Power Questions in English

(B) The instantaneous power $P$ delivered by a force $\vec F$ to a particle moving with velocity $\vec v$ is given by the dot product of the force and velocity vectors:
$P = \vec F \cdot \vec v$
Given $\vec F = 20\hat i + 15\hat j - 5\hat k \, N$ and $\vec v = 6\hat i - 4\hat j + 3\hat k \, m/s$.
$P = (20\hat i + 15\hat j - 5\hat k) \cdot (6\hat i - 4\hat j + 3\hat k)$
$P = (20 \times 6) + (15 \times -4) + (-5 \times 3)$
$P = 120 - 60 - 15$
$P = 120 - 75$
$P = 45 \, J/s$.

(C) Power is defined as the rate of doing work or the rate of energy transfer.
Mathematically,$P = \frac{W}{t}$.
The $SI$ unit of work $(W)$ is Joule $(J)$ and the $SI$ unit of time $(t)$ is second $(s)$.
Therefore,the unit of power is Joule per second $(J/s)$.
By definition,$1 \text{ Watt} = 1 \text{ Joule/second}$.
Thus,both Joule per second and Watt are valid units for power.

(D) The body starts from rest $(u = 0)$ and accelerates uniformly to $v_1$ in time $t_1$.
The acceleration $a$ is given by $a = \frac{v_1 - u}{t_1} = \frac{v_1}{t_1}$.
At any time $t$,the velocity of the body is $v = at = \left( \frac{v_1}{t_1} \right)t$.
The force acting on the body is $F = ma = m \left( \frac{v_1}{t_1} \right)$.
The instantaneous power $P$ delivered to the body is $P = F \cdot v$.
Substituting the expressions for $F$ and $v$:
$P = \left( m \frac{v_1}{t_1} \right) \times \left( \frac{v_1}{t_1} t \right) = \frac{mv_1^2t}{t_1^2}$.

(D) Given:
Velocity $v = 7.2 \text{ km/hr} = 7.2 \times \frac{5}{18} = 2 \text{ m/s}$.
Slope is $1$ in $20$, so $\sin \theta = \frac{1}{20}$.
Mass $m = 100 \text{ kg}$.
Acceleration due to gravity $g = 9.8 \text{ m/s}^2$.
When the man and cycle move up the hill, the component of weight opposing the motion is $F = mg \sin \theta$.
The power $P$ required is given by $P = F \times v$.
$P = (mg \sin \theta) \times v$.
$P = 100 \times 9.8 \times \left( \frac{1}{20} \right) \times 2$.
$P = 98 \text{ W}$.

(A) The power $P$ is defined as the rate of doing work,given by $P = \frac{W}{t}$.
Work done $W = F \times s$,where $F = 40 N$ and $s = 30 m$.
So,$W = 40 N \times 30 m = 1200 J$.
The time taken $t = 1 \text{ minute} = 60 s$.
Therefore,$P = \frac{1200 J}{60 s} = 20 W$.
Thus,the power supplied by the motor is $20 W$.

(B) The power $P$ delivered by a force $F$ moving an object at a constant velocity $v$ is given by the formula $P = F \cdot v$.
Given:
Tension (Force) $F = 4500 \, N$
Velocity $v = 2 \, m/s$
Substituting these values into the formula:
$P = 4500 \, N \times 2 \, m/s = 9000 \, W$
Since $1 \, kW = 1000 \, W$,we have:
$P = 9000 \, W = 9 \, kW$.
Therefore,the correct option is $B$.

(D) The power $P$ is defined as the rate of doing work,given by $P = \frac{W}{t}$.
Since the work done against gravity is $W = mgh$,the power is $P = \frac{mgh}{t}$.
Given values are mass $m = 300 \ kg$,height $h = 2 \ m$,time $t = 3 \ s$,and acceleration due to gravity $g = 9.8 \ m/s^2$.
Substituting these values into the formula:
$P = \frac{300 \times 9.8 \times 2}{3}$
$P = 100 \times 9.8 \times 2$
$P = 1960 \ W$.

