An object of mass 2 kg is held at the top of a triangular wedge, and then released. The reference level for potential energy is at the base of the triangle. Neglect friction and use g=10m/s^2. Height of triangular wedge=3m width of triangular wedge=4m. What is the amount of work done by the gravitational force as the object comes to the bottom of the wedge? Also, what is the speed of the object just before it reaches the bottom of the wedge?

Amount of force done by the grav force : W=mg(4-0)=10*2*4=80 JThe object will have lost 80 J of potential energy, converted into kinetic energy hence at the bottom mv^2/2=80 hence v^2=2*80/2=80The speed at the bottom is thus approximately 8.94 m/s

-- The object's potential energy is            (mass) x (gravity) x (height) =            (2 kg) x (10m/s²) x (3 m) = 60 joules -- As the object falls down the side of the wedge, 60 joules  is exactly the work that gravity does on it.-- Neglecting friction, 60 joules is the kinetic energy it has when it reaches the bottom.                                  Kinetic energy  =  (1/2) x (mass) x (speed)²                                                  60 J  =  (1/2) (2 kg) x (speed)²Divide each side by 1 kg:    60 m²/s²  =  (speed)²Take the sqr-rt of each side:  speed = √60 m²/s² = 7.75 m/s (rounded)You don't need to know the width of the wedge, only the height to which the object is lifted, and through which it falls.  If friction can be neglected, then it makes no difference whether the object falls straight down through the air, or along a straight slope, or through a loop-de-loop on the way down.The kinetic energy it has at the bottom is exactly the potential energy it had at the top, regardless of the route it followed on the way down.