The same toy car and spring from the previous problem sends the car up a slope that rises 0.180 m, still with negligible friction. Taking g = 9.80 m/s^2, what is the car's speed at the top of the slope?

1. Conservation of mechanical energy from release to the top of the slope gives one half k xi squared equal to one half m vf squared plus m g hf. 2. Solving for vf gives vf equal to the square root of [(k xi squared)/m minus 2 g hf]. 3. The first term inside the root, (250.0 N/m)(0.0400 m)^2 / 0.100 kg, equals 4.00 m^2/s^2, since the spring's stored energy is unchanged from before. 4. The second term, 2 g hf, is 2 times 9.80 m/s^2 times 0.180 m, which is 3.528 m^2/s^2. 5. Subtracting gives 4.00 minus 3.528, or 0.472 m^2/s^2, and taking the square root gives vf = 0.687 m/s. 6. 1.20 m/s and 1.88 m/s come from arithmetic slips in combining the two terms, and 0.180 m/s simply reuses the height value as if it were the answer. _Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.4 Conservative Forces and Potential Energy_