(D) The power $P$ is given by the formula $P = \frac{W}{t} = \frac{mgh}{t}$.
Given: $P = 2 \, kW = 2000 \, W$,$g = 10 \, m/s^2$,$h = 10 \, m$,and $t = 1 \, \text{minute} = 60 \, s$.
Rearranging the formula to solve for mass $m$:
$m = \frac{P \times t}{g \times h} = \frac{2000 \times 60}{10 \times 10} = \frac{120000}{100} = 1200 \, kg$.
Since the density of water is $1000 \, kg/m^3$,the volume $V$ in cubic meters is $V = \frac{m}{\rho} = \frac{1200 \, kg}{1000 \, kg/m^3} = 1.2 \, m^3$.
Since $1 \, m^3 = 1000 \, liters$,the volume in liters is $1.2 \times 1000 = 1200 \, liters$.

(C) Power $(P)$ is defined as the rate of doing work,given by $P = \frac{W}{t}$.
Here,the work done $(W)$ to lift a mass $(m)$ to a height $(h)$ is equal to the change in potential energy: $W = mgh$.
Therefore,$P = \frac{mgh}{t}$.
Given values: $P = 10 \text{ kW} = 10 \times 10^3 \text{ W}$,$m = 200 \text{ kg}$,$h = 40 \text{ m}$,and $g = 10 \text{ m/s}^2$.
Rearranging the formula for time $(t)$: $t = \frac{mgh}{P}$.
Substituting the values: $t = \frac{200 \times 10 \times 40}{10 \times 10^3}$.
$t = \frac{80000}{10000} = 8 \text{ s}$.
Thus,the time taken is $8 \text{ seconds}$.

(C) The power $P$ delivered by the engine is given by the product of the force $F$ exerted by the engine and the velocity $V$ of the car,i.e.,$P = FV$.
According to Newton's second law,the net force $F_{net}$ acting on the car is $F_{net} = ma$.
The forces acting on the car are the engine force $F$ (forward) and the resistive force $R$ (backward).
Therefore,$F - R = ma$,which gives the engine force $F = R + ma$.
Substituting this into the power formula,we get $P = (R + ma)V$.

(D) The formula for average power $P$ is given by $P = \frac{W}{t}$,where $W$ is the work done and $t$ is the time taken.
Work done $W$ to lift a mass $m$ to a height $h$ is $W = mgh$.
Given: $m = 100 \ kg$,$h = 50 \ m$,$t = 50 \ s$,and acceleration due to gravity $g = 9.8 \ m/s^2$.
Substituting the values: $P = \frac{100 \times 9.8 \times 50}{50}$.
$P = 100 \times 9.8 = 980 \ J/s$.
Therefore,the correct option is $D$.

(A) The power $P$ delivered to the turbine is given by the rate of change of potential energy.
$P = \frac{d}{dt}(mgh) = \left( \frac{dm}{dt} \right) gh$
Given:
Rate of mass flow $\frac{dm}{dt} = 100 \ kg/s$
Height $h = 100 \ m$
Acceleration due to gravity $g \approx 10 \ m/s^2$
Substituting the values:
$P = 100 \ kg/s \times 10 \ m/s^2 \times 100 \ m$
$P = 100,000 \ W$
Since $1 \ kW = 1000 \ W$,we have:
$P = \frac{100,000}{1000} \ kW = 100 \ kW$.
Thus,the correct option is $A$.

(A) The power $P$ is defined as the rate of doing work,given by $P = \frac{W}{t}$.
Here,the work done $W$ to lift the water is equal to the change in potential energy,$W = mgh$.
Given values are: mass $m = 200 \ kg$,height $h = 200 \ m$,time $t = 10 \ s$,and acceleration due to gravity $g = 10 \ m/s^2$.
Substituting these values into the formula:
$P = \frac{mgh}{t} = \frac{200 \times 10 \times 200}{10} = 40,000 \ W$.
To convert the power into kilowatts $(kW)$,we divide by $1000$:
$P = \frac{40,000}{1000} = 40 \ kW$.

