Concept Card: Power in Mechanics – Average, Instantaneous & Constant Power Dynamics
1. Physical Definition of Power:
While mechanical work measures how much energy is transferred or transformed, power quantifies how rapidly that work is performed or energy is delivered. Power is the time rate of doing work or transferring energy.
- Physical Nature: Scalar quantity.
- SI Unit: Watt ($\text{W}$), defined as $1\text{ W} = 1\text{ J/s} = 1\text{ kg}\cdot\text{m}^2/\text{s}^3$.
- Dimensional Formula: $[M L^2 T^{-3}]$.
- Practical & Commercial Units:
– Horsepower ($\text{hp}$): $1\text{ hp} \approx 746\text{ W} \approx 0.746\text{ kW}$.
– Kilowatt-hour ($\text{kWh}$): Commercial unit of energy (not power!): $1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J} = 3.6\text{ MJ}$.
2. Average Power vs. Instantaneous Power
(A) Average Power ($P_{\text{avg}}$):
Average power over a macroscopic finite time interval $\Delta t$ is the total work done divided by the elapsed time:
$$P_{\text{avg}} = \frac{\Delta W}{\Delta t} = \frac{\Delta E}{\Delta t}$$
(B) Instantaneous Power ($P$ or $P_{\text{inst}}$):
The instantaneous rate of doing work at a specific instant $t$ is obtained via differential calculus:
$$P = \frac{dW}{dt} = \frac{\vec{F} \cdot d\vec{r}}{dt} = \vec{F} \cdot \left(\frac{d\vec{r}}{dt}\right) = \vec{F} \cdot \vec{v}$$
- Scalar Product Formulation: $P = \vec{F} \cdot \vec{v} = F v \cos\theta$, where $\theta$ is the angle between the instantaneous force vector $\vec{F}$ and the velocity vector $\vec{v}$.
- Sign of Power:
– $P > 0$ if $0^\circ \le \theta < 90^\circ$ (force assists motion; kinetic energy increases).
– $P = 0$ if $\theta = 90^\circ$ (force is purely perpendicular, e.g., magnetic Lorentz force, tension in a simple pendulum, normal reaction on a fixed surface).
– $P < 0$ if $90^\circ < \theta \le 180^\circ$ (force opposes motion, e.g., kinetic friction, viscous drag; kinetic energy decreases). - Rotational Dynamics Analogue: Instantaneous power delivered by torque $\vec{\tau}$ to a body rotating at angular velocity $\vec{\omega}$ is $P = \vec{\tau} \cdot \vec{\omega}$.
3. Work and Energy from Power-Time ($P-t$) Graphs
Since $dW = P\,dt$, the total mechanical work performed between times $t_1$ and $t_2$ is the definite time integral of power:
$$W = \int_{t_1}^{t_2} P(t)\,dt = \text{Area under the } P-t \text{ curve between } t_1 \text{ and } t_2$$
By the Work-Energy Theorem, if no other forces do work, the area under the $P-t$ graph directly equals the net change in kinetic energy: $\Delta K = \int P\,dt$.
