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

Answer:

$v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}$

$x = \frac{2m}{c}*v_0^{\frac{3}{2}}$

Explanation:

Given

$f(v) =- cv^\frac{1}{2}$

To start with, we begin with

$F = ma$

Substitute the expression for F

$-cv^\frac{1}{2} = ma$

$-ma = cv^\frac{1}{2}$

Acceleration (a) is:

$a = \frac{dv}{dt}$

So, the expression becomes:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

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Velocity (v) is:

$v = \frac{dx}{dt}$ --- distance/time

$v * \frac{dv}{dx}= \frac{dx}{dt}* \frac{dv}{dx}$

$v * \frac{dv}{dx}= \frac{dv}{dt}$

$\frac{dv}{dt} = v * \frac{dv}{dx}$

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So, we have:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

$mv * \frac{dv}{dx} = -cv^\frac{1}{2}$

Divide both sides by $v^\frac{1}{2}$

$mv^{1-\frac{1}{2}} * \frac{dv}{dx} = -c$

$mv^{\frac{1}{2}} * \frac{dv}{dx} = -c$

Divide both sides by m

$v^{\frac{1}{2}} * \frac{dv}{dx} = -\frac{c}{m}$

$v^{\frac{1}{2}} * dv = -\frac{c}{m} * dx$

Integrate:

$\int\limits^v_{v_0} {v^{\frac{1}{2}}} \, dv = -\frac{c}{m}\int\limits^x_0 {}} \, dx$

$\frac{2}{3}v^{\frac{3}{2}}|\limits^v_{v_0} = -\frac{c}{m}x|\limits^x_0$

$\frac{2}{3}(v^{\frac{3}{2}} - v_0^{\frac{3}{2}} ) = -\frac{cx}{m}$

$v^{\frac{3}{2}} - v_0^{\frac{3}{2}} = -\frac{3cx}{2m}$

$v^{\frac{3}{2}} = v_0^{\frac{3}{2}}-\frac{3cx}{2m}$

$v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}$

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

To do this, we find the derivation by x

$\frac{dv}{dx} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2}{3} - 1} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2-3}{3}} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{3c}{2m} = 0$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{c}{m} = 0$

Divide both sides by $-\frac{c}{m}$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} = 0$

Take cube roots of both sides

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{-1} = 0$

$v_0^{\frac{3}{2}}-\frac{3cx}{2m} = 0$

$\frac{3cx}{2m} = v_0^{\frac{3}{2}}$

$x = \frac{2m}{c}*v_0^{\frac{3}{2}}$