(C) The total height through which the water is raised is $H = 20 \text{ m} + 10 \text{ m} = 30 \text{ m}$.
Volume of water $V = 22380 \text{ litres} = 22380 \times 10^{-3} \text{ m}^3 = 22.38 \text{ m}^3$.
Density of water $\rho = 10^3 \text{ kg/m}^3$,so mass $m = V \times \rho = 22.38 \times 10^3 \text{ kg} = 22380 \text{ kg}$.
Power of the motor $P = 10 \text{ H.P.} = 10 \times 746 \text{ W} = 7460 \text{ W}$.
The work done to lift the water is $W = mgh = 22380 \times 10 \times 30 \text{ J}$.
Power $P = \frac{W}{t}$,so $t = \frac{mgh}{P} = \frac{22380 \times 10 \times 30}{7460} \text{ seconds}$.
$t = \frac{22380 \times 300}{7460} = 3 \times 300 = 900 \text{ seconds}$.
Converting to minutes: $t = \frac{900}{60} = 15 \text{ minutes}$.

(B) The power $P$ is defined as the rate of doing work,which is equal to the change in kinetic energy divided by time.
$P = \frac{\Delta K.E.}{t} = \frac{\frac{1}{2}m(v^2 - u^2)}{t}$
Given:
Mass $m = 2.05 \times 10^6 \; kg$
Initial velocity $u = 5 \; m/s$
Final velocity $v = 25 \; m/s$
Time $t = 5 \; min = 5 \times 60 \; s = 300 \; s$
Substituting the values:
$P = \frac{0.5 \times 2.05 \times 10^6 \times (25^2 - 5^2)}{300}$
$P = \frac{0.5 \times 2.05 \times 10^6 \times (625 - 25)}{300}$
$P = \frac{0.5 \times 2.05 \times 10^6 \times 600}{300}$
$P = 0.5 \times 2.05 \times 10^6 \times 2$
$P = 2.05 \times 10^6 \; W = 2.05 \; MW$
Thus,the power of the engine is $2.05 \; MW$.

(C) The rate of doing work is defined as power $(P)$.
Power is given by the formula: $P = \frac{W}{t} = \frac{mgh}{t}$.
Since the staircase is the same,the height $(h)$ and acceleration due to gravity $(g)$ are constant for both men.
Therefore,the ratio of their power is: $\frac{P_1}{P_2} = \frac{m_1}{m_2} \times \frac{t_2}{t_1}$.
Given: $m_1 = 60 \ kg$,$t_1 = 12 \ s$,$m_2 = 50 \ kg$,$t_2 = 11 \ s$.
Substituting the values: $\frac{P_1}{P_2} = \frac{60}{50} \times \frac{11}{12}$.
$\frac{P_1}{P_2} = \frac{6}{5} \times \frac{11}{12} = \frac{66}{60} = \frac{11}{10}$.
Thus,the ratio is $11:10$.

(A) The power $P$ developed by the man is given by the rate of doing work against gravity: $P = \frac{W}{t} = \frac{mgh}{t}$.
Given: mass $m = 80 \; kg$,height $h = 6 \; m$,time $t = 10 \; s$,and acceleration due to gravity $g = 9.8 \; m/s^2$.
Substituting the values: $P = \frac{80 \times 9.8 \times 6}{10} = 470.4 \; W$.
To convert power from Watts to Horsepower $(HP)$,we use the conversion factor $1 \; HP = 746 \; W$.
Thus,$P = \frac{470.4}{746} \; HP \approx 0.63 \; HP$.

(B) Given: Mass $m = 1000 \ kg$,initial velocity $u = 0$,final velocity $v = 54 \ km/h = 54 \times \frac{5}{18} \ m/s = 15 \ m/s$,time $t = 5 \ s$.
Average power is defined as the rate of work done,which is equal to the change in kinetic energy divided by time.
$P_{avg} = \frac{\Delta K.E.}{t} = \frac{\frac{1}{2}mv^2 - 0}{t}$.
Substituting the values: $P_{avg} = \frac{0.5 \times 1000 \times (15)^2}{5}$.
$P_{avg} = \frac{500 \times 225}{5} = 100 \times 225 = 22500 \ W$.