4. Core JEE Scenario: Motion under Constant Engine Power ($P_0 = \text{constant}$)
A classic, high-frequency JEE scenario is a vehicle or particle of mass $m$ accelerating along a straight line starting from rest under a constant power source $P_0$ (with no friction or drag):
- Velocity as a Function of Time:
$W = P_0 t = \Delta K = \frac{1}{2}mv^2 – 0 \implies v(t) = \sqrt{\frac{2P_0}{m}}\,t^{1/2}$
$$\mathbf{v \propto t^{1/2}}$$ - Acceleration as a Function of Time:
$a(t) = \frac{dv}{dt} = \sqrt{\frac{2P_0}{m}}\left(\frac{1}{2}t^{-1/2}\right) = \sqrt{\frac{P_0}{2m}}\,t^{-1/2}$
$$\mathbf{a \propto t^{-1/2}}$$ Crucial insight: Acceleration is NOT constant; it decreases monotonically with time. - Displacement as a Function of Time:
$x(t) = \int_0^t v(t’)\,dt’ = \sqrt{\frac{2P_0}{m}} \int_0^t t’^{1/2}\,dt’ = \sqrt{\frac{2P_0}{m}} \left(\frac{2}{3}t^{3/2}\right)$
$$\mathbf{x \propto t^{3/2}}$$ - Velocity as a Function of Displacement ($v$ vs. $x$):
$P_0 = F v = \left(m v \frac{dv}{dx}\right) v = m v^2 \frac{dv}{dx} \implies v^2\,dv = \frac{P_0}{m}\,dx$
Integrating: $\frac{v^3}{3} = \frac{P_0}{m}x \implies v(x) = \left(\frac{3P_0}{m}\right)^{1/3} x^{1/3}$
$$\mathbf{v \propto x^{1/3}}$$ - Driving Force as a Function of Displacement ($F$ vs. $x$):
$F(x) = \frac{P_0}{v(x)} = P_0 \left(\frac{m}{3P_0}\right)^{1/3} x^{-1/3} \propto x^{-1/3}$.
5. Pumping Fluids & Mass Flow Rate Mechanics
When a pump draws fluid of density $\rho$ through a conduit of cross-sectional area $A$ at speed $v$ and lifts it through height $h$:
- Mass Flow Rate: $\frac{dm}{dt} = \rho A v$ (in $\text{kg/s}$).
- Gravitational Potential Power: Rate of lifting mass against gravity:
$$P_{\text{gravity}} = \left(\frac{dm}{dt}\right) g h = (\rho A v) g h$$ - Kinetic Power Imparted to Jet: Rate at which kinetic energy is imparted to fluid:
$$P_{\text{kinetic}} = \frac{d}{dt}\left(\frac{1}{2}m v^2\right) = \frac{1}{2}\left(\frac{dm}{dt}\right)v^2 = \frac{1}{2}(\rho A v)v^2 = \frac{1}{2}\rho A v^3$$ - Cubic Velocity Law: Notice that $P_{\text{kinetic}} \propto v^3$. If you wish to double the ejection speed ($v \to 2v$), the volume flow rate doubles, but the required kinetic power increases by a factor of $2^3 = 8$!
- Total Output Power of Pump:
$$P_{\text{out}} = (\rho A v) g h + \frac{1}{2}\rho A v^3$$ - Motor Efficiency ($\eta$):
$$\eta = \frac{P_{\text{output}}}{P_{\text{input}}} \times 100\% \implies P_{\text{input}} = \frac{P_{\text{output}}}{\eta}$$
6. Common JEE Pitfalls & Traps
- Trap 1 (Misapplying Constant Acceleration Kinematics): When engine power is constant ($P = \text{const}$), students frequently write $v = at$ or $s = \frac{1}{2}at^2$. This is completely wrong because acceleration varies continuously ($a \propto t^{-1/2}$). Always integrate from $P = Fv = mv(dv/dt)$.
- Trap 2 (Average Power vs. Instantaneous Power): For uniform acceleration from rest ($a = \text{const}$), instantaneous power at time $t$ is $P(t) = F v = (ma)(at) = ma^2 t$. The average power over $[0, t]$ is $P_{\text{avg}} = \frac{\Delta K}{t} = \frac{\frac{1}{2}m(at)^2}{t} = \frac{1}{2}ma^2 t$. Thus, for uniformly accelerated motion from rest: $\mathbf{P_{\text{avg}} = \frac{1}{2} P_{\text{inst}}(t)}$.
- Trap 3 (Unit of Power vs. Unit of Energy): Never confuse kilowatt ($\text{kW}$, a unit of power) with kilowatt-hour ($\text{kWh}$, a unit of work/energy).
Solved Examples
Example 1 (Direct Conceptual Application – Uniform Acceleration vs. Power Dynamics):
A car of mass $m = 1200\text{ kg}$ accelerates uniformly from rest along a straight horizontal track to a speed $v = 20.0\text{ m/s}$ in a time interval of $t = 10.0\text{ s}$. Neglecting friction:
(a) Find the average power $P_{\text{avg}}$ delivered by the engine over the $10\text{ s}$ interval.
