Work Power & Energy

For bead Q

Bead $Q$ moves along the horizontal straight string $A B$. The length $A B$ is $2 R$ where $R$ is the radius of semicircular path.

Since the string is frictionless and horizontal, there is no component of gravity acting along the direction of motion.

Bead Q maintains a constant horizontal velocity v throughout the motion.

$ t_Q=\frac{A B}{v}=2\left[\frac{R}{v}\right] $

For bead P

Bead P moves along the semi-circular path ACB under the influence of gravity. And along horizontal the net force is zero, i.e., the acceleration along vertical will be non-zero and the acceleration along horizontal will be zero.

At point S , the horizontal component of its velocity is given as v .

As bead P moves from S toward the bottom-most point C , the vertical component of velocity changes but the horizontal component remains same.

When bead P reaches point B then the horizontal displacement is $\mathrm{S}^{\prime} \mathrm{O}+\mathrm{OB}$

$ \mathrm{S}^{\prime} \mathrm{O}=\mathrm{R} \cos 45^{\circ} $

Then the time taken to reach point $B$ is $t_P=\frac{S^{\prime} B}{v}=\frac{R \cos 45^{\circ}+R}{v}=\frac{R}{v}\left[\frac{1}{\sqrt{2}}+1\right]$

So, the ratio of time taken by $Q$ and $P$ to reach $B$ is,

$\frac{t_Q}{t_P}=\frac{2[R / v]}{\left[\frac{1}{\sqrt{2}}+1\right][R / v]}=\frac{2 \sqrt{2}}{\sqrt{2}+1}=[8-2 \sqrt{2}]>1$

$\Rightarrow $ $\frac{\mathrm{t}_{\mathrm{Q}}}{\mathrm{t}_{\mathrm{P}}}>1 \Rightarrow \mathrm{t}_{\mathrm{Q}}>\mathrm{t}_{\mathrm{P}}$

Let the mass of the object be m and the initial speed be u .

The angle of projection is $\theta=60^{\circ}$.

At point A

The kinetic energy at point $A$ is given as $K$.

$ \mathrm{K}_{\mathrm{A}}=\frac{1}{2} \mathrm{mu}^2=\mathrm{K} $

At Point B (Maximum Height)

At the highest point of a projectile's path, the vertical component of velocity becomes zero $\left(v_y=0\right)$. The object only possesses the horizontal component of velocity ( $\mathrm{v}_{\mathrm{x}}$ ).

The horizontal velocity remains constant throughout the motion (assuming no air resistance):

$ \mathrm{v}_{\mathrm{B}}=\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta $

$\Rightarrow $ $ \mathrm{v}_{\mathrm{B}}=\mathrm{u} \cos 60^{\circ}=\mathrm{u}\left(\frac{1}{2}\right)=\frac{\mathrm{u}}{2} $

So, the kinetic energy at B is,

$ \mathrm{K}_{\mathrm{B}}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{B}}\right)^2 $

$\Rightarrow $ $\mathrm{K}_{\mathrm{B}}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^2$

$\Rightarrow $ $\mathrm{K}_{\mathrm{B}}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}^2}{4}\right)$

$\Rightarrow $ $\mathrm{K}_{\mathrm{B}}=\frac{1}{4}\left(\frac{1}{2} \mathrm{mu}^2\right)=\frac{1}{4} \mathrm{~K}_{\mathrm{A}}=\frac{\mathrm{K}}{4}$

$ \Rightarrow \mathrm{K}_{\mathrm{B}}=\frac{\mathrm{K}}{4} $

At Point C

Point C is at the same horizontal level as point A . By the law of conservation of energy, the speed of the projectile when it hits the ground at the same level is equal to the speed of projection.

$ \mathrm{v}_{\mathrm{C}}=\mathrm{u} $

Therefore, the kinetic energy at C is equal to the kinetic energy at A :

$ \mathrm{K}_{\mathrm{C}}=\mathrm{K}_{\mathrm{A}}=\mathrm{K} $

The difference in kinetic energies at B and C :

$ \Delta K_{B C}=\left|K_C-K_B\right|=\left|K-\frac{K}{4}\right|=\frac{3 K}{4} $

So, the ratio of the difference in kinetic energies at points B and C to that at point A is

Ratio $=\frac{\Delta K_{B C}}{K_A}=\frac{(3 / 4) K}{K}=\frac{3}{4}$

Hence, the correct option is (A).

When a mass $m$ is suspended from a spring of constant $k$, the spring stretches.

If the stretch in spring is $\mathrm{x}, \mathrm{kx}=\mathrm{mg} \Rightarrow \mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}}$

The potential energy ( $U$ ) stored in the spring is given by :

$ \mathrm{U}=\frac{1}{2} \mathrm{kx}^2 $

Substituting $\mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}}$ :

$ \mathrm{U}=\frac{1}{2} \mathrm{k}\left(\frac{\mathrm{mg}}{\mathrm{k}}\right)^2=\frac{(\mathrm{mg})^2}{2 \mathrm{k}} $

Mass of block A is $\mathrm{m}_{\mathrm{A}}=100 \mathrm{~g}$.

The spring constant is $\mathrm{k}_{\mathrm{A}}$.

So, the stored energy in spring A is $\mathrm{E}_{\mathrm{A}}=\mathrm{E}$.

Using the formula :

$ E=\frac{\left(m_A g\right)^2}{2 k_A} $

Mass of block B is $\mathrm{m}_{\mathrm{B}}=200 \mathrm{~g}=2 \times 100 \mathrm{~g}=2 \mathrm{~m}_{\mathrm{A}}$

The spring constant of spring B is $\mathrm{k}_{\mathrm{B}}$.

It is given that $4 \mathrm{k}_{\mathrm{A}}=3 \mathrm{k}_{\mathrm{B}} \Rightarrow \mathrm{k}_{\mathrm{B}}=\frac{4}{3} \mathrm{k}_{\mathrm{A}}$.

So, the stored energy in spring B is

$ \mathrm{E}_{\mathrm{B}}=\frac{\left(\mathrm{m}_{\mathrm{B}} \mathrm{~g}\right)^2}{2 \mathrm{k}_{\mathrm{B}}} $

Substituting $\mathrm{m}_{\mathrm{B}}=2 \mathrm{~m}_{\mathrm{A}}$ and $\mathrm{k}_{\mathrm{B}}=\frac{4}{3} \mathrm{k}_{\mathrm{A}}$ :

$ E_B=\frac{\left(2 m_A g\right)^2}{2\left(\frac{4}{3} k_A\right)}$

$\begin{gathered}\Rightarrow \mathrm{E}_{\mathrm{B}}=\frac{4 \times \frac{3}{8} \times\left(\mathrm{m}_{\mathrm{A}} \mathrm{g}\right)^2}{\mathrm{k}_{\mathrm{A}}} \\ \Rightarrow \mathrm{E}_{\mathrm{B}}=\frac{12}{8} \times \frac{\left(\mathrm{m}_{\mathrm{A}} \mathrm{g}\right)^2}{\mathrm{k}_{\mathrm{A}}} \\ \Rightarrow \mathrm{E}_{\mathrm{B}}=\frac{3}{2} \times \frac{\left(\mathrm{m}_{\mathrm{A}} \mathrm{g}\right)^2}{\mathrm{k}_{\mathrm{A}}} \\ \Rightarrow \mathrm{E}_{\mathrm{B}}=3 \times\left(\frac{\left(\mathrm{m}_{\mathrm{A}} \mathrm{g}\right)^2}{2 \mathrm{k}_{\mathrm{A}}}\right) \\ \Rightarrow \mathrm{E}_{\mathrm{B}}=3 \mathrm{E}\end{gathered}$

