A 20lb weight is suspended from the end of a vertical springof negligible mass.
ID: 1663456 • Letter: A
Question
A 20lb weight is suspended from the end of a vertical springof negligible mass. This much weight stretches the spring by 6nches. It is then pulled 2 inches down and released. How is it to determine the y(t) and v(t)? Suppose that the spring weight is pulled 3 inches down andgiven an initial speed of 2 ft/s downward. Determine the positionand speed of the weight as a function of time. Hope to receive an answer.thank you in advance. A 20lb weight is suspended from the end of a vertical springof negligible mass. This much weight stretches the spring by 6nches. It is then pulled 2 inches down and released. How is it to determine the y(t) and v(t)? Suppose that the spring weight is pulled 3 inches down andgiven an initial speed of 2 ft/s downward. Determine the positionand speed of the weight as a function of time. Hope to receive an answer.thank you in advance.Explanation / Answer
Don't have time to do it now, but maybe this will get youstarted. The spring constant k is Force/length. Here, force is theweight. 6in=.5 feet. 20/.5=40. So k=40. The spring equation is F=-kx. x is the displacement, or 2 in(.1667 feet). The questions asks for y(t), so you can write the equation as F=-ky(y is displacement in the vertical direction). The potential energy is U=(1/2)*ky^2. When the spring isreleased the velocity is 0, but when it passes through the centerof its motion (the rest position) it is at maximum velocity, andthis is all converted to kinetic energy. (1/2)*m*v^2. You can solve for v. So at t=0, y=2in, v=0. Not sure what level you're at, but then you have a differentialequation with initial conditions. For the equation. F=m*a So m*v'=-ky The second part is very similar, except that since you have aninitial speed, the block will travel a further distance before itbounces back
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