Work, Power & Energy

$\begin{array}{l} a = \frac {1 0 g - 5 g - 2 5}{1 5} = \frac {1 0 0 - 5 0 - 2 5}{1 5} = \frac {5}{3} \mathrm{ms} ^ {- 2} \\ \text {Since,} v ^ {2} - 0 ^ {2} = 2 a s \\ \Rightarrow \quad v ^ {2} = 2 \left(\frac {5}{3}\right) (2) = \frac {2 0}{3} \\ \Rightarrow \quad v = 2.6 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (A).

Power $P = Fv = \frac{K}{v}v = K = constant$ $\Rightarrow W = Pt = Kt$ Hence, the correct answer is (C).

For month of June (having 30 days), each day motor is operated for 8 hrs. So total time (t) for which motor is operational in June is
$t = (8) (3 0) = 2 4 0 \mathrm{hrs}$
So, total energy consumed in the month of June is equal to work done (1 HP = 746 W)
$W = P t = (2 \times 7 4 6) (2 4 0) \mathrm{watt-hour}$
Since 1 unit of electricity (=1 kw-hr) costs ₹6,
So, bill for the month of June is
$\frac{2 \times 746 \times 240}{1000} \times 6 = 2148Hence, the correct answer is (A).$

Given $\ell = 1\mathrm{m}$
$\begin{array}{r l} \Rightarrow & h = \ell (\cos 3 0 ^ {\circ} - \cos 6 0 ^ {\circ}) \\ \Rightarrow & h = 1 (0.8 6 - 0.5) = 0.3 6 \mathrm{m} \end{array}$
CHAPTER 1

$\begin{array}{r l} & {\mathrm{Since,} v = \sqrt {2 g h} = \sqrt {2 (1 0) (0.3 6)}} \\ & {\Rightarrow v = 2.6 8 \mathrm{ms} ^ {- 1}} \\ & {\mathrm{Also,} P = F v = (m g \sin \theta) v} \\ & {\Rightarrow P = (m g \sin 3 0 ^ {\circ}) v = (1) (1 0) \bigg (\frac {1}{2} \bigg) (2.6 8)} \\ & {\Rightarrow P = 1 3.4 \mathrm{W}} \end{array}$
Hence, the correct answer is (B).

Velocity of ball to just reach the top of the tube should be given by Law of Conservation of Energy, according to which
$\begin{array}{l l} & (U + K) _ {i} = (U + K) _ {f} \\ \Rightarrow & m g h + \frac {1}{2} m v ^ {2} = m g (2 R) + \frac {1}{2} m (0) ^ {2} \\ \Rightarrow & v = \sqrt {2 g (2 R - h)} \end{array}$
Hence, the correct answer is (D).

Since velocity is constant so the kinetic energy does not change. If $W_{F}$ be the work done by external force and $W_{f}$ be the work done by friction, then
$\begin{array}{r l} & W _ {F} + W _ {f} = \Delta K = 0 \\ \Rightarrow & W _ {F} = - W _ {f} \\ \Rightarrow & W _ {F} = 0.5 \mu_ {k} M g x + \mu_ {k} M g x + 0.5 \mu_ {k} M g x \\ \Rightarrow & W _ {F} = 2 \mu_ {k} M g x \end{array}$
Hence, the correct answer is (B).

According to Work Energy Theorem, we have
$\begin{array}{r l} & W _ {\mathrm{total}} = \Delta K \\ \Rightarrow & W _ {\mathrm{ext}} + W _ {n c} = \Delta U + \Delta K \\ \Rightarrow & F x - \mu m _ {1} g x = \frac {1}{2} k x ^ {2} + 0 \\ \Rightarrow & F x - \mu m _ {1} g x - \frac {1}{2} k x ^ {2} = 0 \\ \text {But} k x = \mu m _ {2} g, \text {for just shifting} m _ {2} \\ \Rightarrow & F x - \mu m _ {1} g x = \frac {1}{2} (\mu m _ {2} g) x \\ \Rightarrow & F = \mu \bigg (m _ {1} + \frac {m _ {2}}{2} \bigg) g \\ \Rightarrow & F = 0.4 \bigg (1 + \frac {2}{2} \bigg) (1 0) = 8 \mathrm{N} \end{array}$
Hence, minimum constant force is 8 N.
Hence, the correct answer is (A).

