A particle moves on the positive x axis (x > 0). A force F(x) = + k / x2 pushes

Question

A particle moves on the positive x axis (x > 0). A force F(x) = + k / x2 pushes the particle to larger x. Note that the force decreases to 0 as x approaches infinity.

Suppose the initial conditions at t = 0 are x(0) = x0 and (dx/dt)0 = 0.
Solve for the motion of the particle.

(A) Calculate the time t when v is equal to 0.50 v(infinity).

[Hint:
The position (x) and time (t) can be related by parametric equations
x = x0 cosh2? = (x0/2){ cosh2?+ 1 }
and t = C { sinh2? + 2?}
where C is a constant, which you will need to determine.]

Explanation / Answer

m = 0.51 kg; k = 1.43 Nm2; x0 = 1.7 m

F = mv/dt = mdv/dx dx/dt = mv dv/dx = k/x2

mv dv = k dx/x2

integrate,

mv2/2 = -k/x + C

use x = x0, v = 0, so C = k/x0

mv2 = 2k/x0 * (1 - x0/x)

when x --> , v = vf = (2mk/x0) = 0.9263 m/s

v = vf * (1 - x0/x)

when v = 50%*vf, x0/x = 3/4, x = 4*x0/3 = 

dx/dt = vf * (1 - x0/x)

dx/(1 - x0/x) = vf dt

integrate,

x(1 - x0/x) + (1/2) x0 ln|2x[1 + (1 - x0/x)] - x0| = vf*t + C

use t = 0, x = x0, C = (1/2) x0 ln|x0|

x(1 - x0/x) + (1/2) x0 ln|2x/x0 [1 + (1 - x0/x)] - 1| = vf*t

when x = 4*x0/3,

t = x0/vf * [2/3 + 1/2 * ln(5/3)] = 1.69 s

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