Let a conservative force $\overrightarrow{\mathrm{F}}$ is acting on an object. For infinitesimally small displacement $\mathrm{d} \overrightarrow{\mathrm{r}}=\mathrm{dx} \hat{\mathrm{i}}+\mathrm{dy} \hat{\mathrm{j}}+\mathrm{dz} \hat{\mathrm{k}}$, the work done is,

$ \mathrm{dW}=\overrightarrow{\mathrm{F}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}} $

The work done (W) by a variable force $\vec{F}$ when displacing an object from an initial position vector $\vec{r}_1$ to a final position vector $\overrightarrow{\mathrm{r}}_2$ is the total work done in moving the object,

The work done is, $W=\int\limits_{r_1}^{r_2} \vec{F} \cdot d \vec{r}$

So, this formula is valid for all forces whether it is conservative or non-conservative force. But the statement I states that the work done is $W=-\int_{r_1}^{r_2} \vec{F}$. $d \vec{r}$. Hence, the statement $I$ is false.

Work done by a conservative force is path independent. And the work done by the conservative force is negative of the change in potential energy, i.e., the work done by the conservative force is stored in object as potential energy.

The work done depends only on the initial and final positions.

$ \mathrm{W}=\int\limits_{\overrightarrow{\mathrm{r}}_1}^{\overrightarrow{\mathrm{r}}_2} \overrightarrow{\mathrm{~F}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}}=-\left(\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}\right) $

For a particle going from point A to point B , there are infinitely many possible paths. So, it is a geometrical fact and is correct.

A force is called conservative if $\oint \overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}=0$

That means work done in a closed path is zero.

Hence for motion from A to B :

$ \mathrm{W}_{\mathrm{A} \rightarrow \mathrm{~B}}=\mathrm{U}_{\mathrm{A}}-\mathrm{U}_{\mathrm{B}} $

So work done depends only on initial position and final position and not on the path followed.

But the statement says, "work done changes with path for a conservative force", so the statement is incorrect because for conservative forces, straight path, curved path, zig-zag path all give the same work between the same two points. Hence, the statement II is incorrect.

Therefore, both Statement I and Statement II are false. Hence, the correct option is (C).

Velocity ( $v$ ) is the rate of change of displacement ( $x$ ) with respect to time ( $t$ ) :

$ v(t)=\frac{d x}{d t} $

The displacement function is given as:

$ x(t)=\alpha t^2+\beta t+\gamma $

Differentiating with respect to $t$ :

$ v(t)=\frac{d}{d t}\left(\alpha t^2+\beta t+\gamma\right) $

$\Rightarrow $ $ v(t)=2 \alpha t+\beta $

Substituting the given values $\left(\alpha=1 \mathrm{~m} / \mathrm{s}^2, \beta=1 \mathrm{~m} / \mathrm{s}\right)$ :

$ v(t)=2(1) t+1 $

$\Rightarrow $ $ v(t)=2 t+1 $

Initial velocity $\left(\mathrm{v}_{\mathrm{i}}\right)$ at $\mathrm{t}=2 \mathrm{~s}$ :

$ \mathrm{v}_{\mathrm{i}}=2(2)+1=5 \mathrm{~m} / \mathrm{s} $

Final velocity $\left(\mathrm{v}_{\mathrm{f}}\right)$ at $\mathrm{t}=3 \mathrm{~s}$ :

$ v_f=2(3)+1=7 \mathrm{~m} / \mathrm{s} $

Using Work-Energy Theorem formula,

$\mathrm{W}=\Delta \mathrm{K} \cdot \mathrm{E} \cdot=\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{f}}^2-\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{i}}^2$

$\Rightarrow $ $\mathrm{W}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{f}}^2-\mathrm{v}_{\mathrm{i}}^2\right)$

$\Rightarrow $ $\mathrm{W}=\frac{1}{2} \times 2 \times\left(7^2-5^2\right)$

$\Rightarrow $ $\mathrm{W}=1 \times(49-25)$

$\Rightarrow $ $ \mathrm{W}=24 \mathrm{~J} $

Therefore, the work done on the body during the given time interval is 24 J .

Hence, the correct option is (B).

The force (F) acting on a particle is defined as the negative gradient of its potential energy (V) with respect to position (x) :

$ \mathrm{F}=-\frac{\mathrm{dV}}{\mathrm{dx}} $

Here, $\frac{d V}{d x}$ is the slope of the $V$ versus $x$ graph. So, the magnitude of the force $(|F|)$ is the absolute value of the slope in each region :

$ |\mathrm{F}|=\mid \text { slope of the } \mathrm{V}-\mathrm{x} \text { graph } \mid $

Using the formula of slope

$ \text { Slope }=m=\frac{V_2-V_1}{x_2-x_1} $

For region AB (from $\mathrm{x}=0$ to $\mathrm{x}=1$ ) :

Points are ( 0,0 ) and ( 1,1 )

$ \left|\mathrm{F}_{\mathrm{AB}}\right|=\left|\frac{1-0}{1-0}\right|=1 \text { unit } $

For region BC (from $\mathrm{x}=1$ to $\mathrm{x}=2$ ) :

Points are $(1,1)$ and $(2,4)$

$ \left|\mathrm{F}_{\mathrm{BC}}\right|=\left|\frac{4-1}{2-1}\right|=3 \text { unit } $

For region CD (from $\mathrm{x}=2$ to $\mathrm{x}=4$ ) :

Points are (2,4) and (4,4)

$ \left|\mathrm{F}_{\mathrm{CD}}\right|=\left|\frac{4-4}{4-2}\right|=0 \text { unit } $

For region DE (from $\mathrm{x}=4$ to $\mathrm{x}=6$ ):

Points are $(4,4)$ and $(6,3)$

$ \left|\mathrm{F}_{\mathrm{DE}}\right|=\left|\frac{3-4}{6-4}\right|=0.5 \text { unit } $

The descending order of the magnitude of forces is,

$ 3>1>0.5>0 \Rightarrow \mathrm{~F}_{\mathrm{BC}}>\mathrm{F}_{\mathrm{AB}}>\mathrm{F}_{\mathrm{DE}}>\mathrm{F}_{\mathrm{CD}} $

Therefore, region $B C$ has the greatest slope and thus the strongest force, while region CD has no slope and zero force.

Hence, the correct option is (A).

Work done (W) is defined as the integral of force (F) over a displacement (dx):

$ \mathrm{W}=\int_{\mathrm{x}_{\mathrm{i}}}^{\mathrm{x}_{\mathrm{f}}} \mathrm{Fdx} $

From Newton's Second Law, $\mathrm{F}=\mathrm{ma}$, and acceleration $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}$.

