A 2 000-kg ore car rolls 50.0 m down a frictionless 10.0 $\degree$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 1.00 m?

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The work done by the spring on the ore car 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 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:$$\frac{1}{2}kx^2 = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$Solving for $k$, we get:$$k = \frac{mv_f^2 - mv_i^2}{x^2}$$Plugging in the given values, we get:$$k = \frac{(2000 	ext{ kg})(0 	ext{ m/s})^2 - (2000 	ext{ kg})(10 	ext{ m/s})^2}{(1.00 	ext{ m})^2}$$$$k = 340 	ext{ kN/m}$$

A 2 000-Kg Ore Car Rolls 50 °\degree° InclineIf There Is a Horizontal Spring at the End of Frictionless

A 2 000-Kg Ore Car Rolls 50 $\degree$ InclineIf There Is a Horizontal Spring at the End of Frictionless