Energy conservation with conservative forces: Consider the motion of a 1.00-kg particle that moves with potential energy given by U(x) = (-2.00 J ∙ m)/x + (4.00 J∙m²)/x². Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

Question

Quizplus · Accepted Answer

The total mechanical energy of the particle is conserved, so:$$K_i + U_i = K_f + U_f$$where $K$ is the kinetic energy and $U$ is the potential energy.At $x = 1.00 	ext{ m}$, we have:$$K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(1.00 	ext{ kg})(3.00 	ext{ m/s})^2 = 4.50 	ext{ J}$$$$U_i = U(1.00 	ext{ m}) = \frac{-2.00 	ext{ J}\cdot 	ext{m}}{1.00 	ext{ m}} + \frac{4.00 	ext{ J}\cdot 	ext{m}^2}{(1.00 	ext{ m})^2} = 2.00 	ext{ J}$$So, the total mechanical energy is:$$E = K_i + U_i = 4.50 	ext{ J} + 2.00 	ext{ J} = 6.50 	ext{ J}$$At $x = 5.00 	ext{ m}$, we have:$$U_f = U(5.00 	ext{ m}) = \frac{-2.00 	ext{ J}\cdot 	ext{m}}{5.00 	ext{ m}} + \frac{4.00 	ext{ J}\cdot 	ext{m}^2}{(5.00 	ext{ m})^2} = 0.600 	ext{ J}$$So, the kinetic energy at $x = 5.00 	ext{ m}$ is:$$K_f = E - U_f = 6.50 	ext{ J} - 0.600 	ext{ J} = 5.90 	ext{ J}$$And the speed is:$$v_f = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2(5.90 	ext{ J})}{1.00 	ext{ kg}}} = \boxed{3.67 	ext{ m/s}}$$

Energy Conservation with Conservative Forces: Consider the Motion of a 1.00-Kg