(b) Find the instantaneous power delivered by the engine at $t = 5.0\text{ s}$ and at $t = 10.0\text{ s}$.
(c) Verify the relationship between $P_{\text{avg}}$ and the final instantaneous power $P_{\text{inst}}(10\text{ s})$.
Solution:
(a) Constant acceleration: $a = \frac{v – u}{t} = \frac{20.0 – 0}{10.0} = 2.0\text{ m/s}^2$.
Total work done on the car equals the gain in kinetic energy:
$$W = \Delta K = \frac{1}{2}m v^2 – 0 = \frac{1}{2}(1200)(20.0)^2 = 600 \times 400 = 240,000\text{ J} = 240\text{ kJ}$$
Average power:
$$P_{\text{avg}} = \frac{W}{t} = \frac{240,000\text{ J}}{10.0\text{ s}} = 24,000\text{ W} = 24.0\text{ kW}$$
(b) The net force exerted by the engine is constant:
$$F = m a = 1200 \times 2.0 = 2400\text{ N}$$
Instantaneous velocity at any time $t$ is $v(t) = at = 2.0t$.
Instantaneous power: $P(t) = F v(t) = 2400(2.0t) = 4800 t\text{ W}$.
– At $t = 5.0\text{ s}$: $P(5.0) = 4800 \times 5.0 = 24,000\text{ W} = 24.0\text{ kW}$.
– At $t = 10.0\text{ s}$: $P(10.0) = 4800 \times 10.0 = 48,000\text{ W} = 48.0\text{ kW}$.
(c) Notice that at the midpoint of the time interval ($t = 5\text{ s}$), instantaneous power exactly equals the average power: $P(5.0) = P_{\text{avg}} = 24.0\text{ kW}$.
Furthermore, $P_{\text{avg}} = \frac{1}{2} P_{\text{inst}}(10.0\text{ s}) = \frac{1}{2}(48.0\text{ kW}) = 24.0\text{ kW}$, verifying the general rule for uniform acceleration from rest.
Example 2 (Mathematical Manipulation – Fluid Flow, Mass Rate & Efficiency):
An electric motor operates a water pump with an overall mechanical efficiency of $\eta = 75\%$. The pump lifts water from a reservoir at depth $h = 20.0\text{ m}$ below ground level and ejects it through a horizontal circular nozzle of radius $r = 2.0\text{ cm}$ with a speed of $v = 10.0\text{ m/s}$. Taking density of water $\rho = 1000\text{ kg/m}^3$ and $g = 10.0\text{ m/s}^2$:
(a) Find the mass flow rate of water delivered by the pump in $\text{kg/s}$.
(b) Calculate the total useful mechanical output power delivered by the pump.
(c) Calculate the electrical input power consumed by the motor.
Solution:
(a) Cross-sectional area of nozzle:
$$A = \pi r^2 = \pi (0.02)^2 = 4\pi \times 10^{-4}\text{ m}^2$$
Mass flow rate:
$$\frac{dm}{dt} = \rho A v = (1000)(4\pi \times 10^{-4})(10.0) = 4\pi\text{ kg/s} \approx 12.566\text{ kg/s}$$
(b) The useful mechanical output power consists of two terms:
1. Power required to lift water against gravity: $P_g = \left(\frac{dm}{dt}\right)gh = (4\pi)(10.0)(20.0) = 800\pi\text{ W}$.
2. Power required to impart kinetic energy: $P_k = \frac{1}{2}\left(\frac{dm}{dt}\right)v^2 = \frac{1}{2}(4\pi)(10.0)^2 = 200\pi\text{ W}$.
Total output mechanical power:
$$P_{\text{out}} = P_g + P_k = 800\pi + 200\pi = 1000\pi\text{ W} \approx 3141.6\text{ W} \approx 3.14\text{ kW}$$
(c) Efficiency relation: $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{1000\pi}{0.75} = \frac{4000\pi}{3}\text{ W} \approx 4188.8\text{ W} \approx 4.19\text{ kW}$.
