Respuesta :
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)
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
- 1) [tex]U_G = mg(l - x)sin\theta[/tex]
- 2) [tex]v = \sqrt{2glsin\theta}[/tex]
Generally the equation for the potential energy of the block is mathematically given as
[tex]U_G = mg(l - x)sin\theta[/tex]
2) the speed of the block
[tex]v = \sqrt{2glsin\theta}[/tex]
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