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Answer:
Explanation:
Let y₀ be the initial position of the cylinder when the spring is attached and y its position when it is momentarily at rest.From work-kinetic energy principles, The work done by the spring force + work done by friction + work done by gravity = kinetic energy change of the cylinder
work done by the spring force = ¹/₂k(y₀² - y²)
work done by friction = - f(y - y₀)
work done by gravity = mg(y - y₀)
kinetic energy change of the cylinder = ¹/₂m(v₁² - v₀²)
So ¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(v₁² - v₀²)
Since the cylinder starts at rest, v₀ = 0. Also, when it is momentarily at rest, v₁ = 0
¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(0² - 0²)
¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = 0
¹/₂ky₀² + fy₀ - mgy₀ -¹/₂ky² - fy + mgy = 0
¹/₂ky₀² + fy₀ - mgy₀ = ¹/₂ky² + fy - mgy
Let y₀ = 0, then the left hand side of the equation equals zero. So,
0 = ¹/₂ky² + fy - mgy
¹/₂ky² + fy - mgy = 0
Using the quadratic formula