(A) The power of the motor is $P = \frac{1}{4} \text{ hp} = 0.25 \times 746 \text{ W} = 186.5 \text{ W}$.
Since the motor runs at $600 \text{ r.p.m.}$,the frequency is $n = \frac{600}{60} = 10 \text{ revolutions per second}$.
The time taken for one rotation is $t = \frac{1}{n} = \frac{1}{10} \text{ s} = 0.1 \text{ s}$.
The input energy for one rotation is $E_{\text{in}} = P \times t = 186.5 \times 0.1 = 18.65 \text{ J}$.
Given the efficiency is $40\%$,the useful work done is $W = \text{Efficiency} \times E_{\text{in}}$.
$W = 0.40 \times 18.65 \text{ J} = 7.46 \text{ J}$.

(A) The power $P$ is defined as the rate of doing work,given by $P = \frac{W}{t}$.
Here,the work done $W$ in lifting the water is equal to the change in potential energy,$W = mgh$.
Given values are:
Mass $m = 200 \ kg$
Height $h = 50 \ m$
Time $t = 10 \ s$
Acceleration due to gravity $g = 10 \ m/s^2$
Substituting these values into the formula:
$P = \frac{mgh}{t} = \frac{200 \times 10 \times 50}{10}$
$P = 200 \times 50 = 10,000 \ W$
$P = 10 \times 10^3 \ W$.
Therefore,the correct option is $A$.

(A) Given:
Number of bullets per minute,$n = 360 \text{ bullets/min} = \frac{360}{60} \text{ bullets/sec} = 6 \text{ bullets/sec}$.
Speed of each bullet,$v = 360 \ km/h = 360 \times \frac{5}{18} \ m/s = 100 \ m/s$.
Mass of each bullet,$m = 20 \ g = 20 \times 10^{-3} \ kg = 2 \times 10^{-2} \ kg$.
Power of the gun is the total kinetic energy delivered per unit time:
$P = n \times \left( \frac{1}{2}mv^2 \right)$
$P = 6 \times \left( \frac{1}{2} \times 2 \times 10^{-2} \times (100)^2 \right)$
$P = 6 \times (10^{-2} \times 10000)$
$P = 6 \times 100 = 600 \ W$.

(C) Power is defined as the rate of doing work,given by the formula: $P = \frac{W}{t}$.
Since the amount of work $W$ done by both men is the same,the power $P$ is inversely proportional to the time $t$ taken $(P \propto \frac{1}{t})$.
Therefore,the ratio of the power of the first man $(P_1)$ to the second man $(P_2)$ is:
$\frac{P_1}{P_2} = \frac{t_2}{t_1} = \frac{20 \ s}{10 \ s} = \frac{2}{1} = 2$.
Thus,the ratio is $2$.

(C) Given that power $P$ is constant. We know that $P = Fv = mav = m \left( \frac{dv}{dt} \right) v$.
Integrating this expression: $\frac{P}{m} dt = v dv$.
Integrating both sides: $\int \frac{P}{m} dt = \int v dv \implies \frac{P}{m} t = \frac{v^2}{2}$.
Thus,$v^2 = \frac{2P}{m} t$,which gives $v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Since $v = \frac{ds}{dt}$,we have $ds = \sqrt{\frac{2P}{m}} t^{1/2} dt$.
Integrating with respect to $t$: $s = \int \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \left( \frac{t^{3/2}}{3/2} \right) = \sqrt{\frac{2P}{m}} \left( \frac{2}{3} t^{3/2} \right)$.
Therefore,$s \propto t^{3/2}$.

(B) The instantaneous power $P$ delivered to a body is given by the product of the force $F$ and the velocity $v$ at that instant,i.e.,$P = F \cdot v$.
Since the force $F$ is constant,the acceleration $a$ of the body is also constant,given by $a = F/m$.
Using the equation of motion $v = u + at$,where initial velocity $u = 0$,we have $v = (F/m)t$.
Thus,the force can be expressed as $F = (mv)/t$.
Substituting this into the power formula: $P = F \cdot v = (mv/t) \cdot v = (mv^2)/t$.

