Power Questions in English

(N/A) The kinetic energy $K$ of an object of mass $m$ moving with velocity $v$ is given by $K = \frac{1}{2}mv^2$.
The rate of change of kinetic energy with respect to time $t$ is given by $\frac{dK}{dt} = \frac{d}{dt}(\frac{1}{2}mv^2)$.
Assuming mass $m$ is constant,we apply the chain rule: $\frac{dK}{dt} = \frac{1}{2}m \cdot 2v \cdot \frac{dv}{dt} = mv \cdot \frac{dv}{dt}$.
Since acceleration $a = \frac{dv}{dt}$,we have $\frac{dK}{dt} = mva$.
By Newton's second law,$F = ma$,so $\frac{dK}{dt} = Fv$.
Power $P$ is defined as the rate of doing work,$P = \frac{dW}{dt} = F \cdot v$.
Thus,the rate of change of kinetic energy is equal to the power delivered by the force,i.e.,$\frac{dK}{dt} = P$.

(N/A) The time rate of doing work is known as power.
Power is defined as the work done per unit time or the time rate at which work is done or energy is transformed.
If $\Delta W$ is the work done in time interval $\Delta t$,the average power in time interval $\Delta t$ is $\langle P \rangle = \frac{\Delta W}{\Delta t}$.
Therefore,the instantaneous power at time $t$ is:
$P = \lim_{\Delta t \rightarrow 0} \frac{\Delta W}{\Delta t} = \frac{dW}{dt} \quad (1)$
If $dW$ is the work done by a force $\vec{F}$ during the displacement $d\vec{r}$,then $dW = \vec{F} \cdot d\vec{r}$.
Substituting the value of $dW$ in equation $(1)$:
$P = \frac{d}{dt}(\vec{F} \cdot d\vec{r}) = \vec{F} \cdot \frac{d\vec{r}}{dt} \quad [\because \vec{F} \text{ is constant}]$
$P = \vec{F} \cdot \vec{v} \quad [\because \frac{d\vec{r}}{dt} = \vec{v}]$
Power is a scalar quantity. Its dimensional formula is $M^{1} L^{2} T^{-3}$. The $SI$ unit of power is $J s^{-1}$. This unit is named as $watt$ in honor of the inventor of the steam engine,James Watt.
$1 W = 1 J s^{-1}$.
Larger units of power include:
$1 kW = 10^{3} W, 1 MW = 10^{6} W$.
To measure the power of vehicles and water pumps in practice,horsepower $(hp)$ is used,which is a unit of the British system.
$1 hp = 746 W$.

(FALSE) The statement is false.
$1\,kWh$ (kilowatt-hour) is a unit of energy,not power.
Power is defined as the rate of doing work,measured in watts $(W)$ or kilowatts $(kW)$.
Energy is defined as the product of power and time,which is why $1\,kWh = 1\,kW \times 1\,h = 1000\,W \times 3600\,s = 3.6 \times 10^6\,J$.

(N/A) When the elevator is descending,electrical power is required to control the speed of the lift and prevent it from entering a state of free fall under gravity.
Additionally,as the total weight inside the elevator increases,the gravitational force acting on it increases,which would cause the speed of descent to increase uncontrollably. Therefore,there must be a limit on the number of passengers to ensure the braking system can safely manage the load.

(A) Given:
Mass,$m = 100 \, kg$
Height,$h = 10 \, m$
Time,$t = 20 \, s$
Acceleration due to gravity,$g = 9.8 \, m/s^2$
The work done $(W)$ by the crane is equal to the potential energy gained by the mass: $W = mgh$.
Power $(P)$ is defined as the rate of doing work: $P = \frac{W}{t} = \frac{mgh}{t}$.
Substituting the values:
$P = \frac{100 \times 9.8 \times 10}{20}$
$P = \frac{9800}{20}$
$P = 490 \, W$.

(0.6 W) The work done per beat $= 0.5 \, J$.
Total work done for $72$ beats is given by:
$W = 72 \times 0.5 \, J = 36 \, J$.
The time taken for $72$ beats is $1 \, \text{minute} = 60 \, s$.
Power is defined as the rate of doing work:
$P = \frac{W}{t} = \frac{36 \, J}{60 \, s} = 0.6 \, W$.
Thus,the power used by the heart is $0.6 \, W$.

(C) The power $P$ supplied to the particle is constant. Since power is the rate of change of kinetic energy,we have $\frac{dK}{dt} = P$. Integrating this with respect to time,we get $K = Pt$ (assuming initial kinetic energy is zero).
Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = Pt$,which implies $v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Since velocity $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = \sqrt{\frac{2P}{m}} t^{1/2}$.
Integrating with respect to time,$s = \int \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2}$.
Thus,$s \propto t^{3/2}$.
The graph of $s$ versus $t$ for $s \propto t^{3/2}$ is a curve that is concave upwards (increasing slope),which corresponds to Graph $C$.

