Work, Power & Energy

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).

Since the car is moving with a constant velocity, so
Hence, the correct answer is (A).

When it is lowered gradually then to attain the equilibrium, we have
$m g = k d$...(i)
Further when it is allowed to fall suddenly, and if x is the maximum stretching in the spring, then by Law of Conservation of Energy
$\begin{array}{l} \left( \begin{array}{c} \text {Loss in Gravitational} \\ \text {Potential Energy} \\ \text {of Body} \end{array} \right) = \left( \begin{array}{c} \text {Gain in Elastic} \\ \text {Potential Energy} \\ \text {of Spring} \end{array} \right) \\ \Rightarrow m g x = \frac {1}{2} k x ^ {2} \\ \Rightarrow m g = \frac {1}{2} k x \end{array}.....(ii)$
Equating (1) and (2), we get
x = 2 d
Hence, the correct answer is (B).

By Law of Conservation of Energy
$\begin{array}{l} \frac {1}{2} m v ^ {2} = \frac {1}{2} k x ^ {2} \\ \Rightarrow \quad \frac {1}{2} (4) (1.6) ^ {2} = \frac {1}{2} (2) x ^ {2} \\ \Rightarrow \quad x = 2.2 6 \mathrm{m} \\ \text { Hence,the correct answer is(C). } \end{array}$

$a = g (\sin 3 0 - \mu \cos \theta)$
$ \Rightarrow a = 9.8 \left(\frac {1}{2} - 0.2 \frac {\sqrt {3}}{2}\right)$
$\Rightarrow a = 9.8 (0.3 2 6 8)$
$\Rightarrow a = 3. 2 \mathrm{ms} ^ {- 2}$
$ \text {Since} v ^ {2} - (1 0) ^ {2} = 2 (3.2) (1 0) $
$\Rightarrow v ^ {2} = 1 6 4 $
$\Rightarrow v \cong 1 3 \mathrm{ms} ^ {- 1} $
Hence, the correct answer is (B).

Speed will be maximum at equilibrium, i.e., when
$\begin{array}{r l} & F = k x \\ \Rightarrow & x = \frac {F}{k} \\ \text {Applying MWET, we get} \\ & W _ {e x t} = \Delta U + \Delta K \\ \Rightarrow & F x = \frac {1}{2} k x ^ {2} + \frac {1}{2} m v ^ {2} \\ \Rightarrow & F \left(\frac {F}{k}\right) = \frac {1}{2} k \left(\frac {F}{k}\right) ^ {2} + \frac {1}{2} m v ^ {2} \\ \Rightarrow & \frac {1}{2} m v ^ {2} = \frac {F ^ {2}}{2 k} \\ \Rightarrow & v = \frac {F}{\sqrt {m k}} \end{array}$
Hence, the correct answer is (A).

$\begin{array}{r l} & {\binom {\mathrm{LossinKinetic}} {\mathrm{Energy}} = \binom {\mathrm{GaininGravitational}} {\mathrm{PotentialEnergy}}} \\ {\Rightarrow} & {\frac {1}{2} m V _ {0} ^ {2} - \frac {1}{2} m V ^ {2} = m g h} \\ {\Rightarrow} & {V ^ {2} = V _ {0} ^ {2} - 2 g h} \\ {\Rightarrow} & {V = \sqrt {V _ {0} ^ {2} - 2 g h}} \end{array}$
Hence, the correct answer is (C).

Potential energy of cube at position 1,
$U _ {1} = M g (4 R) = 4 M g R$
Applying Law of Conservation of Energy between 1 and 2, we get
$\begin{array}{r l} & K _ {1} + U _ {1} = K _ {2} + U _ {2} \\ \Rightarrow & 4 M g R = 2 M g R + \frac {1}{2} M v ^ {2} \\ \Rightarrow & 4 M g R = M v ^ {2} \\ \Rightarrow & v ^ {2} = 4 g R \\ \mathrm{Since,} N = \frac {M v ^ {2}}{R} - M g \\ \Rightarrow & N = \frac {M}{R} (4 R g) - M g = 3 M g \end{array}$
Hence, the correct answer is (A).

Let $l$ be the length of spring at the situation shown in Figure.
$\begin{array}{r l} \Rightarrow & \cos (3 7 ^ {\circ}) = \frac {h}{l} \\ \Rightarrow & l = \frac {h}{\cos (3 7 ^ {\circ})} = \frac {5 h}{4} \end{array}$
Extension $x$ in spring is
$x = l - h = \frac {5 h}{4} - h = \frac {h}{4}$
By Law of Conservation of Energy, we have
$\begin{array}{r l} & {\binom{\text {Loss in EPE}}{\text {of springs}} = \binom{\text {Gain in KE}}{\text {of Bead}}} \\ {\Rightarrow} & {\frac {1}{2} k x ^ {2} + \frac {1}{2} k x ^ {2} = \frac {1}{2} m v ^ {2}} \\ {\Rightarrow} & {2 \bigg (\frac {1}{2} (1 0 0 0) \frac {h ^ {2}}{1 6} \bigg) = \frac {1}{2} (5) v ^ {2}} \\ {\Rightarrow} & {v ^ {2} = 2 5 h ^ {2}} \\ {\Rightarrow} & {v = 5 h} \end{array}$
Hence, the correct answer is (A).

