(1)

Displacement of particle, $S=(2t-1)$ SI unit

Force, $F=5\text{N}$

Instantaneous power is given by, $P=\frac{dW}{dt}$

where, W is the work done on the particle.

$P=\frac{d}{dt}(\overrightarrow{F}·\overrightarrow{S})=F·\frac{dS}{dt}=5\times \frac{d}{dt}(2t-1)=5\times 2$

$=10$ $\text{SI }$$\mathrm{Unit}$

(4)

Power is scalar product of torque times velocity.

$P=\overrightarrow{F}·\overrightarrow{v} ; P=(10\hat{i}+10\hat{j}+20\hat{k})·(5\hat{i}-3\hat{j}+6\hat{k})$

$P=50-30+120,$$P=140$$\text{W}$

(3)

Given that,

Mass of lift + passengers, $m=2000\text{kg}$

Speed, $v=1.5\text{m/s}$

Frictional force, $f=3000\text{N}$

Power delivered, $P=\text{Force}\times \text{velocity}$                ...(i)

Force acting

$F=mg+f$

$F=2000\times 10+3000$

$F=23000$$\text{N}$

Using value of $\text{'}F\text{'}$ in equation (i),

$P=23000\times 1.5=34500$$\text{W}$

(3)

Given, $h=60$ $m$

Now, water falls at the rate of 15 kg is

i.e., $\frac{m}{t}=15\text{kg/s}$

$g=10{\text{m/s}}^2$

As loss due to friction is 10%, therefore only 90% of input energy is used to generate power.

$\therefore$   $P=gh\frac{m}{t}\times \frac{90}{100}$

$P=10\times 60\times 15\times \frac{90}{100}=8100\text{W}\Rightarrow P=8.1\text{kW}$

(2)

Here, $\overrightarrow{F}=(2t\hat{i}+3t^2\hat{j})\text{N},$$m=1\text{kg}$

Acceleration of the body, $\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{(2t\hat{i}+3t^2\hat{j})\text{N}}{1\text{kg}}$

Velocity of the body at time $t$,

         $\overrightarrow{v}=\int \overrightarrow{a}dt=\int (2t\hat{i}+3t^2\hat{j}) dt=t^2\hat{i}+t^3\hat{j}$${\text{ms}}^{-1}$

$\therefore$ Power developed by the force at time $t$,

      $P=\overrightarrow{F}·\overrightarrow{v}=(2t\hat{i}+3t^2\hat{j})·(t^2\hat{i}+t^3\hat{j})\text{W}=(2t^3+3t^5)\text{W}$

(3)

Here, Volume of blood pumped by man's heart,

$V=5 \text{litres}=5\times {10}^{-3} {\text{m}}^3 (\because 1\text{litre}={10}^{-3} {\text{m}}^3)$

Time in which this volume of blood pumps,

            $t=1\text{min}=60\text{s}$

Pressure at which the blood pumps,

$P=150\text{mm of Hg}=0.15\text{m of Hg}$

$=(0.15\text{m})(13.6\times {10}^3{\text{kg/m}}^3)\left(10{\text{m/s}}^2\right)$

$=20.4\times {10}^3{\text{N/m}}^2$

$\therefore$ Power of the heart $=\frac{PV}{t}$

        $=\frac{(20.4\times {10}^3{\text{N/m}}^2)(5\times {10}^{-3}{\text{m}}^3)}{60\text{s}}=1.70\text{W}$

(3)

Constant power acting on the particle of mass $m$ is $k$ watt.

or     $P=k;\frac{dW}{dt}=k; dW=k dt$

Integrating both sides, ${\int }_0^{W}dW={\int }_0^{t}k$$dt$

$\Rightarrow W=kt$                                                      ...(i)

Using work-energy theorem,

$W=\frac{1}{2}m{v}^2-\frac{1}{2}m{\left(0\right)}^2\Rightarrow kt=\frac{1}{2}m{v}^2 \text{[Using (i)]}$

$v=\sqrt{\frac{2kt}{m}}$

Acceleration of the particle, $a=\frac{dv}{dt}$

$a=\frac{1}{2}\sqrt{\frac{2k}{m}}$$\frac{1}{\sqrt{t}}=\sqrt{\frac{k}{2mt}}$

Force on the particle, $F=ma=\sqrt{\frac{mk}{2t}}=\sqrt{\frac{mk}{2}}{t}^{-1/2}$