Using the chain rule, $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v} \frac{\mathrm{dv}}{\mathrm{dx}}$.

$ W=\int_{x_i}^{x_f}\left[m\left(v \frac{d v}{d x}\right)\right] d x $

$\Rightarrow $ $W=\int_{v_i}^{v_f} m v d v=m \int_{v_i}^{v_f} v d v=m\left[\frac{v^2}{2}\right]_{v_i}^{v_f}$

$\Rightarrow $ $\mathrm{W}=\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{f}}^2-\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{i}}^2$

$\Rightarrow $ $ \mathrm{W}=\Delta \mathrm{K} . \mathrm{E} . $

The mass of the body is $\mathrm{m}=1 \mathrm{~kg}$

The velocity of body is given as function of position, $\mathrm{v}=2 \mathrm{x}^2$

Initial position is $\mathrm{x}_{\mathrm{i}}=0 \mathrm{~m}$

Final position is $\mathrm{x}_{\mathrm{f}}=5 \mathrm{~m}$

At $\mathrm{x}_{\mathrm{i}}=0$ :

$ \mathrm{v}_{\mathrm{i}}=2(0)^2=0 \mathrm{~m} / \mathrm{s} $

At $\mathrm{x}_{\mathrm{f}}=5$ :

$ v_f=2(5)^2=2(25)=50 \mathrm{~m} / \mathrm{s} $

On substituting these values into the Work-Energy equation :

$ \mathrm{W}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{f}}^2-\mathrm{v}_{\mathrm{i}}^2\right) $

$\Rightarrow $ $\mathrm{W}=\frac{1}{2} \times 1 \times\left(50^2-0^2\right)$

$\Rightarrow $ $\mathrm{W}=\frac{1}{2} \times(2500)$

$\Rightarrow $ $ \mathrm{W}=1250 \mathrm{~J} $

Therefore, the work done by the body during displacement from $\mathrm{x}=0$ to $\mathrm{x}=5 \mathrm{~m}$ is 1250 J .

Hence, the correct option is (C).

For the 6 kg mass $(\mathrm{M})$ moving down and the 2 kg mass $(\mathrm{m})$ moving up, let the acceleration of system is a and tension in the string is T .

Using Newton's second law ( $\mathrm{F}=\mathrm{ma}$ ) of motion for M

$\mathrm{Mg}-\mathrm{T}=\mathrm{Ma} \ldots (i) $

For m ,

$ \mathrm{T}-\mathrm{mg}=\mathrm{ma} \ldots (ii)$

Adding the two equations :

$ (M-m) g=(M+m) a $

$\Rightarrow $ $ a=\frac{(M-m)}{M+m} g $

Substituting $\mathrm{M}=6, \mathrm{~m}=2$, and $\mathrm{g}=10$ :

$ \mathrm{a}=\frac{6-2}{6+2} \times 10=\frac{4}{8} \times 10=5 \mathrm{~m} / \mathrm{s}^2 $

Using the kinematic equation for constant acceleration, starting from rest ( $u=0$ ) :

$ \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} $

$\Rightarrow $ $ \mathrm{v}=\sqrt{2 \mathrm{as}} $

Where s is the height $\mathrm{h}=6 \mathrm{~m}$.

$ v=\sqrt{2 \times 5 \times 6} $

$\Rightarrow $ $ \mathrm{v}=\sqrt{60} \approx 7.745 \mathrm{~m} / \mathrm{s} $

Therefore, the velocity is approximately $7.74 \mathrm{~m} / \mathrm{s}$.

The correct option is (A).

The Work-Energy Theorem states that the total work done by all forces (conservative and non-conservative) on a particle is equal to the change in its kinetic energy.

$ \mathrm{W}_{\text {total }}=\Delta \mathrm{K} $

When rain drop is falling, it experiences gravitational force (vertically downward) and resistive force which opposes motion of rain drop (vertically upward).

$ \mathrm{W}_{\text {gravity }}+\mathrm{W}_{\text {resistive }}=\mathrm{K}_{\text {final }}-\mathrm{K}_{\text {initial }} $

Work done by gravity when an object falls a height h is:

$ \mathrm{W}_{\mathrm{g}}=\mathrm{F}_{\mathrm{g}} \mathrm{~h}=\mathrm{mgh} $

The mass of the rain drop is $\mathrm{m}=1 \mathrm{~g}=1 \times 10^{-3} \mathrm{~kg}$

The height of drop is $\mathrm{h}=1 \mathrm{~km}=1000 \mathrm{~m}$

And the acceleration due to gravity is $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$

Putting the values,

$ W_g=\left(10^{-3}\right) \times(10) \times(1000)=10 \mathrm{~J} $

The initial velocity is $u=0$ as it started from rest, so the initial kinetic energy is $K_{\text {initial }}=\frac{1}{2} m u^2=0$.

The final velocity when rain drop hits the ground is $\mathrm{v}=5 \mathrm{~m} / \mathrm{s}$.

Thus, the final kinetic energy of the rain drop is

$ \mathrm{K}_{\mathrm{final}}=\frac{1}{2} \mathrm{mv}^2 $

$\Rightarrow $ $K_{f i n a l}=\frac{1}{2} \times\left(10^{-3}\right) \times(5)^2$

$\Rightarrow $ $ \mathrm{K}_{\text {final }}=\frac{1}{2} \times 10^{-3} \times 25=12.5 \times 10^{-3} \mathrm{~J}=0.0125 \mathrm{~J} $

Using the work-energy theorem equation,

$ \mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{r}}=\mathrm{K}_{\mathrm{final}}-\mathrm{K}_{\mathrm{initial}} $

$\Rightarrow $ $10 \mathrm{~J}+\mathrm{W}_{\mathrm{r}}=0.0125 \mathrm{~J}$

$\Rightarrow $ $\mathrm{W}_{\mathrm{r}}=0.0125-10$

$\Rightarrow $ $ \mathrm{W}_{\mathrm{r}}=-9.9875 \mathrm{~J} $

Therefore, the magnitude of the work done by the resistive force is approximately 9.98 J .

Thus, the correct option is (D).

Since the inclined plane is moving with constant velocity, there is no acceleration of the block with respect to ground due to the motion of the plane.

The block remains at rest on the incline, so along the plane the net force must be zero.

For the $1\,\text{kg}$ block on a $30^\circ$ incline:

• Component of weight along the plane:

$ mg\sin 30^\circ = 1 \times 10 \times \frac{1}{2} = 5\,\text{N} $

So, friction must act upward along the plane with magnitude

$ f = 5\,\text{N} $

Now the whole assembly moves upward with constant velocity $4\,\text{m/s}$ for $2\,\text{s}$.

Hence, displacement of the block is

$ s = vt = 4 \times 2 = 8\,\text{m} $

The friction force is along the incline, and the displacement is vertical upward.

So we need the angle between friction and displacement.

Since the incline makes $30^\circ$ with the horizontal, friction acts along the incline, i.e. at $30^\circ$ to horizontal.

Displacement is vertical, i.e. at $90^\circ$ to horizontal.