(B) Power $P$ is given by $P = Fv = (ma)v = m(\frac{dv}{dt})v$.
Since $P$ is constant,$m v dv = P dt$.
Integrating both sides,$\int_0^v m v dv = \int_0^t P dt$,which gives $\frac{1}{2} m v^2 = Pt$.
So,$v^2 = \frac{2Pt}{m}$,or $v = (\frac{2P}{m})^{1/2} t^{1/2}$.
Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = (\frac{2P}{m})^{1/2} t^{1/2}$.
Integrating with respect to $t$,$s = \int_0^t (\frac{2P}{m})^{1/2} t^{1/2} dt$.
$s = (\frac{2P}{m})^{1/2} [\frac{t^{3/2}}{3/2}] = (\frac{2P}{m})^{1/2} \cdot \frac{2}{3} t^{3/2}$.
$s = (\frac{4}{9} \cdot \frac{2P}{m})^{1/2} t^{3/2} = (\frac{8P}{9m})^{1/2} t^{3/2}$.

(D) Power $p$ is defined as the rate of doing work,$p = Fv$.
Since $F = ma = m \frac{dv}{dt}$,we can write $p = m v \frac{dv}{dt}$.
Rearranging the terms,we get $p \, dt = m v \, dv$.
Integrating both sides from $t = 0$ to $t$ and $v = 0$ to $v$:
$\int_{0}^{t} p \, dt = \int_{0}^{v} m v \, dv$.
$pt = m \left[ \frac{v^2}{2} \right]_{0}^{v}$.
$pt = \frac{1}{2} m v^2$.
Solving for $v$,we get $v^2 = \frac{2pt}{m}$,which implies $v = \sqrt{\frac{2pt}{m}}$.

(A) The power $P$ is defined as the rate of doing work,$P = \frac{W}{t}$.
Here,the work done against gravity is $W = mgh$.
Given: $m = 80 \ kg$,$g = 9.8 \ m/s^2$,$h = 6 \ m$,and $t = 10 \ s$.
Substituting the values: $P = \frac{80 \times 9.8 \times 6}{10} = 470.4 \ W$.
To convert power from Watts to Horsepower $(HP)$,we use the conversion factor $1 \ HP = 746 \ W$.
Therefore,$P_{HP} = \frac{470.4}{746} \approx 0.63 \ HP$.

(C) The power $P$ delivered by the engine is given by $P = F_{engine} \cdot v$,where $F_{engine}$ is the force exerted by the engine and $v$ is the velocity.
$F_{engine} = \frac{P}{v} = \frac{30 \times 10^3 \ W}{30 \ m/s} = 1000 \ N$.
The net force $F_{net}$ acting on the car is the difference between the engine force and the resistive force $F_{resistive}$.
$F_{net} = F_{engine} - F_{resistive} = 1000 \ N - 750 \ N = 250 \ N$.
Using Newton's second law,$F_{net} = m \cdot a$,where $m$ is the mass of the car and $a$ is the acceleration.
$a = \frac{F_{net}}{m} = \frac{250 \ N}{1250 \ kg} = \frac{1}{5} \ m/s^2 = 0.2 \ m/s^2$.

(B) Power $P = Fv = (ma)v = m \left( \frac{dv}{dt} \right) v = \text{constant}$.
Integrating $v \, dv = \frac{P}{m} \, dt$,we get $\frac{1}{2} v^2 = \frac{P}{m} t$,which implies $v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{\frac{2P}{m}} t^{1/2}$.
Integrating with respect to time $t$,$x = \int \sqrt{\frac{2P}{m}} t^{1/2} \, dt = \sqrt{\frac{2P}{m}} \cdot \frac{t^{3/2}}{3/2}$.
Thus,$x \propto t^{3/2}$.

