Question

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).
1- For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.
Express your answer in terms of g and the variables m, l, x, and θ.
(U^G=?)
2-
Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)
Express your answer in terms of g and the variables m, l, and θ.

Answer 1

Answer:

potential enrgy   U = m g L sin θ

speed     V = √(2g L sin θ)

Explanation:

The expression for the gravitational potential energy of a body is

                    U = mg Y - mg Yo

Where Y give us a constant initial energy from which the differences are measured, for general simplicity it is selected as zero, Yo= 0

What we find an expression for height, let's use trigonometry

                       sin  θ= Y / L

                       Y = l sin θ

   We substitute in the power energy equation

                     U = m g L sin θ

                         

2. The mechanical energy of the system is conserved, so we will write the mechanical energy at two points the highest and the lowest

Highest           Em = U

Lower              Em = K

                         U = K

                         m g L sin θ = ½ m v²

                         V = √(2g L sin θ)