When the brick just reaches at the required height, the man saves the excess kinetic energy possessed by the brick.
So, percentage energy saved is
$\begin{array}{l} \frac {K E}{K E + P E} \times 100 = \frac {\frac {1}{2} m v ^ {2}}{\frac {1}{2} m v ^ {2} + m g h} \times 100 \% \\ \% \text {saved} = \frac {v ^ {2}}{v ^ {2} + 2 g h} \times 100 \% \\ \% \text {saved} = \frac {(10) ^ {2}}{(10) ^ {2} + (2)(10)(5)} \times 100 = 50 \% \end{array}$
Hence, the correct answer is (D).

$\text {Work done = Change in potential energy = \Delta U}$
$\Rightarrow \Delta U = m g L - m g L \cos \theta $
$\Rightarrow \Delta U = m g L (1 - \cos \theta) $
Hence, the correct answer is (A).

In the critical case, velocity at topmost point should be zero i.e.,
v = 0

Using Law of Conservation of Energy, we get
$\begin{array}{r l} & {\frac {1}{2} m u ^ {2} - \frac {1}{2} m v ^ {2} = m g h} \\ {\Rightarrow} & {v ^ {2} = u ^ {2} - 2 g h} \\ {\Rightarrow} & {0 = u ^ {2} - 2 g \ell} \\ {\Rightarrow} & {u = \sqrt {2 g \ell}} \end{array}$
$\left\{\because h = \ell \right\}$
Hence, the correct answer is (B).

For a simple pendulum, when string is vertical, tension in the string is
$T _ {1} = m g + \frac {m v ^ {2}}{L}$
In extreme position (v = 0), so tension is
$T _ {2} = m g \cos \theta$
Given that, $T_{1}=2T_{2}$
$\Rightarrow m g + \frac {m v ^ {2}}{L} = 2 m g \cos \theta$
The speed (v) at the lowermost point is given by
$\begin{array}{r l} & v = \sqrt {2 g L (1 - \cos \theta)} \\ \Rightarrow & m g + \frac {m}{L} [ 2 g L (1 - \cos \theta) ] = 2 m g \cos \theta \\ \Rightarrow & 1 + 2 - 2 \cos \theta = 2 \cos \theta \\ \Rightarrow & \cos \theta = \frac {3}{4} \end{array}$
Hence, the correct answer is (D).

Total mechanical energy = mgH
$\begin{array}{l} \text { Given that } \frac {P E}{K E} = \frac {1}{2} \\ \Rightarrow P E = \frac {1}{3} m g H \text { and } K E = \frac {2}{3} m g H \end{array}$
So, height from the ground at this instant is $h=\frac{H}{3}$ and speed of particle is
$v = \sqrt {2 g \left(\frac {2 H}{3}\right)} = 2 \sqrt {\frac {g H}{3}}$
Hence, the correct answer is (C).

Decrease in potential energy = Increase in kinetic energy
$\begin{array}{r l} \Rightarrow & \frac {1}{2} m v ^ {2} = m g \left(\ell - \frac {\ell}{2}\right) \\ \Rightarrow & v = \sqrt {g \ell} \end{array}$
Hence, the correct answer is (B).

If $x_{i}$ be the compression in the spring initially, then, in equilibrium,
$\begin{array}{r l} & F + m _ {1} g = k x _ {i} \\ \Rightarrow & x _ {i} = \frac {F + m _ {1} g}{k} \end{array}\tag{... (1}$
Let $x_{f}$ be the final elongation in spring when lower block just lifts. For this to happen, we have
$\begin{array}{r l} & k x _ {f} = m _ {2} g \\ \Rightarrow & x _ {f} = \frac {m _ {2} g}{k} \end{array}\tag{... (2}$
Now by Law of Conservation of Energy
$\begin{array}{l} \left( \begin{array}{c} \text {Increase in} \\ \text {Gravitational} \\ \text {Potential} \\ \text {Energy of m} _ {1} \end{array} \right) = \left( \begin{array}{c} \text {Decrease in} \\ \text {Elastic Potential} \\ \text {Energy of Spring} \end{array} \right) \\ \Rightarrow \quad m _ {1} g (x _ {i} + x _ {f}) = \frac {1}{2} k (x _ {i} ^ {2} - x _ {f} ^ {2}) \end{array}$
$x _ {i}$
$x _ {f}$
(1) and (2), we get
$F = (m _ {1} + m _ {2}) g$
Hence, the correct answer is (C).

