Work Energy And Power – NEET UG PhysicsPractice Questions & PYQs

Power describes the rate at which work is performed, and for the heart pumping blood against resistance, it is calculated as $Power = \frac{Pressure \times Volume}{Time}$. To find the pressure, we use the hydrostatic pressure formula $P = h \rho g$. By converting 150 mm to 0.15 m and using the given density of 13.6 x 10^3 kg/m^3 and gravity of 10 m/s^2, the pressure exerted by the blood is determined to be 20.4 x 10^3 N/m^2.

The volume of blood is 5 litres, which converts to 5 x 10^-3 m^3, and the time interval is 60 s. Substituting these values into the power formula, we multiply 20.4 x 10^3 N/m^2 by 5 x 10^-3 m^3 and divide the product by 60 s. This operation yields a final power output of 1.70 W.

The power delivered to a particle is defined as the product of the acting force and its velocity, represented by $P = Fv$. Since force is the product of mass and acceleration, and acceleration is the time derivative of velocity, the constant power $k$ can be expressed as $k = m v \frac{dv}{dt}$. By integrating this relationship to solve for velocity, we obtain $\int v \, dv = \int \frac{k}{m} \, dt$, which simplifies to $\frac{v^2}{2} = \frac{kt}{m}$ for a particle starting from rest, resulting in $v = \sqrt{\frac{2kt}{m}}$. To find the acceleration, we differentiate the velocity expression with respect to time, giving $a = \frac{dv}{dt} = \sqrt{\frac{2k}{m}} \left( \frac{1}{2} t^{-1/2} \right) = \sqrt{\frac{k}{2m}} t^{-1/2}$. Multiplying this acceleration by the mass $m$ yields the final expression for the force as $F = m \sqrt{\frac{k}{2m}} t^{-1/2} = \sqrt{\frac{mk}{2}} t^{-1/2}$.

Imagine being at the location of particle $A$ and observing particle $B$ approach. For the two particles to collide, the relative velocity vector of $B$ with respect to $A$, given by $\overset{\rightarrow}{v_2} - \overset{\rightarrow}{v_1}$, must point directly along the line of sight from $B$ to $A$, represented by the relative position vector $\overset{\rightarrow}{r_1} - \overset{\rightarrow}{r_2}$. This geometric requirement is satisfied when the unit vectors of these relative quantities are equal, leading to the condition $\frac{\overset{\rightarrow}{r_1}-\overset{\rightarrow}{r_2}}{\left|\overset{\rightarrow}{r_1}-\overset{\rightarrow}{r_2}\right|} = \frac{\overset{\rightarrow}{v_2}-\overset{\rightarrow}{v_1}}{\left|\overset{\rightarrow}{v_2}-\overset{\rightarrow}{v_1}\right|}$.

During an inelastic collision where energy is absorbed for excitation, the law of conservation of energy dictates that the total initial kinetic energy of the system equals the sum of the final kinetic energy and the energy required for excitation. Because the particles start with an initial kinetic energy sum of $\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2$, this total energy must account for both the final kinetic energy and the absorbed value $\varepsilon$. Consequently, the balance of energy is described by:
$\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\varepsilon$
Rearranging this energy balance to isolate the initial kinetic energy shows that subtracting the excitation energy from the initial state leaves the final kinetic energy, which is represented by the expression $\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2-\varepsilon=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$.

The energy stored in a stretched spring depends directly on the spring constant $K$ when the displacement is fixed, but inversely when the applied force is fixed. In the first scenario, both springs are extended by the same displacement $x$, which means we determine the work done using the formula $W = \frac{1}{2} K x^2$. Since the displacement $x$ is identical for both, the work done is directly proportional to the stiffness $K$. Because spring P has a larger spring constant $K_P$ than spring Q, it stores more energy, leading to the relationship $W_P > W_Q$.

When the springs are stretched by an identical force $F$ in the second scenario, it is more effective to express the work done as $W = \frac{F^2}{2 K}$. In this condition, the work done becomes inversely proportional to the spring constant $K$. Consequently, the spring with the larger constant $K_P$ results in a smaller denominator, and thus a lower total work done compared to the spring with the smaller constant $K_Q$. This relationship reverses the order observed previously, resulting in $W_Q > W_P$.

