Question

A boy of mass 40kg slides down a chute inclined at 30 degrees to the horizontal. If the chute is smooth and the boy starts from rest, with what velocity does he pass a point 5 m from the starting point?

Answer 1

$mg\sin\alpha=ma$

$a=g\sin\alpha$

$S=\frac{v^2}{2a} \to v=\sqrt{2gS\sin\alpha}=$

$=\sqrt{2\cdot 9.81\cdot 5\cdot\sin30°}=7$ m/s. Answer