Name: Inertial Frames of Reference
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Imagine if you are on a bus or a train, can you say you are in an inertial frame of reference? If yes, how can you predict it? Let us explore more in this vi...

Question

A small body of mass $m$ is located on a horizontal plane at the point $O$ . The body acquires a horizontal velocity $v_{0}$ . Find,

$(a)$ The mean power developed by the friction force during the whole time of motion, if the friction coefficient $k = 0.27, m = 1.0 kg$ , and $v_{0} = 1.5 m s^{- 1}$ ,

$(b)$ The maximum instantaneous power developed by the friction force, if the friction coefficient varies as $k = α x$ , where $α$ is a constant, and $x$ is the distance from the point $O$ .

Accepted Answer

Let us assume that the mass $m$ has an initial velocity $v_{0}$ in the positive $x$ -axis direction from the point $(a)$ .

The only force acting in horizontal direction is kinetic friction $f = k N (- \hat{i}) = - (k m g) \hat{i}$ .

$(a)$ Now, at the time $t$ its velocity and acceleration $(w)$ will be, $w = - \frac{f}{m} = - \frac{k N}{m} = - \frac{k m g}{m} = - k g \hat{i}$

$\vec{v} = \vec{v_{0}} + \vec{w} t = (v_{0} - k g t) \hat{i} . . . . . . (1)$

By the definition of instantaneous power,

$P = \vec{F} \cdot \vec{v} = (- k m g \hat{i}) \cdot (v_{0} - k g t) \hat{i} \Rightarrow P = - k mg (v_{0} - k g t)$

This is the instantaneous power at any time $t$ .
Now, from the equation $(1)$ , the total time taken $α$ by mass to come to rest will be,

$0 = v_{0} - k g t$

$\Rightarrow τ = \frac{v_{0}}{k g}$ .

Hence, the asked average power during the time of motion will be,

$< P > = \frac{\int P d t}{\int d t} = \frac{\int_{0}^{τ} - k m g (v_{0} - k g t) d t}{τ}$

$\Rightarrow < P > = \frac{- k m g (v_{0} τ - \frac{k g τ^{2}}{2})}{τ} \Rightarrow < P > = - k m g (v_{0} - \frac{v_{0}}{2}) = - \frac{k m g v_{0}}{2}$

on substitution,

$< P > = - \frac{0.27 \times 1 \times 9.8 \times 1.5}{2} = - 1.9845 W$

$(b)$ We know force $F_{x} = m w_{x}$ ,

Kinetic friction force $= m w_{x} = m \frac{d v_{x}}{d x} \times \frac{d x}{d t}$ .

$\Rightarrow - k m g = m v_{x} \frac{d v_{x}}{d x}$

$\Rightarrow v_{x} d v_{x} = - k g d x ≐ - α g x d x$ .

To find $v (x)$ , let us integrate the above equation,

$\int_{v_{0}}^{v} v_{x} d v_{x} = - α g \int_{0}^{x} x d x$

$\Rightarrow v^{2} = v_{0}^{2} - α g x^{2} . . . . . . (1)$

Now, $\vec{P} = \vec{F} \cdot \vec{v} = - m α x g \sqrt{v_{0}^{2} - α g x^{2}}$ (kinetic friction and velocity are in opposite direction)

$\Rightarrow \vec{P} = - m α g \sqrt{v_{0}^{2} x^{2} - α g x^{4}} . . . . . . . (2)$

For the maximum power,

$\frac{d}{d t} (v_{0}^{2} x^{2} - α g x^{4}) = 0$ ,

$\Rightarrow 2 v_{0}^{2} x - 4 α g x^{3} = 2 x (v_{0}^{2} - 2 α g x^{2}) = 0$

Which yields $x = \frac{v_{0}}{\sqrt{2 α g}}$ .

Substituting the above value of $x$ , in the equation $(2)$ , we get,

$P_{\max} = - m α g (\frac{v_{0}}{\sqrt{2 α g}}) \sqrt{v_{0}^{2} - α g {(\frac{v_{0}}{\sqrt{2 α g}})}^{2}} = - \frac{1}{2} m v_{0}^{2} \sqrt{α g}$ .

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