In an elastic collision, kinetic energy is strictly conserved, meaning the initial kinetic energy of the moving block must equal the combined final kinetic energies of both blocks. For two blocks of identical mass $M$, the conservation equation is written as:

$\frac{1}{2} M v^2 = \frac{1}{2} M \left( \frac{v}{3} \right)^2 + \frac{1}{2} M v'^2$

By cancelling the common factors of $\frac{1}{2} M$ from every term, the equation simplifies to $v^2 = \frac{v^2}{9} + v'^2$. Subtracting $\frac{v^2}{9}$ from both sides allows us to isolate the square of the second block's speed as $v'^2 = \frac{8}{9} v^2$. Finally, calculating the square root yields $v' = \sqrt{\frac{8}{9} v^2}$, which simplifies to $\frac{2 \sqrt{2}}{3} v$.

The principle of conservation of linear momentum dictates that because the total initial momentum of the body at rest is zero, the vector sum of the momenta of the three resulting pieces must also be zero. The two pieces of mass $m$ moving perpendicularly with speed $v$ produce a combined momentum magnitude of $mv\sqrt{2}$. To balance this, the third piece, with mass $2m$ found by subtracting the other masses from the initial 4m, must have an equal momentum magnitude in the opposite direction.

The kinetic energy for each of the two smaller pieces is $\frac{1}{2}mv^2$. The third piece has a kinetic energy determined by its momentum, calculated as:
$KE_3 = \frac{(mv\sqrt{2})^2}{2(2m)} = \frac{2m^2v^2}{4m} = \frac{1}{2}mv^2$
Summing the kinetic energies of all three fragments, the total energy generated by the explosion is $\frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}mv^2$, which simplifies to $\frac{3}{2}mv^2$.

The motion of a particle is governed by the principle of conservation of energy, where the total energy $E$ is the sum of its kinetic energy $K.E.$ and potential energy $U(x)$, expressed as $E = K.E. + U(x)$. For a particle to move within a particular region, its kinetic energy must be greater than zero. This requirement implies that $E - U(x) > 0$, which necessitates that the potential energy must satisfy the condition $U(x) < E$.

Power describes how rapidly work is accomplished, defined mathematically as the ratio of work done to the time elapsed. Because both individuals are lifting an identical suitcase to the same vertical height, the work done in each case remains constant, meaning power is inversely proportional to the time taken. Expressing this relationship as $\frac{P_1}{P_2} = \frac{t_2}{t_1}$, we can determine the ratio of their powers by using the given durations. By converting the first coolie's time of 1 minute into 60 seconds and using the second coolie's time of 30 seconds, the calculation becomes $\frac{30}{60}$, which simplifies to a ratio of $1:2$.

Work done by a constant force is defined by the scalar dot product of the force vector and the displacement vector. To find the displacement, calculate the difference between the final position vector and the initial position vector, which results in the expression $\vec{d} = (4 - 2)\hat{i} + (3 - 0)\hat{j} + (-1 - 1)\hat{k} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ meters. The work done is determined by the formula $W = \vec{F} \cdot \vec{d}$, where the force is given as $\vec{F} = 3\hat{i} + \hat{j}$. Substituting these vectors into the formula yields the dot product $W = (3)(2) + (1)(3) + (0)(-2) = 6 + 3 = 9$ Joules.

Equilibrium in a physical system occurs where the net force is zero, which corresponds to the point where the rate of change of potential energy with respect to distance is zero. By differentiating $U=\frac{A}{r^2}-\frac{B}{r}$ with respect to $r$, we find the force expression:
$\frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2}$
Setting this derivative equal to zero identifies the equilibrium condition $-\frac{2A}{r^3} + \frac{B}{r^2} = 0$. Rearranging terms to equate $\frac{2A}{r^3} = \frac{B}{r^2}$ allows us to solve for the distance, providing $r = \frac{2A}{B}$ as the location of stable equilibrium.

The collision is governed by the law of conservation of linear momentum, which requires that the total momentum vector remains unchanged in both the x and y directions. Before the collision, sphere B carries all of the system's momentum along the x-axis, represented as $m_2 v$, while the initial momentum along the y-axis is zero. After the collision, sphere B moves with a velocity of $v/2$ along the y-axis, meaning its x-momentum component is zero and its y-momentum component is $m_2 v/2$.

To conserve the system's total momentum, sphere A must have an x-momentum component equal to the initial $m_2 v$ and a y-momentum component that exactly cancels the y-momentum gained by sphere B. Balancing these components gives $m_1 v_A \cos \theta = m_2 v$ along the x-axis and $m_1 v_A \sin \theta = -m_2 v/2$ along the y-axis. Taking the ratio of the y-momentum equation to the x-momentum equation, the mass and velocity terms cancel, resulting in $\tan \theta = -1/2$, which places the direction of sphere A at $\theta = \tan^{-1}(-1/2)$ relative to the x-axis.

