Work, Power & Energy

From 0 to $t_{1}$ , slope is increasing, so speed is increasing. Hence, work done is positive.
From $t_{1}$ to $t_{2}$ , slope is decreasing, so speed is decreasing. Hence, work done is negative.
From $t_{2}$ to $t_{3}$ , slope of s-t graph is zero, so speed is zero. Hence, work done is zero.
From $t_{3}$ to $t_{4}$ , slope is constant, so speed is constant. Hence, work done is zero.
Hence, (A), (B) and (C) are correct.

As soon as the block hits the wall, the suspension point $B$ comes to a stop, whereas the particle $C$ keeps moving with a velocity $v_{0}$ towards left. In order that it complete a full circle, it must have kinetic energy enough to take it to the top of the circle.
$\begin{array}{r l} \Rightarrow & \frac {1}{2} m v _ {0} ^ {2} = m g (2 l) \\ \Rightarrow & v _ {0} = \sqrt {4 g l} \end{array}$
Hence, (A) and (C) are correct.

Conceptual (B), (C) and (D) are correct.

$ E. \propto t$
$\Rightarrow \frac {1}{2} m v ^ {2} = k t$
${\Rightarrow v = \sqrt {\frac {2 k}{m}} t ^ {1 / 2}}$
${\Rightarrow a = \frac {d v}{d t} = \sqrt {\frac {2 k}{m}} \left(\frac {1}{2 \sqrt {t}}\right)}$
${\text {Since} F = m a}$
${\Rightarrow F \propto \frac {1}{\sqrt {t}}}$
${\text {From (1), we get}}$
${\qquad \frac {1}{\sqrt {t}} = \sqrt {\frac {2 k}{m}} \frac {1}{v}}$
${\Rightarrow a = \sqrt {\frac {2 k}{m}} \left(\frac {1}{2} \sqrt {\frac {2 k}{m}} \frac {1}{v}\right) = \frac {2 k}{m} \left(\frac {1}{2}\right) \left(\frac {1}{v}\right)}$
$\Rightarrow a = \frac {k}{m v}$
$\Rightarrow F \propto \frac {1}{v}$
Hence, (B) and (D) are correct.

Motion of block is non-uniform circular motion, so, acceleration is not constant throughout.
At B:
$v ^ {2} = 2 g R$
$\Rightarrow a = \frac {v ^ {2}}{R} = 2 g$
Hence, (A) and (C) are correct.

$\vec {F} = - \vec {\nabla} U $<br /> $\Rightarrow \vec {F} = 7 \hat {i} - 2 4 \hat {j}$
$\Rightarrow m \vec {a} = 7 \hat {i} - 2 4 \hat {j}$
$\Rightarrow \vec {a} = \frac {7}{5} \hat {i} - \frac {2 4}{5} \hat {j} $
$\Rightarrow | \vec {a} | = 5 \mathrm{ms} ^ {- 2}$
$ \text {Also,} \vec {a} \cdot \vec {u} = 0$
$ \Rightarrow \vec {a} \perp \vec {u}$
$\text {Further we have}$
$\vec {v} = \vec {u} + \vec {a t}$
$ \Rightarrow \vec {v} = (1 4.4 \hat {i} + 4. 2 \hat {j}) + (\frac {7}{5} \hat {i} - \frac {2 4}{5} \hat {j}) 4 $
$\Rightarrow \vec {v} = (1 4. 4 \hat {i} + 4.2 \hat {j}) + (\frac {2 8}{5} \hat {i} - \frac {9 6}{5} \hat {j})$
$\Rightarrow \vec {v} = (1 4. 4 \hat {i} + 4. 2 \hat {j}) + (5. 6 \hat {i} - 1 9. 2 \hat {j})$
$ \Rightarrow \vec {v} = 2 0 \hat {i} - 1 5 \hat {j}$
$\Rightarrow | \vec {v} | = \sqrt {2 0 ^ {2} + 1 5 ^ {2}} = 2 5 \mathrm{ms} ^ {- 1}$
Hence, (A), (B) and (C) are correct.

By Law of Conservation of Energy, we have
$\begin{array}{r l} & \frac {1}{2} m u ^ {2} + 0 = 0 + m g \ell (1 - \cos \theta) \\ \text { Since } & 1 - \cos \theta = 2 \sin^ {2} \left(\frac {\theta}{2}\right) \\ \Rightarrow & u ^ {2} = 2 g \ell \left[ 2 \sin^ {2} \left(\frac {\theta}{2}\right) \right] \\ \Rightarrow & \frac {\theta}{2} = \sin^ {- 1} \left(\frac {u}{2 \sqrt {g ^ {\ell}}}\right) \end{array}$
$\Rightarrow \theta = 2 \sin^ {- 1} \left(\frac {u}{2 \sqrt {g ^ {\ell}}}\right)$
Hence, (B) and (C) are correct.