(D) Given: Mass $m = 2\, kg$,Power $P = 1\, J/s$,Time $t = 9\, s$,Initial velocity $u = 0$.
Power is defined as $P = F \cdot v = (m \cdot a) \cdot v = m \cdot \frac{dv}{dt} \cdot v$.
Rearranging the terms: $m \cdot v \cdot dv = P \cdot dt$.
Integrating both sides from rest ($v=0$ at $t=0$): $\int_{0}^{v} m \cdot v \cdot dv = \int_{0}^{t} P \cdot dt$.
$m \cdot \frac{v^2}{2} = P \cdot t \Rightarrow v^2 = \frac{2Pt}{m} \Rightarrow v = \sqrt{\frac{2Pt}{m}}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{\frac{2P}{m}} \cdot t^{1/2}$.
Integrating to find distance $x$: $\int_{0}^{x} dx = \sqrt{\frac{2P}{m}} \int_{0}^{t} t^{1/2} dt$.
$x = \sqrt{\frac{2P}{m}} \cdot \frac{t^{3/2}}{3/2} = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} \cdot t^{3/2}$.
Substituting the values $m = 2$,$P = 1$,$t = 9$: $x = \sqrt{\frac{2 \cdot 1}{2}} \cdot \frac{2}{3} \cdot 9^{3/2} = 1 \cdot \frac{2}{3} \cdot (3^2)^{3/2} = \frac{2}{3} \cdot 27 = 18\, m$.

(B) Power $P$ is constant, so $P = F \cdot V = \text{constant}$.
Since $F = m \cdot a = m \frac{dV}{dt}$, we have $m \frac{dV}{dt} \cdot V = \text{constant}$.
This implies $V \frac{dV}{dt} \propto 1$, which integrates to $\frac{V^2}{2} \propto t$, or $V^2 \propto t$.
Therefore, $V \propto t^{1/2}$.
Since $V = \frac{dx}{dt}$, we have $\frac{dx}{dt} \propto t^{1/2}$.
Integrating both sides with respect to time $t$, we get $x \propto \int t^{1/2} dt$.
Thus, $x \propto t^{3/2}$.

(C) Given that the power $P$ delivered by the machine is constant.
The work done by the machine in time $t$ is $W = P \cdot t$.
According to the work-energy theorem,the work done is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} m v^2$.
Equating the two,we get $\frac{1}{2} m v^2 = P \cdot t$.
This implies $v^2 \propto t$,or $v \propto t^{1/2}$.
Since velocity $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = k \cdot t^{1/2}$ (where $k$ is a constant).
Integrating with respect to time $t$ to find the distance $x$:
$x = \int k \cdot t^{1/2} dt = k \cdot \frac{t^{1/2 + 1}}{1/2 + 1} = k \cdot \frac{t^{3/2}}{3/2}$.
Therefore,$x \propto t^{3/2}$.

(A) Power $P$ is given by $P = Fv = (ma)v = m \left(\frac{dv}{dt}\right)v$.
Since $P$ is constant,$P = mv \frac{dv}{dt}$.
Integrating with respect to time: $\int_{0}^{v} mv \, dv = \int_{0}^{t} P \, dt$.
$\frac{1}{2}mv^2 = Pt \implies v = \sqrt{\frac{2Pt}{m}} = \left(\frac{2P}{m}\right)^{1/2} t^{1/2}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \left(\frac{2P}{m}\right)^{1/2} t^{1/2}$.
Integrating with respect to position and time: $\int_{0}^{x} dx = \int_{0}^{t} \left(\frac{2P}{m}\right)^{1/2} t^{1/2} dt$.
$x = \left(\frac{2P}{m}\right)^{1/2} \left[ \frac{t^{3/2}}{3/2} \right] = \left(\frac{2P}{m}\right)^{1/2} \cdot \frac{2}{3} t^{3/2}$.
$x = \left( \frac{4}{9} \cdot \frac{2P}{m} \right)^{1/2} t^{3/2} = \left( \frac{8P}{9m} \right)^{1/2} t^{3/2}$.

(B) The power output $P$ is given by $50 \%$ of the potential energy per second of the falling water.
Power $P = 0.50 \times \left( \frac{mgh}{t} \right)$.
Since mass $m = \rho V$,where $\rho$ is the density of water $(1000 \,kg/m^3)$ and $V$ is the volume,we have:
$P = 0.50 \times \left( \frac{V}{t} \right) \rho g h$.
Given $P = 1.00 \times 10^9 \,W$,$g = 10 \,ms^{-2}$,$\rho = 1000 \,kg/m^3$,and $h = 500 \,m$.
Rearranging for the flow rate $\frac{V}{t}$:
$\frac{V}{t} = \frac{P}{0.50 \times \rho \times g \times h}$.
$\frac{V}{t} = \frac{1.00 \times 10^9}{0.50 \times 1000 \times 10 \times 500}$.
$\frac{V}{t} = \frac{1.00 \times 10^9}{2.5 \times 10^6} = 400 \,m^3/s$.