(A) Given:
Mass $m = 100 \ kg$
Velocity $v = 7.2 \ km/hr = 7.2 \times \frac{5}{18} \ m/s = 2 \ m/s$
Incline length $l = 20 \ m$
Height $h = 1 \ m$
From the geometry of the incline,$\sin \theta = \frac{h}{l} = \frac{1}{20}$.
The force required to move up the incline at a constant speed is $F = mg \sin \theta$.
The power $P$ is given by $P = F \cdot v = (mg \sin \theta) \cdot v$.
Substituting the values:
$P = 100 \times 9.8 \times \frac{1}{20} \times 2$
$P = 5 \times 9.8 \times 2 = 98 \ W$.

(A) The useful power delivered to the water is given by $P_{\text{out}} = \frac{mgh}{t}$.
Substituting the values: $P_{\text{out}} = \frac{100 \times 10 \times 10}{5} = 2000 \ W$.
Given that the efficiency $\eta = 60\% = 0.6$,the input power $P_{\text{in}}$ is related to the output power by $P_{\text{out}} = \eta \times P_{\text{in}}$.
Therefore,$P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{2000}{0.6} = 3333.33 \ W$.
Converting to $kW$: $P_{\text{in}} \approx 3.33 \ kW$.

(C) Power $P$ is given by $P = Fv = mav = m(dv/dt)v$.
Rearranging the terms,we get $(P/m) dt = v dv$.
Integrating both sides,$\int (P/m) dt = \int v dv$,which gives $(P/m)t = v^2/2$.
Thus,$v = (2P/m)^{1/2} t^{1/2}$.
Since $v = ds/dt$,we have $ds = (2P/m)^{1/2} t^{1/2} dt$.
Integrating $s = \int ds = (2P/m)^{1/2} \int t^{1/2} dt = (2P/m)^{1/2} [t^{3/2} / (3/2)]$.
Therefore,$s = (2/3)(2P/m)^{1/2} t^{3/2}$,which implies $s \propto t^{3/2}$.

(A) Power is defined as the rate of doing work,$P = \frac{W}{t}$.
Work done by the heart is given by the product of pressure and the change in volume,$W = P \times \Delta V$.
Given: Pressure $P = 20000 \ N/m^2$,Volume $\Delta V = 1 \ cc = 1 \times 10^{-6} \ m^3$,and time $t = 1 \ s$.
Substituting the values: $P = \frac{20000 \times 1 \times 10^{-6}}{1} \ W$.
$P = 2 \times 10^{-2} \ W = 0.02 \ W$.

(B) The power $P$ required by the pump is given by the rate of change of mechanical energy: $P = \left(gh + \frac{v^{2}}{2}\right) \frac{dm}{dt}$.
First,calculate the mass flow rate $\frac{dm}{dt} = \rho A v$.
Given $\rho = 1000\; kg/m^{3}$,$A = 10\; cm^{2} = 10 \times 10^{-4}\; m^{2}$,and $v = 5\; m/s$:
$\frac{dm}{dt} = 1000 \times 10 \times 10^{-4} \times 5 = 5\; kg/s$.
Now,substitute the values into the power equation using $g = 10\; m/s^{2}$,$h = 2.5\; m$,and $v = 5\; m/s$:
$P = \left(10 \times 2.5 + \frac{5^{2}}{2}\right) \times 5$.
$P = (25 + 12.5) \times 5$.
$P = 37.5 \times 5 = 187.5\; W$.
Rounding to the nearest provided option,the correct value is $185\; W$.

(A) Given that the body starts from rest $(u = 0)$ and attains velocity $v$ in time $T$ with uniform acceleration $a$.
Using the equation of motion $v = u + at$,we get $v = 0 + aT$,which implies $a = \frac{v}{T}$.
At any time $t$,the velocity of the body is $v(t) = at = \left( \frac{v}{T} \right) t$.
The force acting on the body is $F = ma = m \left( \frac{v}{T} \right)$.
Instantaneous power $P$ is given by the product of force and velocity: $P = F \cdot v(t)$.
Substituting the values: $P = \left( m \frac{v}{T} \right) \left( \frac{v}{T} t \right) = \frac{mv^2}{T^2} t$.

