Work, Energy and Power

$\vec{a} = \frac{\vec{F}}{m} = 2t\hat{i} + 3t^2\hat{j}$ (m = 1 kg)
$d\vec{v} = (2t\hat{i} + 3t^2\hat{j})dt \Rightarrow \vec{v} = t^2\hat{i} + t^3\hat{j}$
Power $P = \vec{F}\cdot\vec{v} = (2t\hat{i} + 3t^2\hat{j})\cdot(t^2\hat{i} + t^3\hat{j})$
$P = (2t^3 + 3t^5)$ W
Note: the source solution's final line prints $(2t^3 + 4t^5)$ W, which is the keyed option; the dot product actually gives $2t^3 + 3t^5$, which matches none of the options. Please check this question before use.

At the top, the minimum condition is $\frac{mv_1^2}{R} = mg \Rightarrow v_1 = \sqrt{Rg}$
Conservation of energy: $\frac{1}{2}mv^2 = \frac{1}{2}mv_1^2 + mg(2R)$
$\frac{mv^2}{2} = \frac{mgR}{2} + 2mgR = \frac{5mgR}{2}$
$v = \sqrt{5gR}$

Momentum conservation: $\frac{10}{1000}\times400 + 0 = 2\times v_1 + \frac{10}{1000}\times v_2 \Rightarrow 4 = 2v_1 + 0.01v_2$ ...(1)
Work–energy theorem for the block: $2\times10\times0.1 = \frac{1}{2}\times2\times v_1^2 \Rightarrow v_1 = \sqrt{2} = 1.4$ m/s
Putting $v_1$ in (1): $4 = 2\times1.4 + 0.01v_2 \Rightarrow v_2 = 120$ m/s

Since both bodies are identical and the collision is elastic, the velocities are interchanged after collision.
$v_A = -0.3$ m/s and $v_B = 0.5$ m/s

Displacement $\vec{s} = (4\hat{j} + 3\hat{k}) - (-2\hat{i} + 5\hat{j}) = 2\hat{i} - \hat{j} + 3\hat{k}$
$W = \vec{F}\cdot\vec{s} = (4\hat{i} + 3\hat{j})\cdot(2\hat{i} - \hat{j} + 3\hat{k}) = 8 - 3 = 5$ J

Given $K_P > K_Q$.
Case (a), same extension x: $W = \frac{1}{2}Kx^2 \Rightarrow W \propto K$, so $W_P > W_Q$.
Case (b), same force F: $x = \frac{F}{K}$, so $W = \frac{F^2}{2K} \Rightarrow W \propto \frac{1}{K}$, so $W_Q > W_P$.

Work done by the retarding force: $W = -\int_{20}^{30} 0.1x\,dx = -0.1\left[\frac{x^2}{2}\right]_{20}^{30} = -0.05(900 - 400) = -25$ J
Initial KE $= \frac{1}{2}\times 10\times 10^2 = 500$ J
By the work-energy theorem, $KE_{final} = KE_{initial} + W = 500 - 25 = 475$ J

Work done in time t: $W = Pt = kt$
By the work-energy theorem, $\frac{1}{2}mv^2 = kt \Rightarrow v = \sqrt{\frac{2kt}{m}}$
$P = Fv \Rightarrow F = \frac{k}{v} = k\sqrt{\frac{m}{2kt}} = \sqrt{\frac{mk}{2t}}$
$F = \sqrt{\frac{mk}{2}}\,t^{-1/2}$

Total energy is conserved in any collision:
Initial KE = Final KE + excitation energy
$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \varepsilon$
$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 - \varepsilon = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Rebound speed: $v = \sqrt{2gh} = \sqrt{2\times10\times20} = 20$ m/s
Energy just after rebound: $E = \frac{1}{2}mv^2 = 200m$
50% is lost in the collision, so the energy just before collision = 400m.
$\frac{1}{2}mu_0^2 + mgh = 400m \Rightarrow \frac{1}{2}u_0^2 + 200 = 400 \Rightarrow u_0 = 20$ m/s

In an elastic collision kinetic energy is conserved:
$\frac{1}{2}Mv^2 = \frac{1}{2}M\left(\frac{v}{3}\right)^2 + \frac{1}{2}Mv'^2$
$v' = \sqrt{v^2 - \frac{v^2}{9}} = \frac{2\sqrt2}{3}v$

The third piece has mass 2m. By conservation of linear momentum: $2mv_1 = \sqrt2\,mv \Rightarrow v_1 = \frac{v}{\sqrt2}$
Total KE $= \frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}(2m)v_1^2 = mv^2 + \frac{mv^2}{2} = \frac{3}{2}mv^2$

Let each half have length s. Work-energy theorem ($W = \Delta KE$):
$(mg\sin\theta)(2s) - (\mu mg\cos\theta)(s) = 0 - 0$
$\mu = 2\tan\theta$

$W = \vec F\cdot\vec S = (3\hat i + \hat j)\cdot[(4 - 2)\hat i + (3 - 0)\hat j + (-1 - 1)\hat k]$
$= (3\hat i + \hat j)\cdot(2\hat i + 3\hat j - 2\hat k) = 3(2) + 1(3) + 0(-2) = 9$ J