(B) Given:
Input power $(P_{\text{in}})$ = $750 \,W$
Volume of water $(V)$ = $300 \,L$,so mass $(m)$ = $300 \,kg$ (since density of water is $1 \,kg/L$)
Height $(h)$ = $6 \,m$
Time $(t)$ = $1 \,minute = 60 \,s$
Acceleration due to gravity $(g)$ = $10 \,m/s^2$
The useful power output $(P_{\text{out}})$ is the rate of doing work against gravity:
$P_{\text{out}} = \frac{mgh}{t} = \frac{300 \times 10 \times 6}{60} = \frac{18000}{60} = 300 \,W$
The efficiency $(\eta)$ is defined as the ratio of useful power output to input power:
$\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{300}{750} = 0.4$
Converting to percentage:
$\eta = 0.4 \times 100 = 40 \%$

(C) Given: Power $P = 4 \,kW = 4000 \,W$,acceleration due to gravity $g = 10 \,m/s^2$,height $h = 20 \,m$,and time $t = 1 \,minute = 60 \,s$.
The formula for power is $P = \frac{W}{t} = \frac{mgh}{t}$.
Substituting the values: $4000 = \frac{m \times 10 \times 20}{60}$.
Simplifying the equation: $4000 = \frac{200m}{60} = \frac{10m}{3}$.
Solving for $m$: $m = \frac{4000 \times 3}{10} = 1200 \,kg$.
Since the density of water is $1 \,kg/litre$,the volume of water is $1200 \,litres$.

(C) The power $P$ is defined as the rate of doing work,given by $P = \frac{W}{t}$.
Here,the work done $W$ is equal to the potential energy gained by the water,$W = mgh$.
Given:
Mass $m = 500 \,kg$
Height $h = 100 \,m$
Time $t = 10 \,s$
Acceleration due to gravity $g = 10 \,m/s^2$
Substituting the values:
$W = 500 \times 10 \times 100 = 500,000 \,J$
$P = \frac{500,000 \,J}{10 \,s} = 50,000 \,W$
Since $1 \,kW = 1,000 \,W$,we have:
$P = \frac{50,000}{1,000} \,kW = 50 \,kW$.
Therefore,the correct option is $C$.

(A) The power delivered by the engine is defined as the rate of change of kinetic energy.
$P = \frac{dK}{dt}$
Integrating this expression over time $t$ to reach a final kinetic energy $K$ from rest $(K_i = 0)$:
$P \cdot t = K_f - K_i$
Since the final kinetic energy is $K_f = \frac{1}{2} m v^2$ and the initial kinetic energy is $0$:
$P \cdot t = \frac{1}{2} m v^2$
Solving for time $t$:
$t = \frac{m v^2}{2 P}$

(A) The power $P$ delivered to the turbine is given by the rate of change of potential energy of the falling water.
$P = \frac{W}{t} = \frac{mgh}{t}$
Given that the rate of mass flow $\frac{m}{t} = 100 \,kg/s$,the height $h = 100 \,m$,and taking acceleration due to gravity $g \approx 10 \,m/s^2$:
$P = \left(\frac{m}{t}\right) \times g \times h$
$P = 100 \,kg/s \times 10 \,m/s^2 \times 100 \,m$
$P = 100,000 \,W$
Since $1 \,kW = 1000 \,W$,we have:
$P = \frac{100,000}{1000} \,kW = 100 \,kW$.
Therefore,the correct option is $A$.

(D) Given that the force $F$ required to row a boat at constant velocity $v$ is proportional to the square of its speed:
$F \propto v^2$
We know that power $P$ is defined as the product of force and velocity:
$P = F \cdot v$
Substituting the proportionality of force into the power equation:
$P \propto v^2 \cdot v = v^3$
Let $P_1$ be the power required for speed $v$,and $P_2$ be the power required for speed $2v$:
$\frac{P_2}{P_1} = \left( \frac{2v}{v} \right)^3 = 2^3 = 8$
Given $P_1 = 4 \, kW$:
$P_2 = 8 \times P_1 = 8 \times 4 \, kW = 32 \, kW$
Therefore,the power required for a speed of $2v \, km/h$ is $32 \, kW$.

(D) The correct option is $D$.
Power is defined as the rate at which work is done or energy is transferred. Mathematically,power $P$ is given by the formula:
$P = \frac{W}{t}$
where $W$ is the work performed and $t$ is the time interval over which the work is done.
Therefore,to determine the power output of an automobile,one must know the total work performed and the time elapsed during that work.

