question_answer

A rectangular box is sliding on a smooth inclined plane of inclination a. At t = 0, the box starts to move on the inclined plane. A bolt starts to fall from point A. Find the time after which bolt strikes the bottom surface of the box.

Answer:

                As the block starts falling on the inclined plane, let at an instant, when the bolt reaches at E, its velocity be v parallel to PQ. It then falls down due to gravity pull. Taking vertical downward motion perpendicular to plane PQ of the bolt, we have u = 0, \[\text{a}\,\text{=}\,\text{gcos}\alpha \], \[s=l\], t = ? As \[\text{s}=\text{ut}+\frac{1}{2}\text{a}{{\text{t}}^{\text{2}}}\] \[\text{l}=0+\frac{1}{2}\text{g cos}\alpha {{\text{t}}^{\text{2}}}\] or                    \[\text{t}=\sqrt{\frac{2l}{g\cos \alpha }}\]