Answer:
psitionx = xo exp [ (m/2k) (V²-Vo²)]
Step-by-step explanation:
we write the second law of newton
F = ma
-k/x = ma
a = (-k/x) / m = -k/(m x)
Let's look for speed
a = dv / dt
a = dv/dx dx/dt
the speed definition
v = dx / dt
a = dv/dx v
v dv = adx
We integrate between the initial limits X = 0 has speed Vo and end x has speed v
∫ v dv = ∫ (-k/[m x]) dx
½ v2 = (-k/m) ln x
Simplifying and evaluating
V² - Vo² = (-2k/m) (ln x- ln xo)
the position
Ln x = ln xo + (m/2k) (V²-Vo²)
x = xo exp [ (m/2k) (V²-Vo²)]
B) We already found it
V²-Vo² = (-2k/m) (ln x- lnxo)
V² = Vo² - (2k/m) (ln x- lnxo)
Vo =0
V² = - (2k/m) (ln x- lnxo)
Let's calculate the mechanical energy at the point
Initial
Em1 = K + U
The kinetic energy is K = ½ m V²
The potential energy can be calculated from the expression
F = -dU / dx
du = -F dx
dU = - (-k/x) dx
U- Uo = k ln x -ln xo
Em1 = ½ m Vo² + k ln xo
Final point
Em2 = ½ m Vf²+ k ln x
We subtract the energies at the colon
Em2 -Em1 = ½ m Vf² + k ln x - k ln xo
we substitute the speed value
Em2 -Em1 = ½ m [- (2k/m) (ln x- lnxo)] + k ln x - k ln xo
Em2 -Em1 = -k (ln x -ln xo) + k ln x -lnxo
Em2 -Em1 = 0
Em2 = Em1
This variation of mechanical energy is cero, which is in accordance with the law of conservation of energy
We calculate the acceleration
a = k / (m x)
a = (k / m) 1 / {xo exp [ (m/2k) (V²-Vo²)] }
a = (k/mxo) exp (-m/2k) exp (Vo²-V²)
Vo=0
a = (k/mxo) exp (-m/2k) exp (-V²)
a = (k/mxo) exp (-m/2k) 2exp (1/V)
We calculate the acceleration
a = k / (m x)
a = (k / m) 1 / {xo exp [ (m/2k) (V²-Vo²)] }
a = (k/mxo) exp (-m/2k) exp (Vo²-V²)
Vo=0
a = (k/mxo) exp (-m/2k) exp (-V²)
a = (k/mxo) exp (-m/2k) 2exp (1/V)