Work, Power & Energy

Ans. 10
By applying work energy theorem in wedge frame between point P and Q. $\Delta KE = W_g + W_{pseudo force}$ $\Rightarrow$ 0 = -mgR + maR $\Rightarrow$ a = g

Ans. $\frac{mgl}{18}$ Mass of dx = dx × $\frac{m}{L}$ dW = dmgh $dW = dx \times \frac{m}{L}gx$ $\int_{0}^{w} dW = \int_{0}^{L/3} \frac{m}{L}gxdx = \frac{mgL}{18}$

Ans. $-\frac{2}{9}\mu MgL$ $dM = \frac{M}{L}dx$ $\int_{0}^{w} dW = \int_{0}^{2L/3} -\mu \frac{M}{L}dxgx = \left( \frac{-2\mu MgL}{9} \right)$

Ans. (i) $2 + 24t^2 + 72t^4J$ , (ii) 48 t N, (iii) 48t + 288t³W, (iv) 1248J
v = 1 + 6t²
and a = 12t $KE = \frac{1}{2}m\left(1 + 36t^4 + 12t^2\right)$ F = 4(12 t) = 48t
P = F·v

Ans. (a) -30 kW, (b) 19.5 kW
(a) P = F·v = -T × 6
= -5000 × 6
= -30 kW

Ans. 46J
$\begin{array}{r l} & P = F \cdot v = m v \cdot \frac {d v}{d t} \\ & \Rightarrow \quad \frac {1}{2} m (v ^ {2} - u ^ {2}) = \int P d t \\ & \qquad = [ t ^ {3} - t ^ {2} + t ] _ {2} ^ {4} \end{array}$

Ans. (a) $\frac{mv^{2}}{R}$ ; (b) $\frac{\mu mv^{2}}{R}$ ; (c) $-\frac{\mu v^{2}}{R}$ ; (d) $v_{0}e^{-2\pi\mu}$
Top vies

a) $F_{c}=N$ $F_{t}=f$ $N=\frac{mv^{2}}{R}$ and $f=\mu N$
b) $f=\frac{\mu mv^{2}}{R}$
c) $a_{t}=\frac{f_{k}}{m}=\left(\frac{\mu mv^{2}}{R}\right).\frac{1}{m}$ $\vec{a}_{t}=-\frac{\mu v^{2}}{R}$ (acc. is o
d) $v\frac{dv}{ds}=-\frac{\mu v^{2}}{R}$ $v\frac{dv}{ds}=-\frac{\mu v^{2}}{R}$ $\frac{dv}{ds}=-\frac{\mu v}{R}$ $\int_{0}^{2\pi R}-\frac{\mu}{R}ds=\int_{v_{0}}^{v}\frac{1}{v}dv$ $-\frac{\mu}{R}(2\pi R)=\ln\left(\frac{v}{v_{0}}\right)$ $v=v_{0}e^{-2\pi\mu}$
$\vec{a}_t = -\frac{\mu v^2}{R}$ (acc. is opposite to velocity)

Ans. (a) $0.2N$ ; (b) $30^{\circ}$
Given $V = 36 \, km/hr$ $= 36 \times \frac{5}{18} = 10 \, m/s$ R = 50 cm
mass of block = 100 gm = 0.1 kg $\mu = 0.58$
a) Top view

$N = \frac {m v ^ {2}}{R}$
$N = \frac {(0 . 1) (1 0) ^ {2}}{5 0} = 0. 2 n e w t o n$
b) In car frame (top view)

$\frac {m v ^ {2}}{R} \cos \theta = f \quad \dots (1)$
$\frac {m v ^ {2}}{R} \sin \theta = N \quad \dots (2)$
when slipping begins then
$f = \mu N$
$\frac {m v ^ {2}}{R} \cos \theta = \mu . \frac {m v ^ {2}}{R} \sin \theta$
$\cot \theta = \mu$
$\cot \theta = 0. 5 8$
$\cot \theta = \frac {1}{\sqrt {3}}$
$\theta = 60^{\circ}$ from horizontal
$\theta = 30^{\circ}$ from vertical

Ans. (i) $f = \frac{\mu m}{L} x g$ ; (ii) $\sqrt{\mu g L}$ ; (iii) $\frac{5L}{2}$

(i)
$\lambda = m / L$

$f _ {\mathrm{k}} = \mu_ {\mathrm{k}} N$
$= \mu m ^ {\prime} g$
$\boxed {f _ {k} = \mu \frac {m}{L} g x}$
(ii) To lie completely
$W _ {\mathrm{net}} = \Delta K E$

