Work Power and Energy

Velocity at B when dropped from A where AC = S $v^{2} = 0 + 2g(S - x)$ or $v^{2} = 2g(S - x)$ … (i)
Potential energy at B = mgx … (ii)
∵ Kinetic energy
= 3 × potential energy
∴ $\frac{1}{2}m \times 2g(S - x) = 3 \times mgx$ $\Rightarrow S - x = 3x$ or S = 4x or x = S/4
From (i),

$v ^ {2} = 2 g (S - x) = 2 g \left(S - \frac {S}{4}\right) = \frac {2 g \times 3 S}{4} = \frac {3 g S}{2}$
$\Rightarrow v = \sqrt {\frac {3 g S}{2}}. \therefore x = \frac {S}{4} \text { and } v = \sqrt {\frac {3 g S}{2}}$

Vertical height = h = l cos $30^{\circ}$ Loss of potential energy = mgh
= mgl cos $30^{\circ} = \frac{\sqrt{3}}{2}$ mgl
∴ Kinetic energy gained = $\frac{\sqrt{3}}{2}$ mgl.

Work done in raising water = mgh
∴ W = (volume × density) gh = (9×1000)×10×10
$⇒ W = 9×10^5 J$
∴ Useful power = $\frac{\text{work}}{\text{time}}$ = $\frac{9\times 10^{5}}{5\times 60}$ = 3kW
∴ Efficiency = $\frac{3}{10}$ = 30%.

Mass of fragments are as 2:3
Total mass = 20 kg
∴ Larger fragment = 12 kg
∴ Smaller fragment = 8 kg
Momentum is conserved
∴ $8 \times 6 = 12 \times v \Rightarrow v = 4 = velocity of larger fragment.$
∴ Kinetic energy = $\frac{1}{2}mv^{2}=\frac{1}{2}\times12\times(4)^{2}=96J$ .

Let a nucleus of mass $M$ splits into two nuclear parts having masses $M_{1}$ and $M_{2}$ and radii $R_{1}$ and $R_{2}$ and densities $\rho_{1}$ and $\rho_{2}$ .
$\therefore M_{1} = \rho_{1}\frac{4}{3}\pi R_{1}^{3}$ and $M_{2} = \rho_{2}\frac{4}{3}\pi R_{2}^{3}$
Given: $\rho_{1} = \rho_{2}$
$\therefore \frac {M _ {1}}{M _ {2}} = \left(\frac {R _ {1}}{R _ {2}}\right) ^ {3}$
According to law conservation of linear momentum,
$M \times 0 = M _ {1} v _ {1} + M _ {2} v _ {2} \text {or} \frac {M _ {1}}{M _ {2}} = - \frac {v _ {2}}{v _ {1}}$
-ve sign show that both the parts are move in opposite direction in order to conserve the linear momentum
$\therefore \frac {v _ {1}}{v _ {2}} = \frac {M _ {2}}{M _ {1}} \mathrm{or} \frac {v _ {1}}{v _ {2}} = \left(\frac {R _ {2}}{R _ {1}}\right) ^ {3}$
${\frac {v _ {1}}{v _ {2}}} = \left({\frac {2}{1}}\right) ^ {3} = {\frac {8}{1}}$
$\left[ \text {Given} \frac {R _ {1}}{R _ {2}} = \frac {1}{2} \right]$

Maximum height reached $H_{max} = \frac{u^{2}}{2g} = \frac{(5)^{2}}{2 \times 10} = 1.25 m$ Work done = $\vec{F} \cdot \vec{d} = F d \cos \theta$ $= mg \times (H_{\text{max}}) \times \cos(180^{\circ})$ $= 0.1 \times 10 \times 1.25 \times (-1) = -1.25 J$

Area of acceleration-displacement curve gives change in KE per unit mass $\frac{1}{2}m(v^{2}-u^{2})=F.S=\frac{mdv}{dt}\times s$ $\therefore \frac{\text{Change in } KE}{\text{Mass}}=\frac{dv}{dt}\times s.$

At a given height the half of the kinetic energy of the body is equal to its potential energy.
Initial kinetic energy of the body $=\frac{1}{2}mv^{2}=\frac{1}{2}m(4)^{2}=8m$ Let at height h, the kinetic energy reduces to half, i.e., it becomes 4m. It is also equal to potential energy.
Hence, mgh = 4m or $h = \frac{4}{g} = \frac{4}{10} = 0.4m$ .

A

B

$P = v\cos \theta = mgv\cos 90^{\circ} = 0.$

Power delivered to the body $P = F \cdot v = m a v$ Since, body undergoes one dimensional motion and is initially at rest, so $v = 0 + a t$ $\therefore P = m a^2 t$ or $P \propto t$ .

Linear momentum of water striking per second to the wall $P_{i} = mv = Av\rho v = Av^{2}\rho$ , similarly linear
momentum of reflected water per second $P_{r}=Av^{2}\rho$ .

Now making components of momentum along x-axes and y-axes. Change in momentum of water per second
$= P_{i}\cos \theta +P_{r}\cos \theta$
$= 2Av^{2}\rho \cos \theta$
By definition of force, force exerted on the Wall = $2Av^{2}\rho \cos\theta$ .