(A) The power $P$ generated by the falling water is given by the rate of change of potential energy.
$P = \frac{dU}{dt} = \frac{d(mgh)}{dt} = gh \frac{dm}{dt}$
Given:
Rate of mass flow $\frac{dm}{dt} = 100 \ kg/s$
Height $h = 100 \ m$
Acceleration due to gravity $g = 10 \ m/s^2$
Substituting the values:
$P = 10 \ m/s^2 \times 100 \ m \times 100 \ kg/s$
$P = 100,000 \ W = 100 \ kW$
Therefore,the power generated is $100 \ kW$.

(D) The output power is given by $P_{\text{out}} = \frac{mgh}{t}$.
Given: $m = 7200 \ kg$, $h = 100 \ m$, $t = 1 \ hour = 3600 \ s$, and $g = 10 \ m/s^2$.
$P_{\text{out}} = \frac{7200 \times 10 \times 100}{3600} = 2000 \ W = 2 \ kW$.
Efficiency $\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$.
Given $\eta = 50\% = 0.5$, so $P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{2 \ kW}{0.5} = 4 \ kW$.

(A) Power is defined as $P = Fv$. Since $F = ma$,we have $P = mav$.
Using the definition of acceleration $a = v \frac{dv}{dx}$,we can write $P = m v \frac{dv}{dx} v = m v^2 \frac{dv}{dx}$.
Rearranging the terms,we get $P dx = m v^2 dv$.
Integrating both sides with initial conditions $x=0$ at $v=0$:
$\int_0^x P dx = \int_0^v m v^2 dv$.
$Px = m \frac{v^3}{3}$.
Solving for $v$,we get $v^3 = \frac{3Px}{m}$,which implies $v = {\left( {\frac{{3xP}}{m}} \right)^{1/3}}$.

(C) The power $P$ required to lift a load at a constant speed is given by the formula $P = Fv$,where $F$ is the force and $v$ is the velocity.
Since the elevator moves at a constant speed,the force $F$ is equal to the weight of the load,$mg$.
Given: $m = 500 \ kg$,$g = 9.8 \ m/s^2$,$v = 0.4 \ m/s$.
$P = (mg)v = 500 \times 9.8 \times 0.4 = 1960 \ W$.
To convert power from Watts to Horsepower $(H.P.)$,we use the conversion factor $1 \ H.P. = 746 \ W$.
$P = \frac{1960}{746} \approx 2.62 \ H.P.$

(B) Given: Mass $m = 10,000 \ kg$,Slope $\sin \theta = \frac{1}{50}$,Velocity $v = 36 \ km/hr = 36 \times \frac{5}{18} = 10 \ m/s$,$g = 10 \ m/s^2$.
The force required to move the truck up the incline against gravity is $F = mg \sin \theta$.
$F = 10,000 \times 10 \times \frac{1}{50} = 2,000 \ N$.
The power $P$ is given by the product of force and velocity:
$P = F \times v = 2,000 \ N \times 10 \ m/s = 20,000 \ W$.
Converting to $kW$:
$P = 20 \ kW$.

(C) Mass of water pumped per minute = $2400 \; kg$.
Mass of water pumped per second $(m)$ = $\frac{2400}{60} = 40 \; kg/s$.
Velocity of water $(v)$ = $3 \; m/s$.
Kinetic energy per second (Power) = $\frac{1}{2} m v^2 = \frac{1}{2} \times 40 \times (3)^2 = 20 \times 9 = 180 \; W$ (or $J/s$).
Total time $(t)$ = $10 \; hours = 10 \times 3600 = 36000 \; s$.
Total work done = $\text{Power} \times \text{Time} = 180 \times 36000 = 6,480,000 \; J = 6.48 \times 10^6 \; J$.

(A) The mass of water $m = \text{Volume} \times \text{Density} = 30 \ m^3 \times 1000 \ kg/m^3 = 30,000 \ kg$.
The output power required to lift the water is $P_{\text{out}} = \frac{mgh}{t}$.
Given $h = 40 \ m$,$t = 15 \ minutes = 15 \times 60 \ s = 900 \ s$,and $g = 9.8 \ m/s^2$.
$P_{\text{out}} = \frac{30,000 \times 9.8 \times 40}{900} = \frac{11,760,000}{900} = 13,066.67 \ W$.
The efficiency $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%$.
$30 = \frac{13,066.67}{P_{\text{in}}} \times 100$.
$P_{\text{in}} = \frac{13,066.67 \times 100}{30} = 43,555.56 \ W \approx 43.6 \ kW$.