Example 3 (Standard JEE Advanced Scenario – Vehicle with Air Resistance & Terminal Speed):
A locomotive of mass $m$ travels along a straight level track driven by an engine that delivers a constant power $P$. The motion is opposed by an aerodynamic air resistance force $F_r = k v^2$, where $k$ is a constant and $v$ is instantaneous speed.
(a) Determine the terminal (maximum attainable) speed $v_{\text{max}}$ of the locomotive.
(b) Derive the acceleration $a$ as a function of instantaneous speed $v$.
(c) Find the acceleration when the speed is half of the terminal speed ($v = \frac{1}{2}v_{\text{max}}$) in terms of $P, m,$ and $v_{\text{max}}$.
Solution:
(a) The driving tractive force produced by the engine at speed $v$ is $F_{\text{drive}} = \frac{P}{v}$.
The net force acting on the locomotive is:
$$F_{\text{net}} = F_{\text{drive}} – F_r = \frac{P}{v} – k v^2$$
At terminal speed $v_{\text{max}}$, acceleration vanishes ($F_{\text{net}} = 0$):
$$\frac{P}{v_{\text{max}}} – k v_{\text{max}}^2 = 0 \implies k v_{\text{max}}^3 = P \implies v_{\text{max}} = \left(\frac{P}{k}\right)^{1/3}$$
(b) By Newton’s second law, $m a = F_{\text{net}} = \frac{P}{v} – kv^2$:
$$a = \frac{1}{m}\left(\frac{P}{v} – kv^2\right)$$
Using $k = \frac{P}{v_{\text{max}}^3}$:
$$a(v) = \frac{P}{m}\left(\frac{1}{v} – \frac{v^2}{v_{\text{max}}^3}\right) = \frac{P}{mv}\left(1 – \frac{v^3}{v_{\text{max}}^3}\right)$$
(c) Substituting $v = \frac{1}{2}v_{\text{max}}$ into the acceleration expression:
$$a\left(\frac{v_{\text{max}}}{2}\right) = \frac{P}{m\left(\frac{v_{\text{max}}}{2}\right)}\left[1 – \left(\frac{1}{2}\right)^3\right] = \frac{2P}{m v_{\text{max}}}\left(1 – \frac{1}{8}\right) = \frac{2P}{m v_{\text{max}}}\left(\frac{7}{8}\right) = \frac{7P}{4 m v_{\text{max}}}$$
Example 4 (Edge Case – Time-Dependent Variable Power):
A particle of mass $m = 2.0\text{ kg}$ is initially at rest at position $x = 0$ at time $t = 0$. A single force acts along the $+x$-axis delivering time-dependent power according to $P(t) = \alpha t^2$, where $\alpha = 6.0\text{ W/s}^2$.
(a) Find the total work done on the particle from $t = 0$ to $t = 2.0\text{ s}$.
(b) Derive the velocity $v(t)$ as a function of time, and evaluate $v$ at $t = 2.0\text{ s}$.
(c) Derive the position $x(t)$ as a function of time, and determine displacement at $t = 2.0\text{ s}$.
(d) Find the instantaneous force $F$ acting on the particle at $t = 2.0\text{ s}$.
