Question

The 1.5-kg collar C starts from rest at A and slides with negligible friction on the fixed rod in the vertical plane. Determine the velocity v with which the collar strikes end B when acted upon by the 14.4-N force, which is constant in direction. Neglect the small dimensions of the collar.

Answer 1

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

$U_{A-B}=\bigtriangleup T\\\\F\cos 30°\times h_A - F\sin30°\times h_A + Wh_A=\frac{1}{2}m(v^2_B-v^2_A)\\\\v_B=\sqrt{\frac{2Fh_A(\cos 30° - \sin30°)+mgh_A}{m+v^2_A}}$

Here, F is the force acting on the collar, $h_A$ is the height of the collar at position A, m is the mass of the collar C, g is the acceleration due to gravity, $v_B$ is the velocity of the collar at position B, and $v_A$ is the velocity of the collar at A

Substitute 14.4N for F, 0.2m for $h_A$, 1.5kg for m, $9.81m/s^2$ for g and 0 for $v_A$

$v_B=\sqrt{\frac{2(14.4\times 0.2(\cos 30° - \sin30°)+1.5\times 9.81\times 0.2}{1.5+0}}\\\\=\sqrt{\frac{6.618}{1.5}}\\\\=4.412m/s$

Therefore, the velocity at which the collar strikes the end B is 4.412m/s