$\int \left(\mu \frac {m g}{L} x\right) d x (- 1) = 0 - \frac {1}{2} m v _ {0} ^ {2}$
$- \frac {\mu m g}{L} \int_ {0} ^ {L} x d x = - \frac {1}{2} m v _ {0} ^ {2}$
$\frac {\mu m g}{L} \left[ \frac {L ^ {2}}{2} \right] = \frac {1}{2} m v _ {0} ^ {2}$
$v _ {0} ^ {2} = \mu g L$
$\boxed {v _ {0} = \sqrt {\mu g L}}$
(iii) $v = 2\sqrt{\mu g L}$
$W _ {\mathrm{Net}} = \Delta K E$
$W _ {\mathrm{mg}} + W _ {\mathrm{N}} + W _ {\mathrm{fk}} = \Delta K E$
$W _ {\mathrm{fk}} = K E _ {\mathrm{f}} - K E _ {\mathrm{i}}$
$\int_ {0} ^ {L} \left(\frac {\mu m g}{L} x\right) d x (- 1) + [ \mu m g ] x (- 1) = 0 - \frac {1}{2} m \left(2 \sqrt {g L \mu}\right) ^ {2}$
$\frac {\mu m g}{L} \left[ \frac {L ^ {2}}{2} \right] + \mu m g x = \frac {m \times 4}{2} (g L \mu)$
$\frac {L}{2} + x = 2 L; \boxed {x = \frac {3 L}{2}}$

Total distance travelled
$D = L + \frac {3 L}{2}; \boxed {D = \frac {5 L}{2}}$

Ans. Graph-1: For all x, Graph-2: $x < a \& x > b$ , Graph-3: $-\frac{b}{2} < x < -\frac{a}{2}$ & $\frac{a}{2} < x < \frac{b}{2}$
$E$ (total mechanical energy) $= KE + PE$
$\because \quad KE\geq 0$
$E - P E \geq 0$
$\boxed {E \geq P E}$
E must always greater than PE for particle to be exist
For graph $-1 \Rightarrow$ for $E < PE$
$x \in R$
For graph $-2 \Rightarrow$ for $E < PE$
x < a \text { and } x > b
For graph - 3 $\Rightarrow$ for $E > PE$
$x \in \left(\frac {- b}{2}, \frac {- a}{2}\right) U \left(\frac {a}{2}, \frac {b}{2}\right)$

Ans. $u = \sqrt{gL\left(2 + \frac{3\sqrt{3}}{2}\right)}$

At point A, v = u
At point B, v = v at $\theta$ with horizontal
$\frac {1}{2} m u ^ {2} = m g (L + L \cos \theta) + \frac {1}{2} m v ^ {2}$
$u ^ {2} = 2 g L (1 + \cos \theta) + v ^ {2}$
$v ^ {2} = u ^ {2} - 2 g L \left(1 + \cos \theta\right)$
$\because \quad \text{Range}, R = \frac{v^2 \sin^2 \theta}{g}$
Also, $\frac{R}{2} = L\sin \theta -\frac{L}{8}$ ... (1)
At $B, T = 0$
$m g \cos \theta = \frac {m v ^ {2}}{L}$
$\boxed {v ^ {2} = L g \cos \theta}$
$u ^ {2} - 2 g L (1 + \cos \theta) = L g \cos \theta$
$u ^ {2} = 2 g L + 2 g L \cos \theta + L g \cos \theta$
$\boxed {\frac {u ^ {2} - 2 g L}{3 g L} = \cos \theta \quad \dots (2)}$
From eq (1)
$\frac {R}{2} = L \sin \theta - \frac {L}{8}$
$\frac {2 v ^ {2} \sin \theta \cos \theta}{2 g} = L \sin \theta - \frac {L}{8}$
$\frac {(L g \cos \theta) \sin \theta \cos \theta}{g} = L \sin \theta - \frac {L}{8}$
$L \sin \theta \cos^ {2} \theta = L \sin \theta - \frac {L}{8}$
$\sin \theta \cos^ {2} \theta = \sin \theta - \frac {1}{8}$
$\sin \theta (\cos^ {2} \theta - 1) = - \frac {1}{8}$
$+ \sin \theta \times \sin^ {2} \theta = + \frac {1}{8}$
$\sin^ {3} \theta = \frac {1}{8}$
$\sin \theta = \left(\frac {1}{8}\right) ^ {\frac {1}{3}} = \frac {1}{2}$
$\boxed {\theta = 3 0 ^ {0}}$
From eq (2)
$\frac {u ^ {2} - 2 g L}{3 g L} = \cos 3 0 ^ {\circ}$
$u ^ {2} - 2 g L = \frac {\sqrt {3}}{2} \times 3 g L$
$u ^ {2} = \left(\frac {3 \sqrt {3}}{2} + 2\right) g L$
$\boxed {u = \sqrt {g L \left(\frac {3 \sqrt {3}}{2} + 2\right)}}$