A \( 2000-\mathrm{kg} \) ore car rolls 50.0 m down a frictionless \( 10.0^{\circ} \) 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 ? 1) \( 340 \mathrm{kN} / \mathrm{m} \) 2) \( 681 \mathrm{kN} / \mathrm{m} \) 3) \( 980 \mathrm{kN} / \mathrm{m} \) 4) \( 1960 \mathrm{kN} / \mathrm{m} \) 5) \( 2100 \mathrm{kN} / \mathrm{m} \)

We start by noting that the ore car slides down a frictionless incline and converts gravitational potential energy into kinetic energy. When it reaches the bottom of the incline, all of this energy is available to be stored in the horizontal spring as elastic potential energy. (We assume here that at the bottom, the ore car is on a horizontal surface so that its kinetic energy is entirely available for compressing the spring.) Step 1. Determine the vertical drop, h, along the incline:   h = (length of incline) × sin(θ)     = 50.0 m × sin(10.0°)     ≈ 50.0 m × 0.17365 ≈ 8.68 m Step 2. Calculate the gravitational potential energy lost:   ΔU = mgh = (2000 kg)(9.8 m/s²)(8.68 m)     ≈ 2000 × 9.8 × 8.68 ≈ 170,117 J Step 3. Relate this energy to the energy stored in the spring. When the spring is compressed a distance x (here, 1.00 m), the spring stores energy   U_spring = (1/2) k x². Setting the two energies equal:   mgh = (1/2) k x² Step 4. Solve for k:   k = (2 mgh) / x² = (2 × 2000 kg × 9.8 m/s² × 8.68 m) / (1.00 m)²     = (2 × 2000 × 9.8 × 8.68) Calculate:   First, 2000 × 9.8 = 19,600.   Then, 19,600 × 8.68 ≈ 170,128.   Multiply by 2: 2 × 170,128 ≈ 340,256 N/m   Which is approximately 340 kN/m. Thus, the required spring constant is about 340 kN/m, which corresponds to answer choice (1).