Therefore angle between friction and displacement is

$ 90^\circ - 30^\circ = 60^\circ $

Work done by friction:

$ W = fs\cos 60^\circ $

$ W = 5 \times 8 \times \frac{1}{2} = 20\,\text{J} $

So, the correct answer is

$ \boxed{20\,\text{J}} $

Hence, Option A is correct.

The correct answer is:

$\textbf{Option A: Frictional force}$

Here's why:

Conservative vs. Non-Conservative Forces:

Conservative Forces: These are forces for which the work done is independent of the path taken; they only depend on the initial and final positions. For conservative forces, you can define a potential energy function. Examples include Coulomb’s force, gravitational force, and the restoring force of a spring.

Non-Conservative Forces: These forces depend on the path taken and usually result in energy dissipation (often as heat). Friction is a typical example, and hence, it cannot be described solely by a potential energy function.

Example Explanations:

Coulomb’s Force: It is a conservative force because the work done in moving a charge in an electric field only depends on the difference in electric potential between two points.

Gravitational Force: It is also conservative. The potential energy associated with gravitational force is given by gravitational potential energy, for example, $U = mgh$ near the Earth’s surface.

Restoring Force: For a spring, the force is given by Hooke’s law: $F = -kx$ and the corresponding potential energy is $U = \frac{1}{2}kx^2.$ This is a conservative force.

Since friction is a non-conservative force that dissipates energy, it cannot be expressed in terms of potential energy.

Therefore, the correct answer is Option A: Frictional force.

Given Information:

Force: $\vec{F} = (2t \hat{i} + 3t^2 \hat{j}) \, \text{N}$

Mass of the object, $m = 1000 \, \text{g} = 1 \, \text{kg}$

Determine Acceleration:

Using Newton's second law, $\overrightarrow{F} = m \overrightarrow{a}$.

Thus, the acceleration, $\overrightarrow{a} = 2t \hat{i} + 3t^2 \hat{j}$.

Velocity Calculation:

To find velocity, integrate acceleration with respect to time:

$ \frac{d\overrightarrow{v}}{dt} = 2t \hat{i} + 3t^2 \hat{j} $

Integrate to find $\overrightarrow{v}$:

$ \overrightarrow{v} = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt = t^2 \hat{i} + t^3 \hat{j} + \vec{C} $

(Assuming initial velocity is zero, $\vec{C} = 0$, hence $\overrightarrow{v} = t^2 \hat{i} + t^3 \hat{j}$).

Calculate Power:

Power is defined as the dot product of force and velocity vectors:

$ P = \overrightarrow{F} \cdot \overrightarrow{v} = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j}) $

Computing the dot product gives:

$ P = (2t \cdot t^2) + (3t^2 \cdot t^3) = 2t^3 + 3t^5 $

The power generated at time $t$ is $P = (2t^3 + 3t^5) \, \text{W}$.

Block travels with uniform velocity

So $\quad \mathrm{a}=0 \Rightarrow \mathrm{~F} \cos 45^{\circ}=$ friction

$\begin{aligned} & \frac{\mathrm{F}}{\sqrt{2}}=\mu\left[\mathrm{mg}-\frac{\mathrm{F}}{\sqrt{2}}\right] \\ & \frac{\mathrm{F}}{\sqrt{2}}=0.25\left[25 \times 9.8-\frac{\mathrm{F}}{\sqrt{2}}\right] \\ & \Rightarrow \quad 1.25 \frac{\mathrm{~F}}{\sqrt{2}}=61.25 \\ & \mathrm{~F}=\frac{61.25 \times \sqrt{2}}{1.25}=49 \sqrt{2} \\ & \mathrm{~W}_{\mathrm{ext}}=\mathrm{FS} \cos 45^{\circ} \\ &=49 \sqrt{2} \times 5 \times \frac{1}{\sqrt{2}}=245 \mathrm{~J} \end{aligned}$

Using the equation of motion for velocity at height $ x $:

$ V^2 = 0 + 2g(S - x) $

Simplifies to:

$ V^2 = 2g(S - x) $

Potential Energy at height $ x $:

$ \text{Potential Energy} = mgx $

Relation between kinetic and potential energy:

Given that the kinetic energy is three times the potential energy,

$ mgx = 3 \times \frac{1}{2} mV^2 $

Simplifying gives:

$ gx = \frac{3}{2} \times 2g(S - x) $

$ gx = 3g(S - x) $

$ 4x = S $

Solving for $ x $:

$ x = \frac{S}{4} $

Finding the speed $ V $ at this instant:

$ V = \sqrt{2g \left( S - \frac{S}{4} \right)} = \sqrt{2g \times \frac{3S}{4}} = \sqrt{\frac{3gS}{2}} $

Thus, the height from the surface of the Earth is $ \frac{S}{4} $ and the speed of the particle at that height is $ \sqrt{\frac{3gS}{2}} $.

Given, ${{dm} \over {dt}} \propto \sqrt v $

So, ${{dm} \over {dt}} = k\sqrt v $ .... (i)

where, k = constant

We know, Power = $P = FV$

$ \Rightarrow P = {{dp} \over {dt}}v$ (As $F = {{dp} \over {dt}}$ where, p = linear momentum)

$ \Rightarrow P = {d \over {dt}}(mv)v$

$ \Rightarrow P = v{{dm} \over {dt}}\,.\,v$ (As speed is constant i.e. v = constant)

$ \Rightarrow P = {v^2}(k\sqrt v )$ [From (i)]

$ \Rightarrow P = k{v^{5/2}}$

by squaring both sides,

$ \Rightarrow {P^2} = {k^2}{v^5}$

$ \Rightarrow {P^2} \propto {v^5}$ (As $\mathrm{k^2}$ = constant)

Hence, option 4 is correct.

$\begin{aligned} & \langle\mathrm{p}\rangle=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})}{4}=6 \\ & \overrightarrow{\mathrm{a}}=\left(\frac{\overrightarrow{\mathrm{F}}}{\mathrm{~m}}=\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \\ & \overrightarrow{\mathrm{v}} \text { at } \mathrm{t}=4 \mathrm{sec}=\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \times 4=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \\ & P_{\text {ins }}=(2 \hat{\mathrm{i}}+3)(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=13 \\ & \frac{\langle\mathrm{P}\rangle}{P_{\text {ins }}}=\frac{6}{13} \end{aligned}$

Note: Given data is not matching.

$\begin{aligned} & S=u t+\frac{1}{2} a t^2 \\ & S=0+\frac{1}{2} \frac{(2 \hat{i}+3 \hat{j})}{4}(4)^2=4 \hat{i}+6 \hat{\mathrm{j}} \end{aligned}$

If $\overrightarrow{\mathrm{r}}_{\mathrm{i}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}$ then $\overrightarrow{\mathrm{r}}_{\mathrm{f}}=7 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$

But Final position given in the question is $(6,10)$.