At the highest point, if $T'$ and $v'$ be the respective tension and velocity, then
$m g + T ^ {\prime} = \frac {m v ^ {\prime 2}}{\ell}$
Since $T' = 2mg$
$\Rightarrow \frac {m v ^ {\prime 2}}{\ell} = m g + 2 m g$
$v ^ {\prime} = \sqrt {3 g \ell}$
By Law of Conservation of Energy
$\frac {1}{2} m v ^ {\prime 2} + m g (2 \ell) = \frac {1}{2} m v ^ {2}$
where v is the velocity at the lowest point
$\Rightarrow v = \sqrt {7 g \ell}$
Hence, (B) and (D) are correct.

$W_{Sp} = U_i - U_f = -\frac{1}{2} k \left( x_2^2 - x_1^2 \right)$
Since $W_{Sp}=\frac{1}{2}kx^{2}$
So, $x_{2}=0$ and $x_{1}=x$ i.e., either the spring was initially stretched by a distance x and finally was in its natural length or the spring was initially compressed by a distance x and finally was in its natural length.
Hence, (A) and (C) are correct.

In equilibrium F = 0
Since, $F = -\frac{dU}{dr}$
$\Rightarrow \quad \frac {d U}{d r} = 0$
Either $U$ is zero or it is a non-zero constant.
So, for equilibrium, we have
Either $F = 0$ , $U = 0$
or $F = 0$ , $U$ is a constant (but not zero).
Hence, (A) and (C) are correct.

In simple harmonic motion, the body is accelerated at all points except at mean position.
At the starting point, spring force is zero. The only force is mg (downwards). Hence, acceleration is maximum, i.e., g (downwards). From mean position to initial position, retardation is due to forces, spring force and force of gravity. When spring is detached only gravity is left i.e., retardation has reduced and so the block rises to a height greater than from where it was released more height.
Hence, (A), (C) and (D) are correct.

$\vec {a} = \frac {\vec {F}}{m} = - \frac {1}{m} \left(\frac {\partial U}{\partial x} \hat {i} + \frac {\partial U}{\partial y} \hat {j}\right)$
$\{m = 1 \mathrm{kg} \}$
$\vec {a} = \frac {\vec {F}}{m} = - (3 \hat {i} + 4 \hat {j}) = \mathrm{constant}$
$\Rightarrow a _ {x} = - 3 \mathrm{ms} ^ {- 2} \text {and} a _ {y} = - 4 \mathrm{ms} ^ {- 2}$
$\Rightarrow a = \sqrt {a _ {x} ^ {2} + a _ {y} ^ {2}} = 5 \mathrm{ms} ^ {- 2} = \mathrm{constant}$
Hence, A) is correct
$v _ {x} = - 3 t \mathrm{and} v _ {y} = - 4 t$
$x = 6 + \frac {1}{2} a _ {x} t ^ {2} = 6 - \frac {3}{2} t ^ {2}$
and $y=4+\frac{1}{2}a_{y}t^{2}=4-2t^{2}$
Particle crosses the y-axis when
$x = 0 \mathrm{or} t = 2 s$
speed of particle at this moment
$v = a t = 5 \times 2 = 1 0 \mathrm{ms} ^ {- 1}$
co-ordinates of particle at t = 1 s are
(x, y) = (4.5, 2)
Hence (C) and (D) are also correct.
Hence, (A), (C) and (D) are correct.

Work done against friction on ice is zero and work done against friction on the road is $(\mu mg)\ell$ . So, average work done is
$\frac {0 + (\mu m g) \ell}{2} = (\mu m g) \frac {\ell}{2}$
Thus, indicating that the effective length of the sledge that has to be dragged so that it just gets completely on the road is $\frac{\ell}{2}$ .
Distance covered by the sledge on the road before
$\frac {v _ {0} ^ {2}}{2 \mu g}. \quad \left\{\because 0 ^ {2} - v _ {0} ^ {2} = 2 (- \mu g) s \right\}$
So total distance moved by the sledge is
$\left(\frac {v _ {0} ^ {2}}{2 \mu g} + \frac {\ell}{2}\right)$
Distance covered by the sledge on the road is
$\ell - \left(\frac {v _ {0} ^ {2}}{2 \mu g} + \frac {\ell}{2}\right) = \left(\frac {v _ {0} ^ {2}}{2 \mu g} - \frac {\ell}{2}\right)$
Hence, (B), (C) and (D) are correct.

