Aparticle of mass m is initially at rest a distance xo from the origin. it then moves along the x axis under the influence of a repulsive force f-k/x (a) find (b) starting from your answer to (a), find the velocity a long time after the particle is released, and confirm that your answer agrees with conservation of energy. (c) starting from your answer to (a), find the initial acceleration, and physically interpret your result.

     psitionx = xo exp [ (m/2k) (V²-Vo²)]

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)

d.7.5cm is your answer

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.