$\begin{array}{r l}&W _ {n c} = W _ {\mathrm{friction}} = \Delta U + \Delta K\\&\Delta U = (\Delta U) _ {A \rightarrow B} + (\Delta U) _ {B \rightarrow C}\\&\Rightarrow \quad \Delta U = U _ {B} - U _ {A} + U _ {C} - U _ {B}\\&\Rightarrow \quad \Delta U = U _ {C} - U _ {A} = 0 - (2) (1 0) (1)\\&\Rightarrow \quad \Delta U = - 2 0 \mathrm{J}\\&\Rightarrow \quad \Delta K = (\Delta K) _ {A \rightarrow B} + (\Delta K) _ {B \rightarrow C}\\&\Rightarrow \quad \Delta K = K _ {B} - K _ {A} + K _ {C} - K _ {B}\\&\Rightarrow \quad \Delta K = K _ {C} - K _ {A} = 0 - 0\\&\Rightarrow \quad \Delta K = 0\end{array}$
$\mathrm{So}, W _ {n c} = - 2 0 \mathrm{J}$
Hence, the correct answer is (B).

Since, $\frac{3}{4}$ th energy is lost i.e., $\frac{1}{4}$ th kinetic energy is left with the particle. Hence, its velocity becomes $\frac{v_0}{2}$ under a retardation of $\mu g$ in time $t_0$ .
$\begin{array}{r l} \Rightarrow & \frac {v _ {0}}{2} = v _ {0} - \mu g t _ {0} \\ \Rightarrow & \mu g t _ {0} = \frac {v _ {0}}{2} \\ \Rightarrow & \mu = \frac {v _ {0}}{2 g t _ {0}} \end{array}$
Hence, the correct answer is (B).

$\begin{array}{r l} & F _ {d o w n} = m g \sin \theta = \frac {3}{5} m g \\ & f _ {\max} = \mu m g \cos \theta = \left(\frac {3}{4}\right) \left(\frac {4}{5}\right) m g = \frac {3}{5} m g \\ & \Rightarrow f _ {\max} = m g \sin \theta = \frac {3}{5} m g \end{array}$
Block will move upwards when
$k x _ {0} = f _ {\mathrm{max}} + m g \sin \theta = \frac {6}{5} m g\tag{... (1}$
By Law of Conservation of Energy, we have
$\begin{array}{r l} & M g x _ {0} = \frac {1}{2} k x _ {0} ^ {2} \\ \Rightarrow & M = \frac {k x _ {0}}{2 g} = \frac {\frac {6}{5} m g}{2 g} = \frac {3 \mathrm{m}}{5} \end{array}$
Hence, the correct answer is (A).

$\begin{array}{l l} \text {Retardation} \propto x \\ \Rightarrow & a = - k x \\ \Rightarrow & v \frac {d v}{d x} = - k x \\ \Rightarrow & v d v = - k x d x \\ \Rightarrow & \int v d v = - k \int x d x \\ \Rightarrow & \frac {v ^ {2}}{2} = - \frac {k x ^ {2}}{2} \\ \Rightarrow & \frac {m v ^ {2}}{2} = K. E. = - k m \frac {x ^ {2}}{2} \\ \Rightarrow & K. E. \propto x ^ {2} \end{array}$
Hence, the correct answer is (B).

Potential energy of particle at $x = \sqrt{\frac{2E}{k}}$ i.e., for x > 0 is zero
So, Total Energy = Kinetic Energy = E $\Rightarrow \frac{1}{2}mv^{2}=E$ $\Rightarrow v=\sqrt{\frac{2E}{m}}$ Hence, the correct answer is (B).

$P = (F_{\text{net}})v$ where $F_{net} = mg + 0.1mg + 0.02mg$ $\Rightarrow \quad F_{net} = 1.12mg$ $\Rightarrow \quad P = (1.12)(1000)(10)(10)$ $\Rightarrow \quad P = 112 kW$ Hence, the correct answer is (A).

