Dynamics - Related Rates

The net force acting on the block is all due to friction, therefore the acceleration is equal to "g" multiplied by the coefficient of friction, which is negative in this case because it acts opposite the direction of the block's motion, which the problem states is positive.

Acceleration can be expressed as dv/dx multiplied by dx/dt; the latter term being equal to the velocity at any given moment.

Equating the initial kinetic energy with the given initial velocity (divided by two) to the kinetic energy involving the "dx/dt" we are looking for and dividing the product of the coefficient of friction with "g" by the result, we are given dv/dx.

Velocity of the block v, when kinetic energy is half the original, is $\dfrac{v(0)}{√2}$ . Hence $\dfrac{dv}{dx}$ = $\dfrac{1}{v}$ $\dfrac{dv}{dt}$ =-(0.3)(√2)=-0.42426

The velocity of the block is given by:

$\displaystyle \frac{1}{2} m \, v(x)^2 = \frac{1}{2} m v_o^2 - \int\limits_{0}^{x} f_r dx$ ...Eq1

Assuming a constant frictional force:

$\displaystyle \int\limits_{0}^{x} f_r dx = \mu m g x$ …Eq2

$\displaystyle \frac{1}{2} m \, v(x)^2 = \frac{1}{2} m v_o^2 - \mu m g x$ ...Eq3

Differentiate both sides of Eq3 with respect to $x$ :

$\displaystyle v \frac{dv}{dx} = -\mu g$ ...Eq4

When the KE is half of the initial KE:

$\displaystyle v^2 = \frac{1}{2} v_o^2 \implies v = \frac{1}{ \sqrt{2} } v_o$ ...Eq5

Sub Eq5 $\to$ Eq4:

$\huge \frac{dv}{dx} = \frac{-\mu g}{ \frac{1}{ \sqrt{2} } v_o } = \frac{- \text{0.3} \cdot \sqrt{2} \cdot \text{10} \, \frac{\text{m}}{ \text{s}^2 } }{ \text{10} \, \frac{\text{m}}{\text{s}}} \approx -0.424 \, \frac{\text{1}}{\text{s}}$

The force of friction slowing the block down is given by $\mu N$ , where $N$ is the normal force on the block, $mg$ . From here, we find using $F = ma$ that the acceleration of the block is $\mu g$ , which is 3 m/ $s^2$

So, applying the equations of motion:

$v_f = v_0 + at$

So, $v_f = 10-3t$

When the kinetic energy of the block is half as much, it is $\Large \frac{mv_0^2}{4}$ .

So to find the time which it takes for the kinetic energy to be lost by half due to friction,

$\Large \frac{m(v_0-3t)^2}{2} = \Large \frac{mv_0^2}{4}$

Plugging in variables and solving for $t$ , we get $t=0.976311$

$\Large \frac{dv}{dx} = \frac{dv/dt}{dx/dt} = \frac{a}{v}$

Substituting the value for the acceleration and velocity at the point of time, we get:

$\Large \frac{-3}{0.706107}$ $=\boxed{-0.42426}$