Work, Power & Energy

$t = 0$
(Before collision) $t$
$\bigoplus_ {v = g t}$
$K = \frac {1}{2} m g ^ {2} t ^ {2}$
$K\propto t^{2}$ : parabolic graph
then during collision kinetic energy first decreases to elastic potential energy and then increases.

$W=\int_0^L Fdx=\int_0^L(ax+bx^2)dx$
$\Rightarrow W=\dfrac{aL^2}{2}+\dfrac{bL^3}{3}$
Hence, the correct answer is (C).

Applying Energy Theorem, we get
$\begin{array}{r l} & m g h = \frac {1}{2} m v ^ {2} \\ \Rightarrow & v = \sqrt {2 g R (\cos \theta - \sin \phi)} \end{array}$
At point Y, we have along radial direction in circular motion
$\frac {m v ^ {2}}{R} = m g \sin \phi$
$\Rightarrow 2mg(\cos \theta -\sin \phi) = mg\sin \phi$
$\Rightarrow$ $2\cos\theta=3\sin\phi$
Hence, the correct answer is (C).

To find KE of spring, we consider an elemental part of spring as shown.
Mass of element is $dm = \frac{m}{\ell} dx$

As spring is uniformly stretched, the speed $v'$ of element is
$v ^ {\prime} = \left(\frac {v}{\ell}\right) x$
Kinetic energy of element is $dK = \frac{1}{2}dmv'^{2}$
$\Rightarrow d K = \frac {1}{2} \left(\frac {m}{\ell} d x\right) \left(\frac {v}{\ell} x\right) ^ {2}$
$\Rightarrow \int d K = \frac {1}{2} \frac {m v ^ {2}}{\ell^ {3}} \int_ {0} ^ {\ell} x ^ {2} d x$
Total kinetic energy of spring is given as
$K = \frac {1}{2} \frac {m v ^ {2}}{\ell^ {3}} \left(\frac {\ell^ {3}}{3}\right) = \frac {1}{6} m v ^ {2}$
Hence, the correct answer is (D).

Wind volume flowing through the cross-sectional area of blades (rotating) of wind mill per unit time is given by
V = A v

If $\rho$ be the density of air, then mass of wind intercepting the blades per second is
$m = \rho A v$
Kinetic energy of wind mass intercepting the section of blades per second is
$\frac {d K}{d t} = \frac {1}{2} m v ^ {2} = \frac {1}{2} \rho A v ^ {3}$
$\Rightarrow$ Electrical power $\propto v^{3}$
Hence, the correct answer is (D).

If plank thickness is $x$ , then, we have

$\begin{array}{r l} & u ^ {2} \left(1 - \frac {1}{n}\right) ^ {2} = u ^ {2} - 2 a x \\ \Rightarrow & 2 a x = u ^ {2} \left[ 1 - \left(\frac {n - 1}{n}\right) ^ {2} \right] \\ \Rightarrow & 2 a x = u ^ {2} \left[ \frac {2 n - 1}{n ^ {2}} \right] \end{array}$
Using constant retardation, we have

$\begin{array}{r l} & 0 = u ^ {2} - 2 a (N x) \\ \Rightarrow & u ^ {2} = N u ^ {2} \bigg [ \frac {2 n - 1}{n ^ {2}} \bigg ] \\ \Rightarrow & N = \frac {n ^ {2}}{2 n - 1} \end{array}$
Hence, the correct answer is (A).

Applying Work-Energy Theorem from point A to C, we get

$\Rightarrow L = \frac {v _ {0} ^ {2}}{2 \mu g} + \frac {h}{\mu}$
Hence, the correct answer is (B).

Let at any instant, the bead make an angle $\theta$ with the vertical, as shown in Figure.

Since $h = R - R\cos \theta$
Applying Law of Conservation of energy, we get
$m g R (1 - \cos \theta) = \frac {1}{2} m v ^ {2}$
According to Newton's Second Law, we have
$\sum F _ {\mathrm{towardsthecentre}} = m a _ {c} = \frac {m v ^ {2}}{R}$
$\Rightarrow m g \cos \theta - N = \frac {m v ^ {2}}{R}$
where, $N$ is the normal force on bead by the wire
$\Rightarrow N = m g \cos \theta - \frac {m v ^ {2}}{R}$
Since $v^{2} = 2gR(1 - \cos \theta)$
$\Rightarrow N = m g (3 \cos \theta - 2)$
Now, $N = 0$ at $\cos \theta = \frac{2}{3}$
The normal force N will act radially outwards on bead, when
$\cos\theta > \frac{2}{3}$ and normal force N will act radially inwards on bead, when
$\cos \theta < \frac {2}{3}$
So, force acting on ring is opposite to normal force acting on bead.
Hence, the correct answer is (C).