Unit vector corresponding to $\hat{i} +\hat{j} +\hat{k}$ is $\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$ So, $\vec{F} = \frac{30}{\sqrt{3}} (\hat{i} +\hat{j} +\hat{k}) = 10\sqrt{3} (\hat{i} +\hat{j} +\hat{k})$ Since, $W = \vec{F}\cdot \Delta \vec{r} = \vec{F}\cdot (\vec{r}_2 - \vec{r}_1)$ $\Rightarrow W = 10\sqrt{3}\bigl (\hat{i} +\hat{j} +\hat{k}\bigr)\cdot (2\hat{i} +\hat{k})$ $\Rightarrow W = 10\sqrt{3}(2 + 1) = 30\sqrt{3} J$ Hence, the correct answer is (B).

Tension is maximum at the lowest point, so
$T _ {L} = \frac {m}{r} \Big (u ^ {2} + g r \Big)$
Tension is minimum at the highest point, so
$T _ {H} = \frac {m}{r} \Big (u ^ {2} - 5 g r \Big)$
Given that $\frac{T_L}{T_H} = 4$
$\begin{array}{r l} & {\Rightarrow \frac {T _ {L}}{T _ {H}} = \frac {\frac {m}{r} (u ^ {2} + g r)}{\frac {m}{r} (u ^ {2} - 5 g r)} = 4} \\ & {\Rightarrow 4 u ^ {2} - 2 0 g r = u ^ {2} + g r} \\ & {\Rightarrow 3 u ^ {2} = 2 1 g r} \\ & {\Rightarrow u = \sqrt {7 g r}} \\ & {\mathrm{Since} v ^ {2} = u ^ {2} - 2 g h} \\ & {\Rightarrow v = \sqrt {7 g r - 2 g (2 r)} = \sqrt {3 g r}} \\ & {\Rightarrow v = \sqrt {3 (1 0) (\frac {1 0}{3})} = 1 0 \mathrm{ms} ^ {- 1}} \end{array}$
Hence, the correct answer is (D).

Velocity of bob at the lowest point just before hitting the pin is
$v ^ {2} = 2 g L (1 - \cos \alpha)$
It will rise further to a height h (say), then by Law of Conservation of Energy, we get
$\begin{array}{c} m g h = \frac {1}{2} m v ^ {2} \\ \Rightarrow h = \frac {v ^ {2}}{2 g} = L (1 - \cos \alpha) \\ \text { Now } h = (L - \ell) (1 - \cos \theta) \\ \Rightarrow (L - \ell) (1 - \cos \theta) = L (1 - \cos \alpha) \\ \Rightarrow \theta = \cos^ {- 1} \left(\frac {L \cos \alpha - \ell}{L - \ell}\right) \end{array}$
Hence, the correct answer is (C).

$W_{ext} = W_{F_0} = \int_{0}^{\frac{\pi}{2}}\vec{F}_0\cdot d\vec{r}$ $\Rightarrow W_{F_0} = \frac{\pi F_0l}{2}$ Applying Law of Conservation of Mechanical Energy, we get
$\frac {\pi F _ {0} l}{2} = \frac {1}{2} m v _ {1} ^ {2} + m g l$
$\Rightarrow v _ {1} = \sqrt {\frac {l}{m} \left(\pi F _ {0} - 2 m g\right)}$
Hence, the correct answer is (A).

In this case, the particle has both tangential and centripetal accelerations. Given that
$a _ {T} = \alpha t$
$\begin{array}{l l} \Rightarrow & \frac {d v}{d t} = \alpha t \\ \Rightarrow & \int_ {0} ^ {v} d v = \alpha \int_ {0} ^ {t} t d t \\ \Rightarrow & v = \frac {\alpha t ^ {2}}{2} \\ \text {Since,} a _ {\text {total}} = \sqrt {a _ {C} ^ {2} + a _ {T} ^ {2}} \end{array}$
Since $a_{total}$ is equally inclined to $a_{c}$ and $a_{T}$ , so $\beta = 45^{\circ}$
$\Rightarrow \frac {\alpha t}{\left(\frac {v ^ {2}}{r}\right)} = \frac {\alpha t \times 4 \times 2}{\alpha^ {2} t ^ {4}} = 1$
At $t = 2\mathrm{s}$ , we have
$\frac {\alpha (2) \times 4 \times 2}{\alpha^ {2} (2) ^ {4}} = 1$
Hence, the correct answer is (B).