When mass is at position A,

Elongation of spring, ${x_1} = 2R - R = R$

(As equilibrium length of spring is R)

${h_1} = 2R$

${v_1} = 0$ (As the bead is released from zero initial speed)

When mass is at position B,

${x_2} = R - R = 0$

${h_2} = R/2,{v_2} = v$

So using energy conservation,

${E_1}^\circ = {E_f}$

$ \Rightarrow {1 \over 2}kx_1^2 + {1 \over 2}mv_1^2 + mg{h_1} = {1 \over 2}kx_2^2 + {1 \over 2}mv_2^2 + mg{h_2}$

$ \Rightarrow {1 \over 2}k{R^2} + mg(2R) = {1 \over 2}k(o) + {1 \over 2}m{v^2} + mg{R \over 2}$

$ \Rightarrow {1 \over 2}k{R^2} + 2mgR - {{mgR} \over 2} = {1 \over 2}m{v^2}$

$ \Rightarrow {1 \over 2}m{v^2} = {1 \over 2}k{R^2} + {{3mgR} \over 2}$

$ \Rightarrow {v^2} = {{k{R^2}} \over m} + 3gR$

$ \Rightarrow v = \sqrt {{{k{R^2}} \over m} + 3gR} $

In a central force field, the work done is independent of the path chosen, this is true because central forces are conservative forces. Every force encountered in mechanics does not have an associated potential energy this is also true because not all forces are conservative. Hence, option 2 is correct.

The work done by a force is given by the integral of the force over the displacement. The force is given by:

$ F(x) = \alpha + \beta x^2 $

To find the work done when the object is displaced from $x = 0$ to $x = 1$, we compute:

$ W = \int_{0}^{1} (\alpha + \beta x^2) \, dx $

Substituting $\alpha = 1 \, \text{N}$:

$ W = \int_{0}^{1} (1 + \beta x^2) \, dx = \int_{0}^{1} 1 \, dx + \int_{0}^{1} \beta x^2 \, dx $

Calculating each integral separately:

$ \int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 $

$ \int_{0}^{1} \beta x^2 \, dx = \beta \left[ \frac{x^3}{3} \right]_{0}^{1} = \beta \left[\frac{1^3}{3} - \frac{0^3}{3}\right] = \frac{\beta}{3} $

Thus, the total work done is:

$ W = 1 + \frac{\beta}{3} $

We are given that the work done is 5 J, so:

$ 1 + \frac{\beta}{3} = 5 $

Subtract 1 from both sides:

$ \frac{\beta}{3} = 4 $

Multiply both sides by 3 to solve for $\beta$:

$ \beta = 12 \, \text{N/m}^2 $

Therefore, the value of $\beta$ is $\boxed{12 \, \text{N/m}^2}$. Thus, Option C $12 \, \text{N/m}^2$ is the correct answer.

Let's break down the problem step by step:

The ball is projected with kinetic energy

$\mathrm{KE} = \frac{1}{2} m v^2,$

where $v$ is the initial speed.

The ball is launched at an angle of $60^\circ$. So, its initial velocity components are:

Horizontal: $v_x = v \cos(60^\circ) = \frac{v}{2}$

Vertical: $v_y = v \sin(60^\circ)$

At the highest point of its flight, the vertical component of the velocity becomes zero (i.e., $v_y = 0$), but the horizontal component remains unchanged.

Therefore, the kinetic energy at the highest point is only due to the horizontal velocity:

$\mathrm{KE}_{\text{top}} = \frac{1}{2} m v_x^2$

Substitute $v_x = \frac{v}{2}$:

$\mathrm{KE}_{\text{top}} = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \frac{v^2}{4} = \frac{1}{8} m v^2$

Notice that the initial kinetic energy is

$\mathrm{KE} = \frac{1}{2} m v^2.$

So we can write:

$\mathrm{KE}_{\text{top}} = \frac{1}{8} m v^2 = \frac{1}{4} \left(\frac{1}{2} m v^2\right) = \frac{1}{4} \mathrm{KE}$

Thus, the kinetic energy at the highest point is $\frac{\mathrm{KE}}{4}$.

Answer: Option C.

To determine the value of $b$ such that the work done on the particle is zero, follow these steps:

The work done by a force on a displacement is given by the dot product:

$\text{Work} = \overrightarrow{F} \cdot \overrightarrow{d}.$

Given the force

$\overrightarrow{F} = 2\hat{i} + b\hat{j} + \hat{k},$

and the displacement

$\overrightarrow{d} = \hat{i} - 2\hat{j} - \hat{k},$

compute the dot product:

$\overrightarrow{F} \cdot \overrightarrow{d} = (2)(1) + (b)(-2) + (1)(-1).$

Simplify the expression:

$\overrightarrow{F} \cdot \overrightarrow{d} = 2 - 2b - 1 = 1 - 2b.$

Since the work done is zero:

$1 - 2b = 0.$

Solve for $b$:

$2b = 1,$

$b = \frac{1}{2}.$

Thus, the value of $b$ is $\frac{1}{2}$, which corresponds to Option B.

To find the total work done by all forces applied on the particle during its displacement, we can use the work-energy theorem which states that the work done by all forces on an object is equal to the change in kinetic energy of the object. So, we first need to find the initial and final kinetic energies of the particle and then calculate the work done.

The velocity of the particle is given by $v = \alpha \sqrt{x}$,

and the kinetic energy $K$ of the particle is given by $K = \frac{1}{2} m v^2$. We can substitute the expression for $v$ into this formula to get the kinetic energy as a function of position $x$:

$K(x) = \frac{1}{2} m (\alpha \sqrt{x})^2 = \frac{1}{2} m \alpha^2 x$

To find the total work done from $x = 0$ to $x = d$, we need to compute the difference in kinetic energy between these two points:

$W = K(d) - K(0)$

At $x = d$,

$K(d) = \frac{1}{2} m \alpha^2 d$

At $x = 0$, since the particle starts from this position,

$K(0) = \frac{1}{2} m \alpha^2 (0) = 0$

So, the work done $W$ is simply the kinetic energy at $x = d$,

$W = \frac{1}{2} m \alpha^2 d - 0 = \frac{1}{2} m \alpha^2 d$

This matches with Option C:

$\frac{m \alpha^2 d}{2}$.

$\begin{aligned} & \mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{Fr}}+\mathrm{W}_{\mathrm{s}}=\Delta \mathrm{KE} \\\\ & 5 \times 10 \times 5-0.5 \times 5 \times 10 \times \mathrm{x}-\frac{1}{2} \mathrm{Kx}^2=0-0 \\\\ & 250=25 \mathrm{x}+50 \mathrm{x}^2 \\\\ & 2 \mathrm{x}^2+\mathrm{x}-10=0 \\\\ & \mathrm{x}=2\end{aligned}$

Given that the bodies A, B, and C have equal kinetic energies, we can use the relationship between kinetic energy ($K.E.$) and linear momentum ($p$) to find the ratio of their momenta. Recall the formula for kinetic energy is $K.E. = \frac{1}{2}mv^2$ and the formula for momentum is $p = mv$, where $m$ is the mass and $v$ is the velocity of the object.

