A smooth 4g marble is held at rest on a smooth plane which is fixed at 30 degrees to a horizontal table. The marble is released from rest - what speed is the marble travelling at 5 seconds after being released? Let g = 9.8ms^2

This question involves forces and kinematics. First begin by drawing the situation and identifying the forces acting on the marble. The weight of the marble acts vertically downwards and is equal to the mass of the marble multiplied by the gravitational acceleration g: for now denote it as mg for simplicity. There is also the reaction force due to contact with the surface - this acts perpendicularly and away from the plane. Due to both the plane and marble being smooth, there are no frictional forces to consider.Resolving forces parallel to the plane gives us mgcos60 or mgsin30 (knowledge of basic SOH CAH TOA trig leads us to this result). This is our resultant force down the plane (component of contact force down the plane is 0). Using F=ma our acceleration down the plane is mgcos60/m=gcos60.Now for the kinematics. Identify which SUVAT equation you need. We have t=5, a=gcos60 and u=0 and we want to know v. Therefore use v=u+at. Substitute in the values to get v = 0 + 9.8cos60*5 = 24.5 m/s