Let A move with speed $v_1$ at angle $\theta$ to the x-axis after the collision.
Momentum along x: $m_1v_1\cos\theta = m_2v$ ...(i)
Momentum along y: $m_1v_1\sin\theta = m_2\frac{v}{2}$ ...(ii) (equal and opposite to B's y-momentum)
Dividing (ii) by (i): $\tan\theta = \frac{1}{2}$
$\theta = \tan^{-1}\left(\frac{1}{2}\right)$ to the x-axis

$F = -\frac{dU}{dr} = -\left(-\frac{2A}{r^3} + \frac{B}{r^2}\right) = \frac{2A}{r^3} - \frac{B}{r^2}$
For equilibrium F = 0: $\frac{2A}{r^3} = \frac{B}{r^2}$
$r = \frac{2A}{B}$
At this r, $\frac{d^2U}{dr^2} > 0$ (U is a minimum), so the equilibrium is stable.

$P_0 = Fv = m\frac{dv}{dt}v$
$v\,dv = \frac{P_0}{m}dt$
Integrating from rest: $\frac{v^2}{2} = \frac{P_0}{m}t$
$v = \sqrt{\frac{2P_0t}{m}} \Rightarrow v \propto t^{1/2}$

The masses stick together, so the collision is perfectly inelastic; linear momentum is conserved.
Initial momentum = $m(v\hat{i}) + 3m(2v\hat{j}) = mv\hat{i} + 6mv\hat{j}$
Final momentum = $(m + 3m)\vec{V} = 4m\vec{V}$
$4m\vec{V} = mv\hat{i} + 6mv\hat{j}$
$\vec{V} = \frac{mv\hat{i} + 6mv\hat{j}}{4m}$
$\vec{V} = \frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$

Power delivered by a force is $P = \vec{F}\cdot\vec{v} = Fv\cos\theta$
Just after projection, the velocity is upward and gravity is downward, so $\theta = 180^\circ$ and the power is negative.
At the highest point, v = 0, so the power is zero.
Just before the body hits the earth, the velocity is downward, i.e., along the force ($\theta = 0^\circ$, $\cos\theta = 1$).
At that instant the speed is maximum and the gravitational force is also maximum (the body is nearest to the earth), so $P = Fv$ is the greatest.
Hence the power is greatest at the instant just before the body hits the earth.

Potential energy is defined only for conservative forces, by the relation $\Delta U = -W_{conservative}$
When the system does work against a conservative force, the work done by that conservative force is negative.
Then $\Delta U = -(\text{negative}) = $ positive, i.e., the potential energy increases (for example, raising a body against gravity or compressing a spring).
When a conservative force does work upon the system, U decreases; work involving non-conservative forces does not get stored as potential energy.
Hence the potential energy increases when work is done by the system against a conservative force.

Work done = area under the force–distance graph.
From the graph, the force is 2 N from d = 3 m to d = 7 m and then falls uniformly to zero at d = 12 m.
Area of the rectangle (3 m to 7 m) = $2\times(7 - 3) = 8$ J
Area of the triangle (7 m to 12 m) = $\frac{1}{2}\times(12 - 7)\times2 = 5$ J
Total work = 8 + 5 = 13 J

By the work–energy theorem, work done on the particle = gain in kinetic energy: $W = \frac{1}{2}MV^2 - 0 = \frac{1}{2}MV^2$
This work is delivered in time T.
Power delivered (average) = $\frac{W}{T}$
$P = \frac{1}{2}\frac{MV^2}{T}$

Mass of water flowing out per second: $\frac{dm}{dt} = (\text{mass per unit length})\times v = 100\times2 = 200$ kg/s
Force exerted by the engine on the water = rate of change of momentum = $v\frac{dm}{dt} = 2\times200 = 400$ N
Power = force × velocity = $Fv = 400\times2$
In symbols, with $\lambda$ = mass per unit length: $P = \lambda v^3 = 100\times2^3$
P = 800 W

Let the masses be m and 2m and the velocities after collision be $v_1$ and $v_2$.
Conservation of momentum: $m\times2 + 2m\times0 = mv_1 + 2mv_2$, so $v_1 + 2v_2 = 2$ ... (i)
Coefficient of restitution: $e = \frac{v_2 - v_1}{u_1 - u_2}$, so $v_2 - v_1 = 0.5\times(2 - 0) = 1$ ... (ii)
Adding (i) and (ii): $3v_2 = 3$, so $v_2 = 1$ m/s
From (ii): $v_1 = v_2 - 1 = 0$
The velocities after collision are 0 and 1 m/s.

The block is released from rest and comes momentarily to rest again at the maximum extension x, so its kinetic energy is zero at both positions.
Loss of gravitational potential energy = gain in elastic potential energy of the spring.
$Mgx = \frac{1}{2}kx^2$
$x = \frac{2Mg}{k}$
(This is twice the equilibrium extension Mg/k, because the block overshoots the equilibrium position and oscillates about it.)