Question

Ablok slides on ahorizonted sur fors which has been lubricated with heavy oil such that the blok soffers a viscous resistance that vories as the square root of the speed f(v)=-cv^1/2 if the initial speed of the blok is v° at time t=0 find the volues of (v) and (x) as afanctions of the time (t)

Answer 1

`v = (v_0^{(3)/(2)}-(3cx)/(2m))^(2)/(3)`

`x = (2m)/(c)*v_0^{(3)/(2)}`

Step-by-step explanation:

Given

`f(v) =- cv^(1)/(2)`

To start with, we begin with

`F = ma`

Substitute the expression for F

`-cv^(1)/(2) = ma`

`-ma = cv^(1)/(2)`

Acceleration (a) is:

`a = (dv)/(dt)`

So, the expression becomes:

`m(dv)/(dt) = -cv^(1)/(2)`

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Velocity (v) is:

`v = (dx)/(dt)` --- distance/time

`v * (dv)/(dx)= (dx)/(dt)* (dv)/(dx)`

`v * (dv)/(dx)= (dv)/(dt)`

`(dv)/(dt) = v * (dv)/(dx)`

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So, we have:

`m(dv)/(dt) = -cv^(1)/(2)`

`mv * (dv)/(dx) = -cv^(1)/(2)`

Divide both sides by
`v^(1)/(2)`

`mv^{1-(1)/(2)} * (dv)/(dx) = -c`

`mv^{(1)/(2)} * (dv)/(dx) = -c`

Divide both sides by m

`v^{(1)/(2)} * (dv)/(dx) = -(c)/(m)`

`v^{(1)/(2)} * dv = -(c)/(m) * dx`

Integrate:

`\int\limits^v_(v_0) {v^{(1)/(2)}} \, dv = -(c)/(m)\int\limits^x_0 {}} \, dx`

`(2)/(3)v^{(3)/(2)}|\limits^v_(v_0) = -(c)/(m)x|\limits^x_0`

`(2)/(3)(v^{(3)/(2)} - v_0^{(3)/(2)} ) = -(cx)/(m)`

`v^{(3)/(2)} - v_0^{(3)/(2)} = -(3cx)/(2m)`

`v^{(3)/(2)} = v_0^{(3)/(2)}-(3cx)/(2m)`

`v = (v_0^{(3)/(2)}-(3cx)/(2m))^(2)/(3)`

Next, is to get the maximum velocity by distance x.

To do this, we find the derivation by x

`(dv)/(dx) = 0`

`(2)/(3)(v_0^{(3)/(2)}-(3cx)/(2m))^{(2)/(3) - 1} * -(3c)/(2m) = 0`

`(2)/(3)(v_0^{(3)/(2)}-(3cx)/(2m))^{(2-3)/(3)} * -(3c)/(2m) = 0`

`(2)/(3)(v_0^{(3)/(2)}-(3cx)/(2m))^{(-1)/(3)} * -(3c)/(2m) = 0`

`(v_0^{(3)/(2)}-(3cx)/(2m))^{(-1)/(3)} * -(c)/(m) = 0`

Divide both sides by
`-(c)/(m)`

`(v_0^{(3)/(2)}-(3cx)/(2m))^{(-1)/(3)} = 0`

Take cube roots of both sides

`(v_0^{(3)/(2)}-(3cx)/(2m))^(-1) = 0`

`v_0^{(3)/(2)}-(3cx)/(2m) = 0`

`(3cx)/(2m) = v_0^{(3)/(2)}`

`x = (2m)/(c)*v_0^{(3)/(2)}`