Change in momentum $= m\vec{v}_2 - m\vec{v}_1 = -mv - mv = -2mv$

Work done by horizontal force
$W = F \times S = F \times l \sin \theta$...(i)
Increment in potential energy of mass M is
$U = M g h = M g (l - l \cos \theta) = M g l (1 - \cos \theta)$...(ii)
From equation (i) and (ii)
$F l \sin \theta = M g l (1 - \cos \theta)$
$\Rightarrow F l \frac {1}{\sqrt {2}} = M g l \left(1 - \frac {1}{\sqrt {2}}\right)$
$[ \mathrm{As} \theta = 4 5 ^ {\circ} ]$

$\therefore F = M g (\sqrt {2} - 1).$

Potential energy $V = \frac{x^4}{4} - \frac{x^2}{2}$
For maximum kinetic energy, potential energy of a particle should be minimum.
For minimum value of $V$ , $\frac{dV}{dx} = 0$ and $\frac{d^2V}{dx^2} > 0$
Force $F = -\left(\frac{dV}{dx}\right) = \frac{4x^3}{4} - \frac{2x}{2} = 0 \Rightarrow x^3 - x = 0$
$\Rightarrow x (x ^ {2} - 1) = 0$
i.e. at x=0, $x=+1$ and x=-1 force on the particle will be zero.
Now $\frac{d^2V}{dx^2} = 3x^2 - 1$
For $x = +1$ and $x = -1$ $\frac{d^2V}{dx^2} > 1$
it means the potential energy of the particle will be minimum at x = 1 and x = -1.
Now substituting these values in expression of potential
energy $V_{\min} = \left[\frac{(1)^4}{4} - \frac{(1)^2}{2}\right]J = \left[\frac{1}{4} - \frac{1}{2}\right]J = -\frac{1}{4}J$
(Kinetic energy) $_{max}$ = Total energy - (Potential energy) $_{min}$ = 2- $\left(-\frac{1}{4}\right)$
$\frac{1}{2} mv_{\max}^2 = \frac{9}{4}\Rightarrow v_{\max}^2 = \frac{9}{2}\Rightarrow v_{\max} = \frac{3}{\sqrt{2}} m / \sec .$

When the ball is released from the top of tower then ratio of distances covered by the ball in first, second and third second $h_{I}: h_{II}: h_{III} = 1:3:5$ [Because $h_{n} \propto (2n-1)$ ]
∴ Ratio of work done $mgh_{I}: mgh_{II}: mgh_{III} = 1:3:5$ .

D

When block of mass $M$ collides with the spring its kinetic energy gets converted into elastic potential energy of the spring.
From the law of conservation of energy $\frac{1}{2}Mv^{2}=\frac{1}{2}KL^{2}\therefore v=\sqrt{\frac{K}{M}}L$ Where $v$ is the velocity of block by which it collides with spring. So, its maximum momentum $P=Mv=M\sqrt{\frac{K}{M}}L=\sqrt{MK}L$ After collision the block will rebound with same linear momentum.

Ball starts from the top of a hill which is 100 m high and finally rolls down to a horizontal base which is 20 m above the ground so from the conservation of energy $mg(h_{1}-h_{2})=\frac{1}{2}mv^{2}$ $\Rightarrow v=\sqrt{2g(h_{1}-h_{2})}=\sqrt{2\times10\times(100-20)}$ $=\sqrt{1600}=40m/s$ .

The heat required for producing 1g of steam = 540 cal
= 540 × 4.2 J = 2268 J
Energy given by immersion heater is
= 1.08 kW = 1080 W
Now time taken to produce 100g of steam
= $\frac{2268 \times 100}{1080}$ = 210 sec.

Here: Energy of one apple $= 21KJ = 21 \times 10^{3} J$ Efficiency of the boy $= 28\% = 0.28$ Mass of the boy $m = 40 kg$ Here the actual energy consumed by the boy is given by as $0.28 \times 21000 = 5880 J$ ... (i)
and the energy consumed by the boy in climbing $h$ meter height is given by $= m g h = 40 \times 9.8 \times h$ ... (ii)
Equating equations (i) and (ii) we get $40 \times 9.8 \times h = 5880$ $h = \frac{5880}{40 \times 9.8} = 15 m$ .

Power = $\frac{workdone}{time}$ = $\frac{pressure \times change in volume}{time}$ = $\frac{20000 \times 1 \times 10^{-6}}{1}$ = $2 \times 10^{-2}$ = 0.02 W.

Power $= \frac{W}{t}$ . If $W$ is constant then $P \propto \frac{1}{t}$ i.e. $\frac{P_1}{P_2} = \frac{t_2}{t_1} = \frac{20}{10} = \frac{2}{1}$ .

As collision is perfectly elastic hence law of conservation of linear momentum and KE follows.

Before collision
After collision
KE before collision = KE after collision
$\frac {1}{2} M v ^ {2} + 0 = \frac {1}{2} M \left(\frac {v}{3}\right) ^ {2} + \frac {1}{2} M v _ {2} ^ {2}$
$v _ {2} = \frac {2 \sqrt {2}}{3} v.$