Chapter 10 - Energy - Exercises and Problems - Page 272: 25

(a) At x = 1.0 m, the particle moves to the right. (b) The maximum speed is 17.3 m/s which occurs at x = 4.0 m (c) The turning points of the motion are at x = 1.0 m and x = 6.0 m

Work Step by Step

(a) At x = 1.0 m, the particle is at rest. Therefore, at x = 1.0 m, the total energy is equal to the potential energy which is 4.0 J. The particle can not go to the left because the potential energy can not go higher than the total energy of 4.0 J. At x = 1.0 m, the particle moves to the right, and the kinetic energy gradually increases as the potential energy gradually decreases. (b) The particle reaches its maximum speed when the kinetic energy is at a maximum. This occurs when the potential energy is at a minimum. At x = 4.0 m, the potential energy is at a minimum of 1.0 J. The kinetic energy is 3.0 J at this point since the total energy is 4.0 J. We can find the speed of the particle at x = 4.0 m. $KE = 3.0~J$ $\frac{1}{2}mv^2 = 3.0~J$ $v^2 = \frac{(2)(3.0~J)}{m}$ $v = \sqrt{\frac{(2)(3.0~J)}{m}}$ $v = \sqrt{\frac{(2)(3.0~J)}{0.020~kg}}$ $v = 17.3~m/s$ The maximum speed is 17.3 m/s which occurs at x = 4.0 m (c) The turning points occur when the potential energy is equal to the total energy of 4.0 J. The potential energy is 4.0 J at x = 1.0 m and x = 6.0 m. The turning points of the motion are at x = 1.0 m and x = 6.0 m.

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