In uniformly accelerated motion
v = u + a t
and $v^{2} = u^{2} + 2as$
$\Rightarrow v = \sqrt {u ^ {2} + 2 a s}$
So, power, $P = Fv = F(u + at)$
also $P = F\sqrt{u^2 + 2as}$
i.e., power varies linearly with time and parabolically with displacement.
So, B) and C) are incorrect
Hence, B) and C) are correct.

Work done by spring on block is
$W = \frac {1}{2} k a ^ {2} - \frac {1}{2} k b ^ {2} = \frac {1}{2} k (a ^ {2} - b ^ {2})$
According to MWET, we have $W_{nc} = \Delta U + \Delta K$
Since $\Delta K = 0$
$\begin{array}{r l} \Rightarrow & - \mu m g (a + b) = \frac {1}{2} k (b ^ {2} - a ^ {2}) \\ \Rightarrow & \mu m g (a + b) = \frac {1}{2} k (a ^ {2} - b ^ {2}) \\ \Rightarrow & \mu = \frac {k (a - b)}{2 m g} \end{array}$
Hence, (B) and (C) are correct.

Work done by spring in parts (A) and (C) is $\frac{1}{2}kx^{2}$ and work done by spring in parts (B) and (D) is $-\frac{1}{2}kx^{2}$ . Hence, (A) and (C) are correct.
$m g x = \frac {1}{2} k x ^ {2}$

At maximum extension, x, we have
$\Rightarrow x = \frac {2 m g}{k}$
When $x'=\frac{x}{2}$ , $F_{net}=0$
When $x' = x$ , then
k x - m g = m a
$\Rightarrow a = g (\text { upwards })$
Hence, (B), (C) and (D) are correct.

$\vec {F} = - \vec {\nabla} U = - \left(\hat {i} \frac {\partial U}{\partial x} + \hat {j} \frac {\partial U}{\partial y} + \hat {k} \frac {\partial U}{\partial z}\right)$
$\Rightarrow \quad \vec {F} = k (y \hat {i} + x \hat {j})$
Work done by this force field is independent of the path followed between any two points and hence this force field is conservative in nature.
Hence, (A) and (C) are correct.

Power, $P = \vec{F} \cdot \vec{v}$
Between O and A angle between $\vec{F}(=m\vec{g})$ and $\vec{v}$ is greater than $90^{\circ}$ . Hence, power is negative. At A angle is $90^{\circ}$ . Hence, power is zero. From A to B angle is less than $90^{\circ}$ therefore, power is positive.

At any instant, $t$
$\vec {F} = - m g \hat {j}$
$\vec {v} = (u \cos \theta) \hat {i} + (u \sin \theta - g t) \hat {j}$
$\Rightarrow P = \vec {F} \cdot \vec {v} = m g (g t - u \cos \theta)$
$\Rightarrow P = m g (g t - u \cos \theta)$
i.e., $P$ varies linearly with $t$
Therefore, OPTIONS (B) and (D) are correct. Hence, (B) and (D) are correct.

Momentum, $p = \sqrt{2mK}$
Therefore, without knowing the mass we cannot compare the momentum.
From Work Energy Theorem, more work will be done to stop B.
Distance travelled before stopping is $s=\frac{v^{2}}{2a}$ , where $a=\mu g$ .
Again, distance cannot be compared unless v is known and for v mass should be known to us.
Hence, (B) and (D) are correct.

In region OA particle is accelerated, in region AB particle has uniform velocity while in region BC particle is decelerating. Therefore, work done is positive in region OA, zero in region AB and negative in region BC.
Hence, (B) and (C) are correct.

For $x > 0$ , $F = -\mathrm{ve}$
For $x < 0$ , $F = +\mathrm{ve}$
From x=3 to x=1, displacement is in negative direction.
From x = -1 to x = -3 also displacement is in negative direction.
Now work done is positive if force and displacement are parallel to each other and work done is negative if they are antiparallel.
Hence, (A) and (D) are correct.

Power Loss= $\frac{\frac{1}{2}(dm)v^{2}}{dt}=\frac{1}{2}\frac{(A dx\rho)v^{2}}{dt}=\frac{1}{2}\rho A v^{3}$
Further, Power = $Fv = \frac{1}{2}\rho A v^{3}$
$\Rightarrow F = \frac {1}{2} \rho A v ^ {2}$
Hence, (A) and (C) are correct.

$Fx_{0} = \frac{1}{2} kx_{0}^{2}$
$\Rightarrow x _ {0} = \frac {2 F}{k}$
Work done by applied force is
$W = F x _ {0}$
Hence, (C) and (D) are correct.

Work done by force of gravity is path independent near the surface of Earth, for one reason that the force of gravity is constant near the surface of Earth. Also work done by force of gravity is path independent at greater height as the force of gravity is a conservative force.
Hence, the correct answer is (D).