(C) Given force: $\vec{F} = t \hat{i} + 3t^2 \hat{j} \ N$ and mass $m = 1 \ kg$.
Using Newton's second law,$\vec{F} = m \vec{a} = m \frac{d\vec{v}}{dt}$.
Since $m = 1 \ kg$,$\frac{d\vec{v}}{dt} = t \hat{i} + 3t^2 \hat{j}$.
Integrating with respect to time $t$ (assuming initial velocity is zero at $t=0$): $\vec{v} = \int (t \hat{i} + 3t^2 \hat{j}) dt = \frac{t^2}{2} \hat{i} + t^3 \hat{j} \ m/s$.
Power $P$ is given by the dot product of force and velocity: $P = \vec{F} \cdot \vec{v}$.
$P = (t \hat{i} + 3t^2 \hat{j}) \cdot (\frac{t^2}{2} \hat{i} + t^3 \hat{j}) = \frac{t^3}{2} + 3t^5$.
At $t = 2 \ s$,$P = \frac{2^3}{2} + 3(2^5) = \frac{8}{2} + 3(32) = 4 + 96 = 100 \ W$.

(C) Given,mass $m = 2 \ kg$ and power $P$ is constant.
Work done by the source in time $t$ is $W = P \times t$.
According to the work-energy theorem,$W = \Delta K = \frac{1}{2} m v^2 - 0$.
So,$\frac{1}{2} m v^2 = P t \implies v = \sqrt{\frac{2 P t}{m}}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{\frac{2 P}{m}} \cdot t^{1/2}$.
Integrating with respect to $t$ from $0$ to $4 \ s$:
$x = \int_0^4 \sqrt{\frac{2 P}{m}} \cdot t^{1/2} dt = \sqrt{\frac{2 P}{m}} \left[ \frac{t^{3/2}}{3/2} \right]_0^4$.
Substituting $m = 2 \ kg$:
$x = \sqrt{\frac{2 P}{2}} \cdot \frac{2}{3} \cdot (4)^{3/2} = \sqrt{P} \cdot \frac{2}{3} \cdot 8 = \frac{16}{3} \sqrt{P}$.
Given displacement is $\frac{1}{3} \alpha^2 \sqrt{P}$.
Comparing $\frac{16}{3} \sqrt{P} = \frac{1}{3} \alpha^2 \sqrt{P}$,we get $\alpha^2 = 16$,so $\alpha = 4$.

(C) The total mass of the elevator system is $M = 1400\,kg$.
The elevator is moving upward at a uniform speed $v = 3\,m/s$. Since the speed is uniform,the acceleration is zero.
The motor must provide a force $F$ to overcome both the gravitational force and the frictional force.
The gravitational force is $F_g = M \times g = 1400\,kg \times 10\,m/s^2 = 14000\,N$.
The frictional force is $f = 2000\,N$.
Therefore,the total force required by the motor is $F = F_g + f = 14000\,N + 2000\,N = 16000\,N$.
The power $P$ used by the motor is given by $P = F \times v$.
$P = 16000\,N \times 3\,m/s = 48000\,W$.
Converting to kilowatts,$P = 48\,kW$.

(D) Average Power is defined as the total work done divided by the total time taken.
$P = \frac{W}{t} = \frac{mgh}{t}$
Given the ratio of powers of two motors is $\frac{P_1}{P_2} = \frac{3 \sqrt{x}}{\sqrt{x}+1}$.
For the first motor: $m_1 = 300 \ kg$,$t_1 = 5 \ minutes = 300 \ s$,$h = 100 \ m$.
For the second motor: $m_2 = 50 \ kg$,$t_2 = 2 \ minutes = 120 \ s$,$h = 100 \ m$.
Calculating the ratio of powers:
$\frac{P_1}{P_2} = \frac{m_1 g h / t_1}{m_2 g h / t_2} = \frac{m_1}{t_1} \times \frac{t_2}{m_2}$
$\frac{P_1}{P_2} = \frac{300}{5} \times \frac{2}{50} = 60 \times 0.04 = 2.4$
Equating the two expressions:
$\frac{3 \sqrt{x}}{\sqrt{x}+1} = 2.4$
$3 \sqrt{x} = 2.4 \sqrt{x} + 2.4$
$0.6 \sqrt{x} = 2.4$
$\sqrt{x} = 4$
$x = 16$

(D) Given force $\overrightarrow{F} = (6t \hat{i} + 6t^2 \hat{j}) \ N$ and mass $m = 2 \ kg$.
Using Newton's second law,$\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{6t \hat{i} + 6t^2 \hat{j}}{2} = (3t \hat{i} + 3t^2 \hat{j}) \ m/s^2$.
To find velocity $\overrightarrow{v}$,we integrate acceleration with respect to time: $\overrightarrow{v} = \int \overrightarrow{a} \ dt = \int (3t \hat{i} + 3t^2 \hat{j}) \ dt = (\frac{3t^2}{2} \hat{i} + t^3 \hat{j}) \ m/s$.
Power $P$ is the dot product of force and velocity: $P = \overrightarrow{F} \cdot \overrightarrow{v} = (6t \hat{i} + 6t^2 \hat{j}) \cdot (\frac{3t^2}{2} \hat{i} + t^3 \hat{j})$.
$P = (6t \cdot \frac{3t^2}{2}) + (6t^2 \cdot t^3) = 9t^3 + 6t^5 \ W$.

