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Question text
A mass of 25kg is attached to a spring (k=650N/m) and is hanged from a ceiling. The mass is released at a height h when the spring is unstretched.
a) What is the speed of the mass when it has fallen a distance of 0.30m?
b) What is the minimum release height (h) if the mass does not strike the floor?
This question is all about the conversion of energy, which is the total energy, the sum of the gravitational potential energy, elastic potential energy, and kinetic energy, remains unchanged. So, the motion can be explained as three stages: the beginning of the release, in the middle of the spring, and the most stretch of the spring. We will get back to it later. First, the falling of 0.30m. PE beginning = KE (0.30m)+ PE (0.30m) +PE (0.30m), which one is gravitational potential energy and the other is elastic potential energy. So, we can get the equation: mgh = 1/2 (mv^2+kx^2) => (25kg)*(10m/s^2)*(0.30m) = 1/2 (650N/m * 0.30m^2 + 25kg * v^2) => v=1.91m/s. Second question, we need to pick only 2 stage of the spring: the beginning of the release and the moment of striking the floor. Because, in this case, both stages has kinetic energy equals to 0. The equation is PE beginning (gravitational)= PE ending (elastic) => (25kg)*(10m/s^2)*h=1/2*650*h^2 => h=10/13=0.77m. You might need to create a graph to help you have a better understanding.