First, from the kinetic energy formula, we can express the velocity in terms of kinetic energy and mass:

$v = \sqrt{\frac{2 \cdot K.E.}{m}}.$

The momentum can then be rewritten using the velocity expression obtained from the kinetic energy equation:

$p = m\sqrt{\frac{2 \cdot K.E.}{m}} = \sqrt{2m \cdot K.E.}.$

Given that the kinetic energies are the same for all three bodies, we can ignore the kinetic energy term when comparing the ratios, simplifying our comparison to the square root of their masses:

$p \propto \sqrt{m}.$

Now, we calculate the ratio of their linear momenta using their masses. Note that the masses should be in consistent units for a valid comparison, so we'll use kilograms for all:

• Mass of A = $400 \mathrm{~g} = 0.4 \mathrm{~kg}$


• Mass of B = $1.2 \mathrm{~kg}$


• Mass of C = $1.6 \mathrm{~kg}$

Thus, the ratio of their momenta will be proportional to the square root of their masses:

$\text{Ratio of momenta} = \sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6} = \sqrt{\frac{4}{10}} : \sqrt{\frac{12}{10}} : \sqrt{\frac{16}{10}} = \sqrt{\frac{2}{5}} : \sqrt{\frac{6}{5}} : \sqrt{\frac{8}{5}}.$

Simplifying these we get:

$\text{Ratio of momenta} = \frac{\sqrt{2}}{\sqrt{5}} : \frac{\sqrt{6}}{\sqrt{5}} : \frac{\sqrt{8}}{\sqrt{5}} = \sqrt{2} : \sqrt{6} : \sqrt{8}.$

Recognizing that $\sqrt{6}$ is equivalent to $\sqrt{2} \cdot \sqrt{3}$ and that $\sqrt{8}$ is equivalent to $\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2} = 2\sqrt{2}$, we see this can also be expressed as:

$\sqrt{2} : \sqrt{2} \cdot \sqrt{3} : 2\sqrt{2}.$

Dividing through by $\sqrt{2}$ to simplify the ratio, the final ratio of their linear momenta is:

$1 : \sqrt{3} : 2,$

which matches Option A $1: \sqrt{3}: 2.$

The relationship between kinetic energy (K.E) and momentum (p) of a body can be expressed through their respective definitions. Kinetic energy is given by $K.E = \frac{1}{2} mv^2$ where $m$ is the mass of the body and $v$ is its velocity. The momentum (p) of a body is given by $p = mv$. To express kinetic energy in terms of momentum, we can manipulate the expression for momentum as follows:

$p = mv \implies v = \frac{p}{m}$

Substituting $v$ in the kinetic energy formula, we get

$K.E = \frac{1}{2} m\left(\frac{p}{m}\right)^2 = \frac{1}{2} \frac{p^2}{m}$

Therefore, we see that kinetic energy is directly proportional to the square of the momentum $(K.E \propto p^2)$.

Now, given that the kinetic energy of a body becomes 36 times its original value, we can set up the proportionality as

$\frac{K.E_{\text{final}}}{K.E_{\text{original}}} = 36$

Since $K.E_{\text{final}} = 36 \times K.E_{\text{original}}$ and knowing $K.E \propto p^2$, we can express this relationship through the squares of the initial and final momentum:

$\frac{p_{\text{final}}^2}{p_{\text{original}}^2} = 36$

Taking the square root of both sides to find the ratio of final to initial momentum, we have

$\frac{p_{\text{final}}}{p_{\text{original}}} = \sqrt{36} = 6$

This indicates that the final momentum is 6 times the original momentum. To find the percentage increase in the momentum, we calculate the increase from the original to the final, subtracting the original momentum (which is considered 1 times itself):

$\text{Percentage increase} = \left(\frac{p_{\text{final}} - p_{\text{original}}}{p_{\text{original}}}\right) \times 100\% = \left(\frac{6p - p}{p}\right) \times 100\% = \left(6 - 1\right) \times 100\% = 5 \times 100\% = 500\%$

Therefore, the correct answer is Option B: 500%.

To find the percentage loss of kinetic energy of the bullet, we first calculate the initial kinetic energy before the bullet hits the plywood and the final kinetic energy after it emerges. The formula for kinetic energy (KE) is given by:

$KE = \frac{1}{2} mv^2$

where $m$ is the mass of the object and $v$ is its velocity.

Let's calculate the initial and final kinetic energies.

Initial Kinetic Energy:

$KE_{\text{initial}} = \frac{1}{2} \times 50 \times (100)^2 = \frac{1}{2} \times 50 \times 10000 = 25 \times 10000 = 250000 \, \text{g.m}^2/\text{s}^2$

Note: To keep units consistent, we used grams and meters per second. We can also convert the mass to kilograms (by dividing by 1000) which would result in the energy being calculated in Joules, but for the purpose of finding the percentage change, the form of units does not matter as long as they are consistent, since it will be a ratio.

Final Kinetic Energy:

$KE_{\text{final}} = \frac{1}{2} \times 50 \times (40)^2 = \frac{1}{2} \times 50 \times 1600 = 25 \times 1600 = 40000 \, \text{g.m}^2/\text{s}^2$

The loss of kinetic energy is then:

$\Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 250000 - 40000 = 210000 \, \text{g.m}^2/\text{s}^2$

Finally, the percentage loss of kinetic energy can be calculated using the formula:

$\text{Percentage loss of KE} = \left( \frac{\Delta KE}{KE_{\text{initial}}} \right) \times 100\%$

$\text{Percentage loss of KE} = \left( \frac{210000}{250000} \right) \times 100\% = 0.84 \times 100\% = 84\%$

Thus, the percentage loss of kinetic energy is 84%, which corresponds to Option C.

The momentum $p$ of a particle is given by the product of its mass $m$ and its velocity $v$, that is, $p = m \cdot v$. For a given momentum, the relationship between mass and velocity can be understood as inversely proportional. This means that as the mass increases, the velocity decreases to maintain the same momentum, and vice versa.

The kinetic energy ($K.E.$) of a particle is given by the formula $K.E. = \frac{1}{2} m v^2$. This equation shows that the kinetic energy depends on both the mass of the particle and the square of its velocity.

Given that four particles $A, B, C, D$ have masses $\frac{m}{2}, m, 2 m, 4 m$, respectively, and all have the same momentum, we can assume the momentum of each particle to be $p$. This common value of momentum allows us to express the velocity of each particle in terms of its mass and the common momentum $p$. The velocity $v$ of each particle will be $v = \frac{p}{m}$.

Thus, for each particle, we can determine the velocity as follows:

• For $A$: $v_A = \frac{p}{\frac{m}{2}} = \frac{2p}{m}$


• For $B$: $v_B = \frac{p}{m}$


• For $C$: $v_C = \frac{p}{2m} = \frac{p}{2m}$


• For $D$: $v_D = \frac{p}{4m}$

Now, substituting these velocities into the kinetic energy formula yields the kinetic energies for each particle:

• $K.E._A = \frac{1}{2} \cdot \frac{m}{2} \cdot \left(\frac{2p}{m}\right)^2 = \frac{1}{2} \cdot \frac{m}{2} \cdot \frac{4p^2}{m^2} = \frac{2p^2}{m}$


• $K.E._B = \frac{1}{2} \cdot m \cdot \left(\frac{p}{m}\right)^2 = \frac{1}{2} \cdot m \cdot \frac{p^2}{m^2} = \frac{p^2}{2m}$


• $K.E._C = \frac{1}{2} \cdot 2m \cdot \left(\frac{p}{2m}\right)^2 = \frac{1}{2} \cdot 2m \cdot \frac{p^2}{4m^2} = \frac{p^2}{4m}$


• $K.E._D = \frac{1}{2} \cdot 4m \cdot \left(\frac{p}{4m}\right)^2 = \frac{1}{2} \cdot 4m \cdot \frac{p^2}{16m^2} = \frac{p^2}{8m}$

Comparing these kinetic energies, we see that the particle $A$ has the maximum kinetic energy, as it is inversely related to mass in this scenario, and $A$ has the least mass but the highest velocity squared component, thus maximizing its kinetic energy. Therefore, the correct answer is:

Option D: A

When a body moves under the influence of a constant power, the relationship between displacement and time can be established through the concept of power. Power (P) is defined as the rate at which work is done, and it can also be expressed in terms of force (F) and velocity (v) as $ P = F \cdot v $.