(C) Given that power $P$ is constant.
Since $P = F \cdot v = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v$,we have $m \cdot v \cdot \frac{dv}{dt} = P$.
Integrating both sides with respect to time: $\int m \cdot v \cdot dv = \int P \cdot dt$.
This gives $\frac{1}{2} m v^2 = P \cdot t$,which implies $v^2 \propto t$,or $v \propto t^{1/2}$.
Since velocity $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} \propto t^{1/2}$.
Integrating with respect to time: $s = \int t^{1/2} dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2}$.
Therefore,displacement $s \propto t^{3/2}$.

(D) Given:
Displacement $x = 2t - 1 \,m$
Force $F = 5 \,N$
Step $1$: Find the velocity of the particle.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(2t - 1) = 2 \,m/s$
Step $2$: Calculate the instantaneous power.
Power $P = F \cdot v$
$P = 5 \,N \times 2 \,m/s = 10 \,W$
Thus, the instantaneous power is $10 \,W$.

(C) The power delivered to the particle is constant,$P = 0.5 \ W$.
Since the initial speed is zero,the initial kinetic energy is $0 \ J$.
The work done by the force in time $t = 5 \ s$ is equal to the change in kinetic energy.
Work done $W = P \times t = 0.5 \ W \times 5 \ s = 2.5 \ J$.
Using the work-energy theorem,$W = \Delta K = \frac{1}{2}mv^2 - 0$.
$2.5 \ J = \frac{1}{2} \times 0.2 \ kg \times v^2$.
$2.5 = 0.1 \times v^2$.
$v^2 = 25$.
$v = 5 \ m/s$.

(B) Given: Mass $m = 4 \ kg$,Force $\overrightarrow{F} = (2 \hat{i} + 3 \hat{j}) \ N$,Displacement $\overrightarrow{d} = \overrightarrow{OQ} - \overrightarrow{OP} = (6-3) \hat{i} + (10-4) \hat{j} = (3 \hat{i} + 6 \hat{j}) \ m$,Time $t = 4 \ s$.
Work done $W = \overrightarrow{F} \cdot \overrightarrow{d} = (2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i} + 6 \hat{j}) = (2 \times 3) + (3 \times 6) = 6 + 18 = 24 \ J$.
Average power $\langle P \rangle = \frac{W}{t} = \frac{24}{4} = 6 \ W$.
Acceleration $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{2 \hat{i} + 3 \hat{j}}{4} = (0.5 \hat{i} + 0.75 \hat{j}) \ m/s^2$.
Velocity at $t = 4 \ s$ is $\overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a}t = 0 + (0.5 \hat{i} + 0.75 \hat{j}) \times 4 = (2 \hat{i} + 3 \hat{j}) \ m/s$.
Instantaneous power $P_{ins} = \overrightarrow{F} \cdot \overrightarrow{v} = (2 \hat{i} + 3 \hat{j}) \cdot (2 \hat{i} + 3 \hat{j}) = 4 + 9 = 13 \ W$.
The ratio of average power to instantaneous power is $\frac{\langle P \rangle}{P_{ins}} = \frac{6}{13}$.

(D) Given force $\vec{F} = (2t \hat{i} + 3t^2 \hat{j}) \ N$ and mass $m = 1000 \ g = 1 \ kg$.
Using Newton's second law,$\vec{F} = m\vec{a}$,we have $\vec{a} = \frac{\vec{F}}{m} = (2t \hat{i} + 3t^2 \hat{j}) \ m/s^2$.
Since $\vec{a} = \frac{d\vec{v}}{dt}$,we integrate with respect to time (assuming initial velocity $\vec{v} = 0$ at $t = 0$):
$\vec{v} = \int \vec{a} \ dt = \int (2t \hat{i} + 3t^2 \hat{j}) \ dt = t^2 \hat{i} + t^3 \hat{j} \ m/s$.
Power $P$ is defined as the dot product of force and velocity: $P = \vec{F} \cdot \vec{v}$.
$P = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j})$.
$P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 \ W$.

(A) The power $P$ exerted by a force $\vec{F}$ on a body moving with velocity $\vec{v}$ is given by the dot product: $P = \vec{F} \cdot \vec{v}$.
Given,$\vec{F} = (4 \hat{i} + \hat{j} - 2 \hat{k}) \text{ N}$ and $\vec{v} = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \text{ m s}^{-1}$.
Substituting these values into the formula:
$P = (4 \hat{i} + \hat{j} - 2 \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} + 3 \hat{k})$
$P = (4 \times 2) + (1 \times 2) + (-2 \times 3)$
$P = 8 + 2 - 6$
$P = 4 \text{ W}$.