Power is fundamentally the product of force and velocity, so we start with the expression $P_0 = Fv$. By substituting force with the product of mass and acceleration, where acceleration is the rate of change of velocity over time, we obtain $P_0 = m v \frac{dv}{dt}$. Separating the variables for time and velocity allows us to integrate both sides of the equation as shown below:

$P_0 \int_{0}^{t} dt = m \int_{0}^{v} v dv$

Carrying out the integration results in $P_0 t = \frac{1}{2} m v^2$. Solving for velocity yields $v = \sqrt{\frac{2 P_0 t}{m}}$, which demonstrates that the velocity of the car is proportional to $t^{1/2}$.

The core principle here is the conservation of mechanical energy, which states that the total initial energy of the moving cylinder is entirely converted into elastic potential energy when it momentarily stops at maximum compression. A solid cylinder in motion possesses both translational kinetic energy, given by $\frac{1}{2}mv^2$, and rotational kinetic energy, given by $\frac{1}{2}I\omega^2$. By substituting the moment of inertia for a solid cylinder $I = \frac{1}{2}mR^2$ and the rolling condition $\omega = \frac{v}{R}$, the total kinetic energy combines to $\frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$, which simplifies to $\frac{3}{4}mv^2$.

Since this energy is fully transferred to the spring at maximum compression $x$, we equate it to the spring's potential energy formula:
$\frac{3}{4}mv^2 = \frac{1}{2}kx^2$
Rearranging this expression to isolate $x$, we find $x = \sqrt{\frac{3mv^2}{2k}}$. Plugging in the mass $m = 3\text{ kg}$, velocity $v = 4\text{ m/s}$, and spring constant $k = 200\text{ N/m}$, the equation becomes $x^2 = \frac{3 \times 3 \times 16}{2 \times 200}$. Calculating the right side gives $x^2 = \frac{144}{400} = 0.36$, which results in a maximum compression of 0.6 m.

The motion of a particle constrained to a parabolic path $y = kx^2$ under gravity exhibits simple harmonic motion for small displacements, as the potential energy $U = mgy = mgkx^2$ generates a restoring force $F = -\frac{dU}{dx} = -2mgkx$. Because the restoring force is directly proportional to the negative displacement, the particle oscillates according to a sinusoidal function, which, starting from rest at a maximum displacement, follows the cosine relationship $x(t) = A \cos(\omega t)$.

Potential energy is defined by the energy stored within a system's configuration in relation to conservative forces. When a system performs work against these conservative forces, it essentially opposes the inherent field, which transfers energy into the system and results in an increase in potential energy. This outcome is specific to conservative interactions, as working against nonconservative forces, such as friction, generally results in energy being lost to the surroundings rather than stored within the system.

Collisions governed by the law of conservation of linear momentum allow us to determine the resultant motion by analyzing the horizontal and vertical vector components independently. Since the two objects stick together, the system conserves the total momentum from both the mass moving along the x-axis and the mass moving along the y-axis.

The first object provides an initial momentum of $mv\hat{i}$, while the second object adds a momentum of $3m(2v)\hat{j}$, resulting in a total system momentum of $mv\hat{i} + 6mv\hat{j}$. As the combined mass is $4m$, we find the final velocity by dividing the total momentum vector by this combined mass. The velocity of the final combination is given by:

$\vec{v'} = \frac{mv\hat{i} + 6mv\hat{j}}{4m} = \frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$

Work done by a force acting on a particle corresponds to the total area enclosed under the Force-distance graph. Because the force varies over the displacement, the total area can be split into a rectangle representing the region from 3 meters to 7 meters and a triangle covering the region from 7 meters to 12 meters. The work done for the rectangular portion is $2 \times (7-3) = 8$ J, and the work done for the triangular portion is $\frac{1}{2} \times 2 \times (12-7) = 5$ J. Adding these values together results in a total work done of 13 J.

Power is determined by the dot product of the force and velocity vectors, expressed as $P = \vec{F} \cdot \vec{v} = Fv \cos \theta$. As the object falls back towards the surface, both the gravitational force and velocity point downward, resulting in an angle of $\theta = 0^{\circ}$ where the cosine term reaches its maximum value. Given that the object attains its highest speed at the very end of its descent, this configuration ensures the peak power output occurs at the instant immediately before impact.

The power required to pump water through a pipe depends on the rate at which kinetic energy is imparted to the water stream. Since the water has a constant mass distribution along the pipe and maintains a steady velocity, the engine must supply energy continuously to sustain this flow. This relationship is expressed by the formula $P = \mu v^3$, where $\mu$ represents the mass per unit length of the water and $v$ is its velocity. By taking the mass per unit length of 100 kg/m and the velocity of 2 m/s, the calculation becomes $100 \times 2^3$, resulting in a power output of 800 W.