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.

We have that for the 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."

it can be said that

Generally the equation for the potential energy of the block is mathematically given as

2)  the speed of the block

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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)

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