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Answer:
k = 652 lb/ft
Explanation:
Given :
Weight of the collar = 1.6 lb
The upstretched length of the spring = 6 in
Speed = 16 ft/s
PA = 8 + 10
= 18 inch
Let the initial elongation be [tex]$\Delta x_i$[/tex]
∴ [tex]$\Delta x_i$[/tex] = 18 - 6
= 12 inch = 1 foot
[tex]$PB = \sqrt{13^2+5^2}$[/tex]
= 13.925 inch
Final elongation in the spring
[tex]$\Delta x_B = 7.928 $[/tex] inch = 0.66 feet
Applying the conservation of the mechanical energy between A and B is
[tex]$K.E_A+P.E_{g,A}+P.E_{sp,A}= K.E_B+P.E_{g,B}+P.E_{sp,B} $[/tex]
[tex]$0+mg_r+\frac{1}{2}k(\Delta x_i)^2=\frac{1}{2}mv_B^2+0+\frac{1}{2}k(\Delta x_B)^2$[/tex]
[tex]$\frac{1}{2}k[(1)^2-(0.66)^2]=\frac{1.6}{2}\times (16)^2-1.6 \times 32 \times \frac{5}{12}$[/tex]
[tex]0.281 \ k =204.8-21.33[/tex]
k = 652 lb/ft
