Consider the \"Freefall Experience\" experiment shown above. A person is in a me

Question

Consider the "Freefall Experience" experiment shown above. A person is in a metal box resting on a launcher (a large spring). The spring is compressed and then released. The box flies up into the air and then lands back on the spring. The box is guided by vertical poles, as shown. Initially, assume the friction from the poles is zero. You know: M = Mass of the box and person yi - Initial position of the box, yf = Final position of the box, y = 0 where F_sprring = 0 M_spcing = 0 (The spring is "ideal" meaning its mass can be neglected) What you want to determine: k = Spring constant that will launch the box as desired. You will use energy concepts to solve this problem that depend on the system. If a force is external to the system, its effect on the energy of the system is "Work". If a force is internal to the system, and is conservative, its energy is "Potential Energy". The relation between "Work" and "Potential Energy" for a force F is DeltaPE_f = - W_f. For each system defined below, find an algebraic solution for k. Show your work. Your answer must include: (i) the total initial energy, (ii) the total final energy, (iii) the work done, (iv) a conservation-of-energy equation, and (v) the solution for k.A) Solve using as your system: The Box (and person) B) Solve using as your system: The Box (and person) and the Spring C) Solve using as your system: The Box (and person), the Spring, and the Earth D) Solve for the spring constant, k, for M = 250 kg, yi = 2 m, y_f - 4 m. You may use A, B, or C above to solve the problem. E) The rails guiding the box exert a friction force: = 0.1 * M g. What is the maximum height the box will reach, using the spring constant from part D? Show your work. You may use any of the systems above, A - C, for your solution]

Explanation / Answer

here conservatioin of energy gravitational potential energy = elastic potential energy

   mgh = 1/2 kx^2 ==> k = 2mgh/x^2

initial position , yi = -2 m, k = 2*250*9.8*(-2)/(-2)^2 = 2450 N/m

now this elastic potential energy will be conveted in to kinetic energy so that the box will move to a position yf

if no friction by by the poles then 1/2 M v^2 = 1/2 kX^2

v^2 = kx^2/M = 2450*(-2)^2/250 => v = sqrt(2450*(-2)^2/250)= 6.26 m/s

s = u^/2g = 6.26^2/2*9.8 = 1.99= 2 m it can reach

if the rails guiding the box exerts the friction force then

maximum height the box can reach is

   force equation on the box is ma = 1/2 kx^2 + Ff

               a = 1/m(1/2 kx^2 + 0.1*mg)
= k/m x^2 + 0.1*g
= (2450/250)*2^2+0.1*9.8
= 40.18 m/s2

now the height reached by box is s = u^2/2a = 6.26/2*40.18 = 0.078 m

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