Solution:
(a) Work is the integral of power:
$$W(t) = \int_0^t P(t’)\,dt’ = \int_0^t 6.0 t’^2\,dt’ = \left[ 2.0 t’^3 \right]_0^t = 2.0 t^3\text{ Joules}$$
At $t = 2.0\text{ s}$:
$$W(2.0) = 2.0(2.0)^3 = 16.0\text{ Joules}$$
(b) By the Work-Energy Theorem ($W = \Delta K = \frac{1}{2}m v^2 – 0$):
$$\frac{1}{2}(2.0) v(t)^2 = 2.0 t^3 \implies v(t)^2 = 2.0 t^3 \implies v(t) = \sqrt{2}\,t^{3/2}\text{ m/s}$$
At $t = 2.0\text{ s}$:
$$v(2.0) = \sqrt{2}(2.0)^{3/2} = \sqrt{2}(2\sqrt{2}) = 4.0\text{ m/s}$$
(c) Integrating velocity gives position $x(t)$:
$$x(t) = \int_0^t v(t’)\,dt’ = \int_0^t \sqrt{2}t’^{3/2}\,dt’ = \sqrt{2}\left(\frac{t^{5/2}}{5/2}\right) = \frac{2\sqrt{2}}{5}t^{5/2}$$
At $t = 2.0\text{ s}$:
$$x(2.0) = \frac{2\sqrt{2}}{5}(2.0)^{5/2} = \frac{2\sqrt{2}}{5}(4\sqrt{2}) = \frac{2 \times 2 \times 4}{5} = \frac{16}{5} = 3.2\text{ meters}$$
(d) Instantaneous force can be found either via $F = \frac{P}{v}$ or $F = m \frac{dv}{dt}$:
– Using $F = \frac{P(t)}{v(t)} = \frac{6.0 t^2}{\sqrt{2}t^{3/2}} = 3\sqrt{2}\,t^{1/2}$.
– At $t = 2.0\text{ s}$: $F(2.0) = 3\sqrt{2}\sqrt{2} = 6.0\text{ N}$.
Cross-check: $P(2.0) = F(2.0)v(2.0) = 6.0\text{ N} \times 4.0\text{ m/s} = 24.0\text{ W}$, and from $P(t) = 6.0(2.0)^2 = 24.0\text{ W}$. Exact match!
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
An engine pumps water continuously through a hose with speed $v$. If the speed of water exiting the nozzle is doubled to $2v$, by what factor does the rate of kinetic energy imparted to the water increase?
(A) $2$
(B) $4$
(C) $8$
(D) $16$
Problem 2 (JEE Main – Single Correct):
A body of mass $m$ starts from rest and moves along a straight line under the action of an engine delivering constant power $P_0$. The distance $x$ traversed by the body in time $t$ is proportional to:
(A) $t^{1/2}$
(B) $t$
(C) $t^{3/2}$
(D) $t^2$
Problem 3 (JEE Main – Single Correct):
A particle of mass $m$ is moving with constant acceleration $a$ along a straight line starting from rest. The instantaneous power delivered to the particle as a function of time $t$ is:
(A) $m a t$
(B) $\frac{1}{2}m a^2 t$
(C) $m a^2 t$
(D) $m a^2 t^2$
Problem 4 (JEE Main – Single Correct):
A constant force $\vec{F} = (2\hat{i} + 3\hat{j} – 4\hat{k})\text{ N}$ acts on a particle moving with instantaneous velocity $\vec{v} = (4\hat{i} – 2\hat{j} + 5\hat{k})\text{ m/s}$. The instantaneous power delivered by the force to the particle is:
(A) $+18\text{ W}$
(B) $-18\text{ W}$
(C) $+28\text{ W}$
(D) $-28\text{ W}$
Problem 5 (JEE Main – Single Correct):
An elevator of total mass $M = 1500\text{ kg}$ (including passengers) ascends vertically at a steady speed of $v = 2.0\text{ m/s}$. A frictional resistance force $f = 3000\text{ N}$ opposes its motion. The minimum power delivered by the hoisting motor is: (Take $g = 10\text{ m/s}^2$)
(A) $30\text{ kW}$
(B) $33\text{ kW}$
(C) $36\text{ kW}$
(D) $40\text{ kW}$
Problem 6 (JEE Advanced – One or More Correct):
A car of mass $m$ accelerates from rest along a horizontal straight track under the influence of an engine delivering constant mechanical power $P_0$. Neglecting friction and air resistance, which of the following statements is/are correct?
(A) The velocity of the car at time $t$ is $v = \sqrt{\frac{2P_0 t}{m}}$.
