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?

- - 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) ²

Divide each side by 1 kg: 60 m²/s² = (speed) ²

Take 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.

Amount of force done by the grav force : W=mg (4-0) = 10*2*4=80 J

The 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=80

The speed at the bottom is thus approximately 8.94 m/s