(A) Given: Mass $m = 1 \ kg$,Force $\vec{F} = t \hat{i} + 2t^2 \hat{j}$.
Using Newton's second law,$\vec{F} = m \vec{a}$,so $\vec{a} = \frac{\vec{F}}{m} = t \hat{i} + 2t^2 \hat{j}$.
Velocity $\vec{v} = \int \vec{a} \ dt = \int (t \hat{i} + 2t^2 \hat{j}) \ dt = \frac{t^2}{2} \hat{i} + \frac{2t^3}{3} \hat{j}$ (assuming initial velocity is zero).
Power $P = \vec{F} \cdot \vec{v} = (t \hat{i} + 2t^2 \hat{j}) \cdot (\frac{t^2}{2} \hat{i} + \frac{2t^3}{3} \hat{j})$.
$P = \frac{t^3}{2} + \frac{4t^5}{3}$.
At $t = 3 \ s$:
$P = \frac{3^3}{2} + \frac{4(3^5)}{3} = \frac{27}{2} + 4(3^4) = 13.5 + 4(81) = 13.5 + 324 = 337.5 \ W$.

(B) Given: Mass $m = 1 \ kg$,initial velocity $u = 0$,time $t = 2 \ s$,final velocity $v = 10 \ m/s$.
First,find the acceleration $a$ using $v = u + at$:
$10 = 0 + a(2) \implies a = 5 \ m/s^2$.
Now,find the velocity $v_1$ at $t = 1 \ s$:
$v_1 = u + a(1) = 0 + 5(1) = 5 \ m/s$.
The power $P$ at any instant $t$ is given by $P = F \cdot v = (ma) \cdot v$.
At $t = 1 \ s$:
$P = (1 \ kg) \times (5 \ m/s^2) \times (5 \ m/s) = 25 \ W$.

(A) Initial kinetic energy $K_1 = \frac{1}{2} mu^2$.
Final kinetic energy $K_2 = \frac{1}{2} m(2u)^2 = \frac{1}{2} m(4u^2) = 2mu^2$.
Work done $W = \Delta K = K_2 - K_1 = 2mu^2 - \frac{1}{2} mu^2 = \frac{3}{2} mu^2$.
Power $P = \frac{W}{t} = \frac{3mu^2}{2t}$.

(A) Mass of water $m = 6 \text{ tonnes} = 6000 \ kg$.
Total height $h$ through which water is lifted = depth of well + height of first floor = $25 \ m + 35 \ m = 60 \ m$.
Time taken $t = 20 \text{ minutes} = 20 \times 60 \ s = 1200 \ s$.
Work done $W = mgh = 6000 \ kg \times 10 \ ms^{-2} \times 60 \ m = 3,600,000 \ J$.
Power $P = \frac{W}{t} = \frac{3,600,000 \ J}{1200 \ s} = 3000 \ W$.
Since $1 \ kW = 1000 \ W$,the power of the pump is $3 \ kW$.

(B) Let the body start from rest $(u = 0)$ and move with a constant acceleration $a$.
At any time $t$,the velocity $v$ of the body is given by the first equation of motion: $v = u + at = 0 + at = at$.
The force $F$ acting on the body is given by Newton's second law: $F = ma$.
Power $P$ delivered to the body is defined as the product of force and velocity: $P = F \cdot v$.
Substituting the expressions for $F$ and $v$: $P = (ma) \cdot (at) = ma^2t$.
Since mass $m$ and acceleration $a$ are constants,the term $ma^2$ is constant.
Therefore,$P \propto t$.

(D) The average power delivered by a force is defined as the work done by the force divided by the time taken.
$P_{av} = \frac{W_g}{t}$
At the highest point, the vertical displacement is the maximum height $H_{\max} = \frac{u^2 \sin^2 \theta}{2g}$.
The work done by gravity $W_g$ as the object moves from the ground to the highest point is $-mgH_{\max}$.
The time taken to reach the highest point is $t = \frac{u \sin \theta}{g}$.
Thus, the magnitude of average power is:
$P_{av} = \frac{mg H_{\max}}{t} = \frac{mg \left( \frac{u^2 \sin^2 \theta}{2g} \right)}{\left( \frac{u \sin \theta}{g} \right)}$
$P_{av} = \frac{mg u \sin \theta}{2}$

(D) The power $P$ delivered by a force $\vec{F}$ to an object moving with velocity $\vec{v}$ is given by the dot product of the force and velocity vectors: $P = \vec{F} \cdot \vec{v}$.
Given:
$\vec{v} = 7 \hat{i} + 2 \hat{j} - 5 \hat{k} \text{ m/s}$
$\vec{F} = 9 \hat{i} + 3 \hat{j} - 3 \hat{k} \text{ N}$
Calculating the dot product:
$P = (9 \hat{i} + 3 \hat{j} - 3 \hat{k}) \cdot (7 \hat{i} + 2 \hat{j} - 5 \hat{k})$
$P = (9 \times 7) + (3 \times 2) + (-3 \times -5)$
$P = 63 + 6 + 15$
$P = 84 \text{ W}$

