Laws of Motion Part 2

Concept:The block moves up the incline against gravity, which provides a constant deceleration along the slope.Explanation:The block is given an initial upward velocity $u$ along the incline at $t=0$.Only the component of gravity $mg \sin \theta$ acts down the slope, opposing the motion.Using Newton's second law: $F = ma$, we get $-mg \sin \theta = ma$.Thus, the acceleration is $a = -g \sin \theta$ (negative because it opposes the upward motion).Initial velocity: $v_i = u$.Final velocity: $v_f = 0$ (the block comes to rest).Apply the third equation of motion: $v_f^2 = v_i^2 + 2aS$.Substitute the values: $0^2 = u^2 + 2(-g \sin \theta)S$.Simplify: $0 = u^2 - 2g \sin \theta \, S$.Rearrange: $2g \sin \theta \, S = u^2$.Solve for $S$: $S = \frac{u^2}{2g \sin \theta}$.The distance travelled before velocity becomes zero is $\frac{u^2}{2g \sin \theta}$.Answer:Option C: $\frac{u^2}{2g \sin \theta}$