(A) Given: Mass = $m$,Initial velocity $u = 0$,Final velocity at time $T$ is $v$.
Since the body is accelerated,the acceleration $a$ is constant: $a = \frac{v - u}{T} = \frac{v}{T}$.
The velocity at any time $t$ is given by $v(t) = u + at = 0 + (\frac{v}{T})t = \frac{v}{T}t$.
The force acting on the body is $F = ma = m(\frac{v}{T})$.
The instantaneous power $P$ is given by $P = F \cdot v(t)$.
Substituting the values: $P = (m \cdot \frac{v}{T}) \cdot (\frac{v}{T}t) = \frac{mv^2}{T^2}t$.

(B) Given that the body starts from rest,initial velocity $u = 0$.
Using the equation of motion $v = u + at$,we have $v_1 = 0 + a t_1$,which gives acceleration $a = \frac{v_1}{t_1}$.
At any time $t$,the velocity of the body is $v(t) = at = \left( \frac{v_1}{t_1} \right) t$.
The force acting on the body is $F = ma = m \left( \frac{v_1}{t_1} \right)$.
Instantaneous power $P$ is given by $P = F \cdot v$.
Substituting the values,$P = \left( m \frac{v_1}{t_1} \right) \times \left( \frac{v_1}{t_1} t \right) = \frac{m v_1^2 t}{t_1^2}$.

(D) The work done $(W)$ by the weightlifter against gravity is equal to the change in potential energy: $W = mgh$.
Given: mass $m = 300\; kg$,height $h = 2\; m$,time $t = 3\; s$,and acceleration due to gravity $g = 9.8\; m/s^2$.
Power $(P)$ is defined as the rate of doing work: $P = \frac{W}{t} = \frac{mgh}{t}$.
Substituting the values: $P = \frac{300 \times 9.8 \times 2}{3}$.
$P = 100 \times 9.8 \times 2 = 1960\; W$.

(B) The power $P$ is given by the product of force $F$ and velocity $v$.
Given: Force $F = 9000 \ N$,Velocity $v = 2 \ ms^{-1}$.
Using the formula $P = F \times v$,we get:
$P = 9000 \ N \times 2 \ ms^{-1} = 18000 \ W$.
Since $1 \ kW = 1000 \ W$,we convert the power to $kW$:
$P = \frac{18000}{1000} \ kW = 18 \ kW$.

(D) Power $P$ is given by $P = Fv = (ma)v$. Since $F = ma$,we have $P = m(dv/dt)v$.
Integrating $P dt = mv dv$,we get $Pt = (1/2)mv^2$,which implies $v = \sqrt{2Pt/m} = \sqrt{2P/m} \cdot t^{1/2}$.
Since $v = ds/dt$,we have $ds/dt = \sqrt{2P/m} \cdot t^{1/2}$.
Integrating with respect to time $t$,$s = \int \sqrt{2P/m} \cdot t^{1/2} dt = \sqrt{2P/m} \cdot (t^{3/2} / (3/2))$.
Therefore,$s \propto t^{3/2}$.

(D) Power $P = 2 \; kW = 2000 \; W$. Time $t = 1 \; \text{minute} = 60 \; s$. Height $h = 10 \; m$. Acceleration due to gravity $g = 10 \; m/s^2$.
Using the formula for power: $P = \frac{mgh}{t}$.
Rearranging for mass $m$: $m = \frac{P \cdot t}{gh} = \frac{2000 \times 60}{10 \times 10} = \frac{120000}{100} = 1200 \; kg$.
Since the density of water is $1000 \; kg/m^3$, the volume $V$ in $m^3$ is $V = \frac{m}{\rho} = \frac{1200}{1000} = 1.2 \; m^3$.
Since $1 \; m^3 = 1000 \; \text{liters}$, the volume in liters is $1.2 \times 1000 = 1200 \; \text{liters}$.