For a constant power P and assuming the force acts in the direction of the velocity, we can analyze how displacement (s) changes with time (t). Since force can also be written as $ F = \frac{d(mv)}{dt} $ for a constant mass m, this simplifies to $ F = m \frac{dv}{dt} $, because mass doesn't change with time for most cases. Integrating force over a distance gives work (W), and power is the rate of doing work, thus we can connect these concepts.

The kinetic energy (K.E) of the body is given by $ K.E = \frac{1}{2}mv^2 $, and the work done by the force is equal to the change in kinetic energy. Considering power is constant, $ P = \frac{dW}{dt} = \frac{d(\frac{1}{2}mv^2)}{dt} $. Rearranging terms to focus on velocity and integrating with respect to time will give us a relation involving velocity and time.

For a constant mass system, and using $ P = F \cdot v = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v $, and knowing that $ P = \text{constant} $, we rearrange to find the relationship between velocity and time.

Given $ P = m \cdot v \cdot \frac{dv}{dt} $, we rearrange to $ \frac{P}{m} dt = v dv $. Integrating both sides where the initial condition is when $ t = 0, v = 0 $, we get $ \frac{P}{m} t = \frac{1}{2} v^2 $, solving for $ v $ gives $ v \propto t^{1/2} $, so $ v = k \cdot t^{1/2} $ for some constant $ k $.

The displacement $ s $ is obtained by integrating the velocity with respect to time, $ s = \int v dt = \int k \cdot t^{1/2} dt = \frac{2}{3}k \cdot t^{3/2} $. Therefore, the displacement $ s $ is proportional to $ t^{3/2} $.

The correct answer is Option B, $ t^{3/2} $.

Work done in both cases is equal to $-m g \Delta h$

$\therefore \quad \text { Ratio }=1: 1$

From energy conservation $\rightarrow$

$\begin{aligned} & m g(R+R \sin 45)=\frac{1}{2} m v^2 \\ & \Rightarrow 10\left(1+\frac{1}{\sqrt{2}}\right) \times 14=\frac{1}{2} v^2 \\ & \Rightarrow 10\left(1+\frac{\sqrt{2}}{2}\right) \times 28=v^2 \\ & \Rightarrow 10(1+0.7) \times 28=v^2 \\ & \Rightarrow v=21.81 \end{aligned}$

To solve this problem, we need to analyze both the energy loss and the initial velocity of the rubber ball before it strikes the ground.

First, let's consider the energy loss. The energy involved here is gravitational potential energy. The initial potential energy of the ball when it is about to fall is given by $U_i = mgh$, where $U_i$ is the initial potential energy, $m$ is the mass of the ball, $g$ is the acceleration due to gravity, and $h$ is the initial height from which the ball falls. After the ball rebounds, it reaches a height of $h/2$. The potential energy at this new height is $U_f = mg \cdot \frac{h}{2}$.

The energy loss can be calculated as the difference between the initial and final potential energies, and to find the percentage energy loss, we divide this difference by the initial energy and multiply by 100:

$\text{Energy loss percentage} = \frac{(U_i - U_f)}{U_i} \times 100$

Substituting the values of $U_i$ and $U_f$ gives:

$\text{Energy loss percentage} = \frac{(mgh - mg\frac{h}{2})}{mgh} \times 100$

By simplifying, we find:

$\text{Energy loss percentage} = \frac{mgh - \frac{1}{2} mgh}{mgh} \times 100 = \frac{1}{2} \times 100 = 50\%$

This tells us that the energy loss percentage is indeed $50\%$.

Next, we'll find the velocity of the ball just before it strikes the ground. The velocity can be determined using the formula for the velocity of an object in free fall:

$v = \sqrt{2gh}$

Here, $v$ is the velocity of the ball just before impact, $g$ is the acceleration due to gravity, and $h$ is the height from which the ball falls. This formula shows that the initial velocity of the ball before it strikes the ground is $\sqrt{2gh}$, not taking into account air resistance and assuming it starts from rest.

Therefore, the correct answer is Option A: $50\%$, $\sqrt{2gh}$.

$\begin{aligned} & \vec{F}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) N \\ & \vec{F}=m \vec{a}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) \\ & \vec{a}=\frac{\vec{F}}{m}=\left(3 t \hat{i}+3 t^2 \hat{j}\right) \\ & \vec{v}=\int_\limits0^t \vec{a} d t=\frac{3 t^2}{2} \hat{i}+t^3 \hat{j} \\ & P=\vec{F} \cdot \vec{v}=\left(9 t^3+6 t^5\right) W \end{aligned}$

Work done again frictional force

$\begin{aligned} & =\mu \mathrm{N} \times 10 \\ & =0.1 \times 5 \times 10=5 \mathrm{~J} \end{aligned}$

By COME

$\begin{aligned} & \mathrm{KE}_{\mathrm{A}}+\mathrm{U}_{\mathrm{A}}=\mathrm{KE}_{\mathrm{B}}+\mathrm{U}_{\mathrm{B}} \\ & 0+\mathrm{mg}(1)=\frac{1}{2} \mathrm{mv}^2+\mathrm{mg} \times 0.5 \\ & v=\sqrt{g}=\sqrt{10} \mathrm{~m} / \mathrm{s} \end{aligned}$

Apply energy conservation between A & B

$\begin{aligned} & \frac{1}{2} \mathrm{mV}_{\mathrm{L}}^2=\frac{1}{2} \mathrm{mV}_{\mathrm{H}}^2+\mathrm{mg}(2 \mathrm{~L}) \\ & \because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}} \end{aligned}$

So, $\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}$

$\frac{(\mathrm{K} . \mathrm{E})_{\mathrm{A}}}{(\mathrm{K} . \mathrm{E})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^2}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^2}=\frac{5}{1}$

$\begin{aligned} & \text { Given } \mathrm{m}=100 \mathrm{~kg} \\ & \mathrm{~s}=10 \mathrm{~m} \\ & \mu=0.4 \\ & \text { As } \mathrm{f}=\mu \mathrm{mg}=0.4 \times 100 \times 10=400 \mathrm{~N} \\ & \text { Now } \mathrm{W}=\mathrm{f} . \mathrm{s}=400 \times 10=4000 \mathrm{~J} \end{aligned}$