(D) Given: Mass $m = 2000 \ kg$,Power $P = 10 \ kW = 10000 \ W$,Time $t = 10 \ s$.
We know that Power $P = F \cdot v = (m \cdot a) \cdot v = m \cdot \frac{dv}{dt} \cdot v$.
So,$P \cdot dt = m \cdot v \cdot dv$.
Integrating both sides from $t = 0$ to $t = 10 \ s$ and $v = 0$ to $v = v$:
$\int_{0}^{10} P \, dt = \int_{0}^{v} m \cdot v \, dv$.
$P \cdot t = \frac{1}{2} \cdot m \cdot v^2$.
$10000 \times 10 = \frac{1}{2} \times 2000 \times v^2$.
$100000 = 1000 \times v^2$.
$v^2 = 100$.
$v = 10 \ m \ s^{-1}$.

(B) Given: Mass of the train $m = 10^6 \ kg$,Speed $v = 108 \ km/h$.
First,convert the speed into $SI$ units: $v = 108 \times \frac{5}{18} \ m/s = 30 \ m/s$.
The frictional force per $100 \ kg$ is $0.5 \ N$.
Total frictional force $F = \frac{10^6 \ kg}{100 \ kg} \times 0.5 \ N = 10^4 \times 0.5 \ N = 5000 \ N$.
Since the train is moving at a constant speed,the driving force must be equal to the frictional force: $F_{drive} = F = 5000 \ N$.
Power $P = F_{drive} \times v = 5000 \ N \times 30 \ m/s = 150,000 \ W = 150 \ kW$.

(D) Power $(P)$ is defined as the rate of doing work, $P = Fv$. Since $P$ is constant, $Fv = \text{constant}$.
Using Newton's second law, $F = ma = m \frac{dv}{dt}$.
Thus, $m \frac{dv}{dt} v = P$, which implies $v dv = \frac{P}{m} dt$.
Integrating both sides, $\int v dv = \int \frac{P}{m} dt$, we get $\frac{1}{2} v^2 = \frac{P}{m} t + C$.
Assuming the body starts from rest at $t=0$, $C=0$, so $v^2 = \frac{2P}{m} t$, or $v = \sqrt{\frac{2P}{m}} t^{1/2}$.
Since $v = \frac{ds}{dt}$, we have $\frac{ds}{dt} = k t^{1/2}$ where $k = \sqrt{\frac{2P}{m}}$.
Integrating with respect to time, $s = \int k t^{1/2} dt = k \frac{t^{3/2}}{3/2} = \frac{2k}{3} t^{3/2}$.
Therefore, $s \propto t^{3/2}$, which means $x = \frac{3}{2}$.

(D) The work done $(W)$ to lift the coal is given by $W = mgh$, where $m = 8000 \ kg$, $g = 10 \ m \ s^{-2}$, and $h = 108 \ m$.
$W = 8000 \times 10 \times 108 = 8.64 \times 10^6 \ J$.
The time taken $(t)$ is $1 \ hour = 3600 \ s$.
The useful power output $(P_{out})$ is $P_{out} = W / t = (8.64 \times 10^6) / 3600 = 2400 \ W = 2.4 \ kW$.
Given the efficiency $(\eta)$ is $80 \% = 0.8$, the input power $(P_{in})$ is $P_{in} = P_{out} / \eta$.
$P_{in} = 2.4 / 0.8 = 3 \ kW$.
Thus, the power of the crane is $3 \ kW$.

(D) Power is defined as the dot product of force and velocity,given by $P = \vec{F} \cdot \vec{v}$.
Since the body is moving with a uniform speed '$v$',the net force acting on the body must be zero according to Newton's first law of motion.
From the figure,the forces $F_1$ and $F_2$ act in opposite directions. Therefore,the net force $F_{net} = F_1 - F_2 = 0$.
The net power $P_{net}$ is the power delivered by the net force:
$P_{net} = F_{net} \cdot v = (F_1 - F_2) \cdot v$.
Since $F_1 = F_2$ for uniform motion,$P_{net} = 0 \cdot v = 0$.

(C) The mass of water pumped per hour is $m = 7560 \text{ kg}$.
The time taken is $t = 1 \text{ hour} = 3600 \text{ s}$.
The depth of the well is $h = 100 \text{ m}$.
The acceleration due to gravity is $g = 10 \text{ m s}^{-2}$.
The useful work done (output energy) per second is the power output $P_{\text{out}} = \frac{mgh}{t}$.
$P_{\text{out}} = \frac{7560 \times 10 \times 100}{3600} = \frac{7560000}{3600} = 2100 \text{ W} = 2.1 \text{ kW}$.
Given the efficiency $\eta = 70 \% = 0.7$, the power input $P_{\text{in}}$ is given by $\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$.
$P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{2.1 \text{ kW}}{0.7} = 3 \text{ kW}$.
Therefore, the power of the pump is $3 \text{ kW}$.