Speed of ball is maximum at equilibrium, so
$x _ {e q} = \frac {m g}{k}$
Hence, the correct answer is (A).

According to Work-Energy Theorem for a non-conservative system, we have
$W _ {c} + W _ {n c} = \Delta k$
{OPTION (C)}
Also, we know that the work done by a conservative force equals the decrease in potential energy, so
$\begin{array}{r l} & W _ {c} = - \Delta U \\ \Rightarrow & W _ {n c} - \Delta U = \Delta k \end{array}\tag{{OPTION (B)}}$
{OPTION (A)}
Hence, (A), (B) and (C) are correct.

Tension in the string will be maximum when mass m is at point B, so
$T = m g + \frac {m v ^ {2}}{\ell}$
By Law of Conservation of Energy,
$\begin{array}{r l} & {\frac {1}{2} m v ^ {2} = m g h = m g \ell (1 - \cos \theta)} \\ {\Rightarrow} & {v ^ {2} = 2 g \ell (1 - \cos (6 0 ^ {\circ}))} \\ {\Rightarrow} & {T = m g + \frac {2 m g \ell}{\ell} \bigg (1 - \frac {1}{2} \bigg)} \\ {\Rightarrow} & {T = 2 m g} \\ {\mathrm{Formass4mtobestationary,} f \geq T} \\ {\Rightarrow} & {\mu (4 m g) \geq 2 m g} \\ {\Rightarrow} & {\mu \geq 0. 5} \end{array}$
Hence, (A) is correct
When m moves from A to B it has vertical displacement downwards, i.e., along the direction of gravitational force.
So, B) is correct.
Tension is always perpendicular to the velocity, so power delivered by the tension is zero.
So, C) is correct.
According to Work Energy Theorem,
$W _ {\mathrm{total}} = \Delta K$
Since work done by tension is zero therefore work is done only by the gravitational force. So, (D) is also correct.
Hence, (A), (B), (C) and (D) are correct.

$P = (\text {Total Frictional Force}) (\text {velocity})$
$\Rightarrow P = (m f) v $
$\Rightarrow P _ {\text {extra}} = m g (\sin \theta) v = \frac {m g h v}{s} $
Hence, (A), (B) and (C) are correct.

Using the concepts of Laws of Motion and Kinematics we can draw following diagrams and then using the definition of work, we can find corresponding work done.

Hence, (A), (B), (C) and (D) are correct.

For a light string pulled by a constant
Force $F = 40 \mathrm{~N}$ , tension (T) in the string is also $40 \mathrm{~N}$ . So, $T = 40 \mathrm{~N}$
Since $W = 40 \mathrm{~J} = Fs$ where $F = 40 - 20 \mathrm{~N} = 20 \mathrm{~N}$
$\begin{array}{r l} \Rightarrow & 4 0 = 2 0 \mathrm{s} \\ \Rightarrow & s = 2 \mathrm{m} \end{array}$
Since acceleration is constant, so $v \propto t$ and hence $P = Fv$
$\Rightarrow P \propto t$
$\mathrm{Also}, v = \sqrt {2 a x}$
$\Rightarrow P \propto \sqrt {x}.$
So, all options (A), (B), (C) and (D) are correct.
Hence, (A), (B), (C) and (D) are correct.

Work done by $\vec{F}_1$ is
$W _ {1} = \int_ {P _ {1}} ^ {P _ {2}} F _ {1} \cos \theta d s$
Where, $ds = (6)d(-2\theta) = -12d\theta$ and $F_{1} = 20 N$
$\Rightarrow W _ {1} = - 2 4 0 \int_ {\pi / 4} ^ {0} \cos \theta d \theta$
$\Rightarrow W _ {1} = 2 4 0 \sin \left(\frac {\pi}{4}\right) = 1 2 0 \sqrt {2} \mathrm{J}$

$\vec{F}_{1}$ is conservative because it is always directed towards a fixed point $P_{2}$ . Therefore, $W_{1}$ can be directly calculated as
$W _ {1} = F _ {1} \left(P _ {1} P _ {2}\right) = (2 0) \left(6 \sqrt {2}\right) = 1 2 0 \sqrt {2} \mathrm{J}$
Similarly, $W_{2} = F_{2}(OP_{2}) = (30)(6) = 180 \mathrm{~J}$
and
$W _ {3} = \int_ {0} ^ {6 (\pi / 2)} F _ {3} d s = \int_ {0} ^ {3 \pi} 1 5 d s$
$W _ {3} = 1 5 s | _ {0} ^ {3 \pi} = 4 5 \pi \mathrm{J}$
Hence, (A), (C) and (D) are correct.