig. 5.28). The system is eased into its equilibrium position, where it stays. (a) Determine the change in elastic potential energy of the system. (b) Determine the system’s change in gravitational potential energy. (a) 0.154 J (b) 65. (a) The component of the weight, mg, of the object along the incline (parallel to the spring) is equal to mg sin . (See Exercise 4-69.) This is the force that stretches the spring. x = = = = 0.0420 m. Us = k x2 k xo2 = (175 N/m)(0.0420 m)2 0 = . (b) The vertical distance the mass moves down is equal to x sin = (0.0420 m) sin 30 = 0.0210 m. Ug = mgy = (1.50 kg)(9.80 m/s2)(0.0210 m 0) = .
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