A 2,000-kg ore car rolls 50.0 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 1.00 m?

Question

Quizplus · Accepted Answer

The gravitational potential energy lost by the ore car is converted into the elastic potential energy stored in the spring. Using energy conservation: mgh = (1/2)kx^2. Calculate h using h = 50.0 m * sin(10.0°), then solve for k using k = 2mgh/x^2. The correct spring constant is approximately 347 kN/m.

A 2,000-Kg Ore Car Rolls 50