1. A 4.6 kg block is at rest against a horizontal spring that is compressed by 0.40 m. The spring has a spring constant of k₁ = 2750 N/m. After leaving the spring, it travels up a 28° incline to a height of 2.8 m. At the top of the hill is a second spring with a spring constant of k₂ = 350 N/m. The horizontal portions are frictionless, but the hill has a coefficient of kinetic friction equal to uk= 0.16. x₁ = 0.40 m ((0))) k₂ N m = 2750- Simplifies to Mk= 0.16 a. How fast does the block leave the spring on the bottom of the hill? X₁ = 0.40 m h₁ = 0 m Vi= 0 m/s 28° k = 2750 N/m m = 4.6 kg h = 2.8 m ½/2kx;² = ½ mvf2 X₁ = 0 m h₁=0 m Vi=? m/s Wnc = 0 J 2 Wnc + mghi +1⁄2kx² +½ mv₁² = mghf +½⁄₂ kxf² +½½ mv₁² vf = √k/m* Xi =√(2750 N/m ÷4.6 kg) * (0.40 m) Vf= 9.78 m/s ())) k₂ = 350 N m

1. A 4.6 kg block is at rest against a horizontal spring that is compressed by 0.40 m. The spring has a spring constant of k₁ = 2750 N/m. After leaving the spring, it travels up a 28° incline to a height of 2.8 m. At the top of the hill is a second spring with a spring constant of k₂ = 350 N/m. The horizontal portions are frictionless, but the hill has a coefficient of kinetic friction equal to uk= 0.16. x₁ = 0.40 m ((0))) k₂ N m = 2750- Simplifies to Mk= 0.16 a. How fast does the block leave the spring on the bottom of the hill? X₁ = 0.40 m h₁ = 0 m Vi= 0 m/s 28° k = 2750 N/m m = 4.6 kg h = 2.8 m ½/2kx;² = ½ mvf2 X₁ = 0 m h₁=0 m Vi=? m/s Wnc = 0 J 2 Wnc + mghi +1⁄2kx² +½ mv₁² = mghf +½⁄₂ kxf² +½½ mv₁² vf = √k/m* Xi =√(2750 N/m ÷4.6 kg) * (0.40 m) Vf= 9.78 m/s ())) k₂ = 350 N m

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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