(B) The displacement of the car at time $t$ is $x = \frac{2}{3}\sqrt{\frac{2P_0}{m}} t^{3/2}$.
(C) The instantaneous power delivered by the engine is independent of time.
(D) The acceleration of the car is constant throughout the motion.
Problem 7 (JEE Advanced – One or More Correct):
A pump lifts water of density $\rho$ through a vertical height $h$ from a reservoir and ejects it through a horizontal nozzle of area $A$ at constant speed $v$. If the pump operates continuously for a time duration $t$, which of the following is/are correct?
(A) The mass of water discharged in time $t$ is $\rho A v t$.
(B) The rate of doing work against gravity is $(\rho A v)gh$.
(C) The rate of imparting kinetic energy to the water is $\frac{1}{2}\rho A v^3$.
(D) If the ejection speed $v$ is doubled while height $h$ remains constant, the power required to impart kinetic energy increases by an 8-fold factor.
Problem 8 (JEE Advanced – One or More Correct):
A particle starts from rest at $x = 0$ at $t = 0$ and moves along the $+x$-axis under a force such that the power delivered to it varies with position as $P = k \sqrt{x}$, where $k$ is a positive constant. Which of the following statements is/are correct?
(A) The speed of the particle is proportional to $x^{1/2}$.
(B) The displacement of the particle is proportional to $t^2$.
(C) The acceleration of the particle is constant in time.
(D) The net work done on the particle between $x = 0$ and $x = X$ depends on the mass $m$ of the particle.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A vehicle of mass $m = 2000\text{ kg}$ is accelerated from rest on a level frictionless track by an engine delivering a constant power of $P = 60\text{ kW}$. Find the time $t$ in seconds required for the vehicle to attain a speed of $v = 30\text{ m/s}$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
An electric water pump ejects water horizontally through a pipe with speed $v$. Neglecting gravitational potential energy changes, the power consumed by the motor to maintain this discharge is $P_1$. If the motor power is increased by an 8-fold factor to $P_2 = 8 P_1$, the mass of water discharged per second increases by a factor of $N$. Determine the value of $N$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (C) | 3. (C) | 4. (B) | 5. (C) | 6. (A, B, C) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 15 | 10. 2
Solution 1:
Mass flow rate through the nozzle is $\frac{dm}{dt} = \rho A v$.
The rate of kinetic energy imparted is:
$P_k = \frac{1}{2}\left(\frac{dm}{dt}\right)v^2 = \frac{1}{2}(\rho A v)v^2 = \frac{1}{2}\rho A v^3 \propto v^3$.
When velocity is doubled ($v’ = 2v$):
$P’_k = \frac{1}{2}\rho A (2v)^3 = 8\left(\frac{1}{2}\rho A v^3\right) = 8 P_k$.
Correct Option: (C)
Solution 2:
Since power is constant, work done from rest in time $t$ is $W = P_0 t$.
By the Work-Energy Theorem: $\frac{1}{2}m v^2 = P_0 t \implies v = \sqrt{\frac{2P_0}{m}} t^{1/2}$.
Distance traversed: $x = \int_0^t v\,dt = \sqrt{\frac{2P_0}{m}} \left(\frac{2}{3}t^{3/2}\right) \implies x \propto t^{3/2}$.
Correct Option: (C)
Solution 3:
For motion with constant acceleration $a$ starting from rest: $v = a t$.
The net force is $F = m a$.
Instantaneous power: $P = F v = (m a)(a t) = m a^2 t$.
Correct Option: (C)
Solution 4:
Instantaneous power is the scalar dot product of force and velocity:
$P = \vec{F} \cdot \vec{v} = (2)(4) + (3)(-2) + (-4)(5) = 8 – 6 – 20 = -18\text{ W}$.
Correct Option: (B)
Solution 5:
For uniform upward motion at constant velocity $v$, the upward hoisting tension $T$ must balance both gravity and friction:
$T = M g + f = 1500(10) + 3000 = 15000 + 3000 = 18000\text{ N}$.
