Question
$\Lambda$ body moves over a plane surface, starting from rest. The frictional force is constant. The tractive force is $(P-K t)$ in gravitational units. The maximum velocity is attained after $t$ seconds. What is the distance covered before it attains the maximum speed if the mass of body is $m$.Solution Forec $-m a-m \frac{d v}{d t}-(p-K t) g \quad \therefore \quad m v-\int(P-K t) g d t-\left(P t-\frac{K t^{2}}{2}\right) g$The maximum velocity is attained when $\frac{d}{d t}\left(P t-\frac{K t^{2}}{2}\right)$ is 'rero.or $P-K t=0$ or $t=(T)=\frac{P}{K}$. But $v=\frac{d s}{d t}$$\therefore \quad m \frac{d s}{d t}-\left(P t-\frac{K t^{2}}{2}\right) g$ or $\quad$ ins $-\left(\frac{P t^{2}}{2}-\frac{K t^{3}}{6}\right) g$$\therefore \quad s-\frac{1}{i n}\left[\frac{P}{2} \frac{P^{2}}{K^{2}} \quad \frac{K}{6} \frac{P^{3}}{k^{-3}}\right] g-\frac{g}{m}\left[\frac{P^{3}}{2 K^{2}} \quad \frac{P^{3}}{6 K^{2}}\right]-\frac{g}{i n} \frac{P^{3}}{3 K^{2}}-\frac{g}{m} \frac{K^{3} T^{3}}{3 K^{2}}-\frac{g K T^{3}}{3 t h}$
Step 1
The force acting on the body is the tractive force minus the frictional force, which is given by $(P-Kt)g$. So, we have the equation $m \frac{dv}{dt} = (P-Kt)g$. Show more…
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Let the body $m$ acquire the horizontal velocity $v_{0}$ along positive $x$ - axis at the point $O$. (a) Velocity of the body $t$ seconds after the begining of the motion, $$ \vec{v}=\overrightarrow{v_{0}}+\vec{w} t=\left(v_{0}-k g t\right) \vec{i} $$ Instantaneous power $P=\vec{F} \cdot \vec{v}=(-k m g \vec{i}) \cdot\left(v_{0}-k g t\right) \vec{i}=-\operatorname{lmg}\left(v_{0}-k g t\right)$ From Eq. (1), the time of motion $\tau=v_{0} / \mathrm{kg}$ Hence sought average power during the time of motion $$ <P>=\frac{\int_{0}^{\tau}-k m g\left(v_{0}-k g t\right) d t}{\tau}=-\frac{k m g v_{0}}{2}=-2 \mathrm{~W} \text { (On substitution) } $$ From $F_{x}=m w_{x}$ or, $$ \begin{aligned} -k m g=m w_{x} &=m v_{x} \frac{d v_{x}}{d x} \\ v_{x} d v_{x}=-k g d x &=-\alpha g x d x \end{aligned} $$To find $v(x)$, let us integrate the above equation $$ \begin{aligned} \int_{v_{0}}^{v} v_{x} d v_{x} &=-\alpha g \int_{0}^{x} x d x \text { or, } v^{2}=v_{0}^{2}-\alpha g x^{2} \\ \text { Now, } \quad \vec{P}=\vec{F} \cdot \vec{v}=-m \alpha x g \sqrt{v_{0}^{2}-\alpha g x^{2}} \end{aligned} $$ For maximum power, $\frac{d}{d t}\left(\sqrt{v_{0}^{2} x^{2}-\lambda g x^{4}}\right)=0$ which yields $x=\frac{v_{0}}{\sqrt{2 \alpha g}}$ Putting this value of $x$, in Eq. (2) we get, $$ P_{\max }=-\frac{1}{2} m v_{0}^{2} \sqrt{\alpha g} $$
Physical Fundamentals Of Mdchanics
Laws of Conservation of Energy, Momemtum, and Angular Momentum
Find the acccleration of the block of mass $M$ assuming friction to be absent. Solution When $m^{\prime}$ moves downward, the block $M$ moves to the right. If $M$ moves to the right by $x, m$ moves $2 x$ and $m^{\prime}$ moves $3 x$ down the plane. Loss of P.E. $-\left(m^{\prime} g\right) 3 x \sin \alpha$ gain in K.E. $-\frac{1}{2} M v^{2}+\frac{1}{2} m(2 v)^{2}+\frac{1}{2} m^{\prime}\left[(3 v)^{2}+v^{2}+2 \times 3 v \cdot v \cos (180-\alpha) \mid\right.$ $-\frac{1}{2} M v^{2} \mid 2 m v^{2}, \frac{m^{\prime}}{2} v^{2}[10 \quad 6 \cos \alpha]-\frac{y^{2}}{2}\left[M|4 m| m^{\prime}(10 \quad 6 \cos \alpha)\right]$ $\therefore \quad m^{\prime} g \cdot 3 x \sin \alpha-\frac{v^{2}}{2}\left[M|4 m| m^{\prime}(10 \quad 6 \cos \alpha)\right]$ $\therefore \quad y^{2}-\frac{2 m^{\prime} g \cdot 3 x \sin \alpha}{M+4 m+m^{\prime}(10-6 \cos \alpha)} \quad$ But $v^{2}-2 a x$, where $a$ is acceleration of $M$. $\therefore \quad a-\frac{3 m^{\prime} g \sin \alpha}{M|4 m| m^{\prime}(10 \quad 6 \cos (x)}$
Work, Energy, Power and Circular Motion
Section A
Case 1. When the body is launched up : Let $k$ be the coefficeint of friction, $u$ the velocity of projection and $l$ the distance traversed along the incline. Retarding force on the block $=m g \sin \alpha+k m g \cos \alpha$ and hence the retardation $=g \sin \alpha+k g \cos \alpha$ Using the equation of particle kinematics along the incline, or, $$ \begin{gathered} 0=u^{2}-2(g \sin \alpha+k g \cos \alpha) l \\ l=\frac{u^{2}}{2(g \sin \alpha+k g \cos \alpha)} \end{gathered} $$ and $$ 0=u-(g \sin \alpha+k g \cos \alpha) t $$ or, $u=(g \sin \alpha+k g \cos \alpha) t$ Using (2) in (1) $l=\frac{1}{2}(g \sin \alpha+k g \cos \alpha) t^{2}$ Case (2). When the block comes downward, the net force on the body $=m g \sin \alpha-k m g \cos \alpha$ and hence its acceleration $=g \sin \alpha-k g \cos \alpha$ Let, $t$ be the time required then, $$ l=\frac{1}{2}(g \sin \alpha-k g \cos \alpha) t^{\prime 2} $$ From Eqs. (3) and (4) $$ \frac{t^{2}}{t^{\prime 2}}=\frac{\sin \alpha-k \cos \alpha}{\sin \alpha+k \cos \alpha} $$ $$ \text { But } \frac{t}{t^{\prime}}=\frac{1}{\eta} $$ (according to the question), Hence on solving we get $$ k=\frac{\left(\eta^{2}-1\right)}{\left(\eta^{2}+1\right)} \tan \alpha=0 \cdot 16 $$
The Fundamental Equation of Dynamics
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