A 2 500-kg ore car rolls 55.0 m down a frictionless 10.0° incline.If there is a horizontal spring at the end of the incline,what spring constant is required to stop the ore car in a distance of 2.10 m?

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Quizplus · Accepted Answer

Explanation:The work done by the spring is equal to the change in kinetic energy of the ore car. The change in kinetic energy is:$$\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$where:* $m$ is the mass of the ore car* $v_i$ is the initial velocity of the ore car* $v_f$ is the final velocity of the ore carThe initial velocity of the ore car is zero, so the change in kinetic energy is:$$\Delta K = \frac{1}{2}mv_f^2$$The work done by the spring is:$$W = \frac{1}{2}kx^2$$where:* $k$ is the spring constant* $x$ is the distance the spring is compressedSetting the work done by the spring equal to the change in kinetic energy, we get:$$\frac{1}{2}kx^2 = \frac{1}{2}mv_f^2$$Solving for $k$, we get:$$k = \frac{mv_f^2}{x^2}$$The final velocity of the ore car is:$$v_f = \sqrt{2gh}$$where:* $g$ is the acceleration due to gravity* $h$ is the height of the inclineThe height of the incline is:$$h = d\sin	heta$$where:* $d$ is the distance the ore car rolls down the incline* $	heta$ is the angle of the inclineSubstituting the expressions for $v_f$ and $h$ into the equation for $k$, we get:$$k = \frac{m(2gh)^2}{x^2}$$Substituting the given values, we get:$$k = \frac{(2500	ext{ kg})(2)(9.81	ext{ m/s}^2)(55.0	ext{ m})^2}{(2.10	ext{ m})^2}$$$$k = 2.8	imes10^6	ext{ kN/m}$$Therefore, the correct answer is C.

A 2 500-Kg Ore Car Rolls 55