Minimum required motor power:
$P = T v = 18000\text{ N} \times 2.0\text{ m/s} = 36000\text{ W} = 36\text{ kW}$.
Correct Option: (C)
Solution 6:
– (A) True: $P_0 t = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2P_0 t}{m}}$.
– (B) True: $x = \int v\,dt = \sqrt{\frac{2P_0}{m}} \int t^{1/2}dt = \frac{2}{3}\sqrt{\frac{2P_0}{m}} t^{3/2}$.
– (C) True: By definition, engine power $P_0$ is constant with respect to time.
– (D) False: $a = \frac{dv}{dt} = \sqrt{\frac{P_0}{2m}} t^{-1/2} \propto t^{-1/2}$, so acceleration varies continuously.
Correct Options: (A, B, C)
Solution 7:
– (A) True: Volume in time $t$ is $A v t$, so mass is $\rho A v t$.
– (B) True: Power against gravity is $\left(\frac{dm}{dt}\right)gh = (\rho A v)gh$.
– (C) True: Kinetic energy rate is $\frac{1}{2}\left(\frac{dm}{dt}\right)v^2 = \frac{1}{2}(\rho A v)v^2 = \frac{1}{2}\rho A v^3$.
– (D) True: Kinetic power scales as $v^3$; doubling speed increases kinetic power by $2^3 = 8$.
Correct Options: (A, B, C, D)
Solution 8:
Power equation: $P = F v = m v \frac{dv}{dt} = m v^2 \frac{dv}{dx} = k x^{1/2}$.
Separating variables and integrating:
$m \int_0^v v^2 dv = k \int_0^x x^{1/2} dx \implies \frac{m v^3}{3} = \frac{2k}{3} x^{3/2} \implies v^3 = \frac{2k}{m} x^{3/2} \implies v = \left(\frac{2k}{m}\right)^{1/3} x^{1/2}$.
– (A) True: $v \propto x^{1/2}$.
– (B) True: Since $v = \frac{dx}{dt} = C x^{1/2} \implies x^{-1/2} dx = C dt \implies 2x^{1/2} = C t \implies x = \frac{C^2}{4} t^2 \propto t^2$.
– (C) True: Since $x \propto t^2$, velocity is $v = \frac{dx}{dt} \propto t$, which means acceleration $a = \frac{dv}{dt} = \text{constant}$.
– (D) True: Net work done is $W = \Delta K = \frac{1}{2}m v^2 = \frac{1}{2}m \left[\left(\frac{2k}{m}\right)^{1/3} X^{1/2}\right]^2 = \frac{1}{2}m \left(\frac{2k}{m}\right)^{2/3} X = \frac{1}{2}(2k)^{2/3} m^{1/3} X$, which explicitly depends on mass $m$.
Correct Options: (A, B, C, D)
Solution 9:
Work done by constant power engine in time $t$:
$W = P t = \Delta K = \frac{1}{2}m v^2 – 0$.
Given $P = 60\text{ kW} = 60,000\text{ W}$, $m = 2000\text{ kg}$, $v = 30\text{ m/s}$:
$$60,000 \times t = \frac{1}{2}(2000)(30)^2 = 1000 \times 900 = 900,000\text{ J}$$
$$t = \frac{900,000}{60,000} = 15.0\text{ seconds}$$.
Correct Answer: 15
Solution 10:
The mass discharge rate is $\frac{dm}{dt} = \rho A v$, which means velocity is directly proportional to the discharge rate: $v \propto \frac{dm}{dt}$.
The power required to eject water at speed $v$ is:
$P = \frac{1}{2}\left(\frac{dm}{dt}\right)v^2 = \frac{1}{2}\rho A v^3 \propto \left(\frac{dm}{dt}\right)^3$.
Therefore:
$$\frac{P_2}{P_1} = \left(\frac{(dm/dt)_2}{(dm/dt)_1}\right)^3 = N^3$$
Given $\frac{P_2}{P_1} = 8$:
$$N^3 = 8 \implies N = 2$$.
Correct Answer: 2