The force acting on a particle can be determined from the potential energy function by using the negative gradient. In three-dimensional space, the force vector F is related to the potential energy function $ U $ by:

$ \mathbf{F} = -\nabla U $

where $ \nabla U $ (the gradient of $ U $) is a vector with components given by the partial derivatives of $ U $ with respect to $ x $, $ y $, and $ z $:

$ \nabla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right) $

Given the potential energy function:

$ U = 2x^2 + 3y^3 + 2z $

Calculate the partial derivative of $ U $ with respect to $ x $ to find the $ x $-component of the force:

Partial Derivative with respect to $ x $:

$ \frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y^3 + 2z) = 4x $

Calculate the $ x $-component of the force:

Since $ F_x = -\frac{\partial U}{\partial x} $, the $ x $-component of the force is:

$ F_x = -4x $

Evaluate at point $ P(1,2,3) $:

Substitute $ x = 1 $ into the expression for $ F_x $:

$ F_x = -4(1) = -4 $

The magnitude of the $ x $-component of the force is:

$ |F_x| = 4 \, \text{N} $

Thus, the correct answer is Option A: 4.

Loss in kinetic energy $=$ Gain in potential energy

$\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos \theta) \\ & \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos \theta) \end{aligned}$

Acceleration at lowest point $=\frac{\mathrm{v}^2}{\ell}$

Acceleration at extreme point $=g \sin \theta$

$\begin{aligned} & \text { Hence, } \frac{\mathrm{v}^2}{\ell}=\mathrm{g} \sin \theta \\ & \therefore \sin \theta=2(1-\cos \theta) \\ & \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}$

$\begin{aligned} & v^2-u^2=2 a S \\ & \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \\ & \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \\ & -\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \ldots(1) \\ & 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \\ & -\frac{4 u^2}{9}=-2 a x \ldots(2) \end{aligned}$

$(1)/(2)$

$\begin{aligned} & \frac{5}{4}=\frac{4 \times 10^{-2}}{\mathrm{x}} \\ & \mathrm{x}=\frac{16}{5} \times 10^{-2} \\ & \mathrm{x}=3 \cdot 2 \times 10^{-2} \mathrm{~m} \\ & \mathrm{x}=32 \times 10^{-3} \mathrm{~m} \end{aligned}$

First, we will use conservation of momentum to find the velocity of the bullet-block system just after the bullet gets embedded into the block.

The initial momentum of the system is given by the momentum of the bullet (as the block is initially at rest), and the final momentum of the system is the combined momentum of the bullet and the block.

Setting initial momentum equal to final momentum:

$m_{\text{bullet}} \cdot v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}$

Solving for ($v_{\text{final}}$):

$v_{\text{final}} = \frac{m_{\text{bullet}} \cdot v_{\text{bullet}}}{m_{\text{bullet}} + m_{\text{block}}}$

Substituting the given values:

$v_{\text{final}} = \frac{0.1 \, \text{kg} \cdot 400 \, \text{m/s}}{0.1 \, \text{kg} + 3.9 \, \text{kg}} = 10 \, \text{m/s}$

Next, we know the block comes to rest after moving 20 m due to friction. The work done by the friction force is equal to the initial kinetic energy of the block (since it comes to rest, the final kinetic energy is 0). The work done by friction is given by the friction force times the distance, and the friction force is equal to the coefficient of friction times the normal force (which is equal to the weight of the block).

So, setting the work done by friction equal to the initial kinetic energy of the block:

$\mu \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot g \cdot d = \frac{1}{2} \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}^2$

Solving for ($\mu$):

$\mu = \frac{\frac{1}{2} \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}^2}{(m_{\text{bullet}} + m_{\text{block}}) \cdot g \cdot d}$

Substituting the given values:

$\mu = \frac{\frac{1}{2} \cdot (0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot (10 \, \text{m/s})^2}{(0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot 10 \, \text{m/s}^2 \cdot 20 \, \text{m}} = 0.25$

When a man lifts a bucket out of a well using a rope, work is done by the man and the gravitational force. The work done by the man is positive as he has to exert an upward force to lift the bucket. The work done by the gravitational force is negative because the direction of the force is opposite to the direction of displacement.

Therefore, the statement (A) "Work done by a man in lifting a bucket out of a well by means of rope tied to the bucket is negative." is incorrect.

Therefore, the statement (B) "Work done by gravitational force in lifting a bucket out of a well by a rope tied to the bucket is negative." is correct.

Work is defined as the product of force and displacement in the direction of the force. When a body slides down an inclined plane, the force of friction acts against the motion of the body, opposing its descent.

The direction of the force of friction is opposite to the direction of the displacement of the body, which is downwards. Hence, the work done by the force of friction is negative.

Therefore, the statement (C) "Work done by friction on a body sliding down an inclined plane is positive" is incorrect.

If the body is moving on a rough horizontal plane, there will be friction present, which will act in the opposite direction to the applied force. The force of friction will oppose the motion of the body, reducing its velocity. As a result, the net work done on the body will not be zero, as the force of friction and the applied force will not cancel each other out completely.

Therefore, the statement (D) "Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is zero." is incorrect.

Statement E: "Work done by the air resistance on an oscillating pendulum is negative."

This statement refers to the work done by the air resistance on an oscillating pendulum, which is a physical system that swings back and forth under the influence of gravity.

As the pendulum oscillates, it experiences air resistance, which opposes its motion and slows it down. The direction of the air resistance force is opposite to the direction of the displacement of the pendulum, which is back and forth.

Hence, the work done by the air resistance force is negative, as the direction of the force and the displacement are opposite.

Therefore, the statement (E) "Work done by the air resistance on an oscillating pendulum is negative" is correct.

$K.E. = E = {1 \over 2}m{v^2}$

at highest point

$K.E' = {1 \over 2}m{v^2}{\cos ^2}\theta $

$ = {1 \over 2}m{v^2}\left( {{1 \over 4}} \right)$

$ = {E \over 4}$

${1 \over 2}m{x^2} = 90$

$ \Rightarrow {1 \over 2} \times 0.2 \times {x^2} = 90$,

${x^2} = 900$

$x = 30$ m/s

${1 \over 2}m{v^2} = 40 \Rightarrow v = {2 \over 3} \times 30 = 20$ m/s

$20 = 30 - a \times 1 \Rightarrow a = - 10$ m/s²

$0 - {x^2} = 2as$

$s = {{{x^2}} \over { - 2a}} = {{30 \times 30} \over {2 \times 10}}$

$ = 45$ m

${{dm} \over {dt}} = 0.5$ kg/s

$v = 5$ m/s

$F = {{vdm} \over {dt}} = 2.5$ kg m/s²

$P = \overline F \,.\,\overline v = (2.5)(5)$ W

$ = 12.5$ W

$\because$ Loss in KE = Gain in spring energy

$ \Rightarrow {1 \over 2}m{v^2} \times 2 = {1 \over 2}kx_m^2$

$ \Rightarrow 2 \times {1 \over 4} \times {v^2} = 2 \times x_m^2$

$ \Rightarrow {x_m} = \sqrt {{{{v^2}} \over 4}} = {v \over 2}$