Ans. (C)
$a _ {B} = \frac {v _ {B} ^ {2}}{R} = 3 g$
$v _ {B} ^ {2} = 3 g R$
Applying COE at A & B
$m g R + \frac {1}{2} m v _ {A} ^ {2} = \frac {1}{2} m v _ {B} ^ {2} = \frac {1}{2} m \times 3 g R$
$\frac {1}{2} m v _ {A} ^ {2} = \frac {m g R}{2}$
$v _ {A} = \sqrt {g R}$

Ans. (C)
Here centripetal force is provided by normal reaction of vertical wall.
(i) Normal reaction of floor $N_{F} = mg$
(ii) Normal reaction of vertical wall $N_{W}=\frac{mv^{2}}{r}$ .

Ans. (ABCD)
w.r.t B :

$\omega_ {A, B} = \frac {v \cos 3 0 ^ {\circ}}{a}$

Here $\frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}$
That's why force is conservative and particle is moving on circle i.e. $x^{2} + y^{2} = a^{2}$
Hence $W = \int_{a}^{0} F_{x} dx + \int_{0}^{a} F_{y} dy$
$= \frac {K}{a ^ {3}} \int_ {a} ^ {0} x d x + \frac {K}{a ^ {3}} \int_ {0} ^ {a} y d y = 0.$

By definition, $P = \frac{dw}{dt} = \frac{dk}{dt} = \text{constant}$ .

(ad) After elastic collision, block speed is $0.5u_{0}$

So when it will come back to equilibrium point its speed will be $u_0$
Displacement $x = A\sin \omega t$
Velocity $v = A\omega \cos \omega t = \frac{\omega A}{2}$
At the time of collision
$\cos \omega t = \frac {1}{2}$
$t = \frac {\pi}{3} \sqrt {\frac {m}{k}}$
$\omega t = \frac {\pi}{3} \Rightarrow t = \frac {\pi}{3 \omega} = \frac {\pi}{3} \sqrt {\frac {m}{k}}$
Time to reach equilibrium position first time= $\frac{2\pi}{3}\sqrt{\frac{m}{k}}$
Second time it will reach at time $= \frac{2\pi}{3}\sqrt{\frac{m}{k}} +\pi \sqrt{\frac{m}{k}}$
$= \frac{5\pi}{3}\sqrt{\frac{m}{k}}$ (So it is correct).

Maximum energy loss = $\frac{P^{2}}{2m} - \frac{P^{2}}{2(m + M)}$
$= \frac {P ^ {2}}{2 m} \left[ \frac {M}{(m + M)} \right] = \frac {1}{2} m v ^ {2} \left\{\frac {M}{m + M} \right\} \quad \left(f = \frac {M}{m + M}\right)$

Using work energy theorem
$MgR\sin 30^{\circ} + W_{f} = \frac{1}{2} MV^{2}$
$2 0 0 - 1 5 0 = \frac {1}{2} V ^ {2}$
$\Rightarrow V = 1 0 m / s.$

$N - \frac{Mg}{2} = \frac{M(10)^{2}}{40} \Rightarrow N = 7.5N.$

Maximum energy loss is
$\frac {1}{2} \frac {M m}{M + m} (v - 0) ^ {2} = \frac {M}{M + m} \left(\frac {1}{2} m v ^ {2}\right)$
So, Statement-I is wrong.
Hence, the correct answer is (D).

If a is upward acceleration of man before leaping, then
$v = \sqrt {2 a h _ {1}} = \sqrt {2 g h _ {2}}$

$\Rightarrow a = \frac {g h _ {2}}{h _ {1}} = \frac {1 0 \times 1}{0.5} = 2 0 \mathrm{ms} ^ {- 2}$
Velocity at take-off is
$v = \sqrt {2 \times 1 0 \times 1} = \sqrt {2 0} \mathrm{ms} ^ {- 1}$
Power delivered by muscles is
P = F v = m a v
$\Rightarrow P = 7 0 \times 2 0 \times \sqrt {2 0}$
$\Rightarrow P = 6.26\times 10^{3}\mathrm{W}$
Hence, the correct answer is (B).

Since, $\vec{r} = \overrightarrow{OP} = x\hat{i} + y\hat{j}$
$\Rightarrow \vec {F} = \frac {k}{(x ^ {2} + y ^ {2}) ^ {\frac {3}{2}}} (x \hat {i} + y \hat {j}) = \frac {k}{r ^ {3}} (\vec {r})$
$\Rightarrow \vec {F} | | \vec {r}$
Since, $\vec{F}$ is along $\vec{r}$ or in radial direction, so work done is zero.
Hence, the correct answer is (D).