(A) The formula for average power is $P_{av} = \frac{W}{t} = \frac{mgh}{t}$.
Given:
Mass $m = 90 \ kg$
Height $h = 600 \ m$
Time $t = 90 \ minutes = 90 \times 60 \ s = 5400 \ s$
Acceleration due to gravity $g = 10 \ ms^{-2}$.
Substituting the values:
$P_{av} = \frac{90 \times 10 \times 600}{90 \times 60} = \frac{540000}{5400} = 100 \ W$.

(D) The power $P$ of the machine gun is defined as the total kinetic energy delivered per unit time.
$P = \frac{\text{Total Kinetic Energy}}{\text{Time}}$
$P = \frac{n \times (\frac{1}{2}mv^2)}{t} = \frac{n}{t} \times \frac{1}{2}mv^2$
Given:
Number of bullets $n = 300$
Time $t = 60 \,s$
Mass of each bullet $m = 4 \,g = 4 \times 10^{-3} \,kg$
Velocity $v = 500 \,ms^{-1}$
Substituting the values:
$P = \frac{300}{60} \times \frac{1}{2} \times (4 \times 10^{-3}) \times (500)^2$
$P = 5 \times \frac{1}{2} \times 0.004 \times 250000$
$P = 5 \times 0.002 \times 250000$
$P = 5 \times 500 = 2500 \,W$
$P = 2.5 \,kW$

(C) Given: Mass $m = 400 \ g = 0.4 \ kg$,Power $P = 1.2 \ W$,Time $t = 6 \ s$,Initial velocity $u = 0 \ m \ s^{-1}$.
Work done by the force is given by $W = P \times t$.
$W = 1.2 \times 6 = 7.2 \ J$.
According to the work-energy theorem,the work done by the force is equal to the change in kinetic energy of the bead.
$W = \Delta K.E. = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$.
Since $u = 0$,$W = \frac{1}{2} m v^2$.
$7.2 = \frac{1}{2} \times 0.4 \times v^2$.
$7.2 = 0.2 \times v^2$.
$v^2 = \frac{7.2}{0.2} = 36$.
$v = \sqrt{36} = 6 \ m \ s^{-1}$.

(C) The rate at which the force does work is defined as power $(P)$.
Given:
Mass $m = 50 \,kg$
Speed $v = 4 \,ms^{-1}$
Applied force $F = 500 \,N$
Angle $\theta = 30^{\circ}$ above the horizontal.
The power delivered by a force is given by the dot product of force and velocity: $P = \vec{F} \cdot \vec{v} = Fv \cos \theta$.
Substituting the values:
$P = 500 \times 4 \times \cos 30^{\circ}$
$P = 2000 \times \frac{\sqrt{3}}{2}$
$P = 1000 \sqrt{3} \,W$
Using $\sqrt{3} \approx 1.732$:
$P = 1000 \times 1.732 = 1732 \,W$.

(C) Given: Force $F = 5 \,N$, initial velocity $u = 0$, time $t = 3 \,s$, and instantaneous power $P = 5 \,W$.
Using Newton's second law, the acceleration $a$ is given by $a = \frac{F}{m} = \frac{5}{m}$.
The velocity $v$ at time $t = 3 \,s$ is $v = u + at = 0 + (\frac{5}{m}) \times 3 = \frac{15}{m} \,m/s$.
The instantaneous power is defined as $P = F \cdot v$.
Substituting the known values: $5 = 5 \times (\frac{15}{m})$.
Solving for $m$: $1 = \frac{15}{m}$, which gives $m = 15 \,kg$.

(D) We know that, $\text{Power} = \frac{\text{Work done}}{\text{Time}}$.
$\Rightarrow P = \frac{W}{t} = \frac{F \times s}{t}$.
$\Rightarrow P = \frac{mgh}{t}$ ... $(i)$.
Here, the total weight of passengers is $mg = 50 \times 600 \,N = 30,000 \,N$.
The height is $h = 100 \,m$.
The power is $P = 15 \,kW = 15,000 \,W$.
Substituting these values into equation $(i)$, we get:
$t = \frac{mgh}{P}$.
$t = \frac{30,000 \times 100}{15,000}$.
$t = \frac{3,000,000}{15,000} = 200 \,s$.

(D) Given,power $P = \frac{1}{4} \ hp = \frac{746}{4} = 186.5 \ W$.
Effective power $P^{\prime} = 186.5 \times \frac{60}{100} = 111.9 \ W$.
Angular velocity $\omega = \frac{2 \pi \times 600}{60} = 20 \pi \ rad/s$.
Time taken for one rotation $t = \frac{2 \pi}{\omega} = \frac{2 \pi}{20 \pi} = 0.1 \ s$.
Work done in one rotation $W = P^{\prime} \times t = 111.9 \times 0.1 = 11.19 \ J$.