Answer:
Explanation:
check attached image for figure, there is supposed to be a figure for this question containing a distance(height of collar at position A) but i will assume 0.2m or 200mm
Consider the energy equilibrium of the system
[tex]U_{A-B}=\bigtriangleup T\\\\F\cos 30°\times h_A - F\sin30°\times h_A + Wh_A=\frac{1}{2}m(v^2_B-v^2_A)\\\\v_B=\sqrt{\frac{2Fh_A(\cos 30° - \sin30°)+mgh_A}{m+v^2_A}}[/tex]
Here, F is the force acting on the collar, [tex]h_A[/tex] is the height of the collar at position A, m is the mass of the collar C, g is the acceleration due to gravity, [tex]v_B[/tex] is the velocity of the collar at position B, and [tex]v_A[/tex] is the velocity of the collar at A
Substitute 14.4N for F, 0.2m for [tex]h_A[/tex], 1.5kg for m, [tex]9.81m/s^2[/tex] for g and 0 for [tex]v_A[/tex]
[tex]v_B=\sqrt{\frac{2(14.4\times 0.2(\cos 30° - \sin30°)+1.5\times 9.81\times 0.2}{1.5+0}}\\\\=\sqrt{\frac{6.618}{1.5}}\\\\=4.412m/s[/tex]
Therefore, the velocity at which the collar strikes the end B is 4.412m/s
