A delicate 0.2 lb instrument is placed in a tray. To protect it during shipping

Question

A delicate 0.2 lb instrument is placed in a tray. To protect it during shipping it is attached to a spring as shown in the figure. Given the spring's experimentally obtained stiffness characteristic, shown below, what is the maximum velocity for the tray and instrument if the force on the delicate instrument is not to exceed 12lb. Assume the tray comes to an abrupt stop. The force can be represented by the following interpolated polynomial function of x. F (x)=3.133x-4.333x^2+5.333x^3-2.667x^4+0.5333x^5.

Explanation / Answer

F(x) = 3.133x - 4.333 x^2 + 5.333x^3 - 2.667 x^4 + 0.5333 x^5

also given F_max = 12 lbs..

so.. to find the x corresponding to F(x) = 12 ..
3.133x - 4.333 x^2 + 5.333x^3 - 2.667 x^4 + 0.5333 x^5 = 12

on solving we get... x = 2.50107 inch

so... elastic energy corresponding to a compression = integration ( F(x) dx ) -- x from 0 to 2.50107

so.. maximum elastic energy = integration ( (3.133x - 4.333 x^2 + 5.333x^3 - 2.667 x^4 + 0.5333 x^5)dx )
= (3.133/2)x^2 - (4.333/3) x^3 + (5.333/4)x^4 - (2.667/5) x^5 + (0.5333/6) x^6 -- from x = 0 to 2.50107

so.. maximum elastic energy = 8.92601


let the maximum speed be v..
sokinetic energy of the block = 0.5*mv^2 = 0.5 * 0.2*v^2 = 0.1v^2

so.. 0.1v^2 = 8.92601
so.. v_max = 9.4478

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