Work Energy And Power – NEET UG PhysicsPractice Questions & PYQs

The instantaneous power delivered to an object is defined as the product of the force applied and the velocity of the object at that exact moment. Given that the particle starts from rest and reaches a speed $V$ in time $T$ under uniform acceleration, the acceleration is determined by the ratio of final velocity to time, expressed as $a = V/T$.

Newton's second law allows us to identify the constant force responsible for this motion as $F = Ma = \frac{MV}{T}$. By applying the power formula $P = Fv$ at the specific instant $T$, where the velocity is $V$, the power is calculated as $P = (\frac{MV}{T}) \cdot V$, which simplifies to $\frac{MV^2}{T}$.

Collisions between objects are governed by the law of conservation of linear momentum and the definition of the coefficient of restitution. Because the total momentum before the impact must equal the total momentum afterward, we account for the mass of the first ball and the stationary second ball to establish the equation $2 = v_1 + 2v_2$. The coefficient of restitution, which describes the ratio of the relative velocities of the objects, dictates that the difference \in velocities after the impact is half the initial relative velocity, represented by the equation $1 = v_2 - v_1$. Solving this system of two linear equations reveals that the first ball comes to a stop at 0 m/s while the second ball moves forward at 1 m/s.

The rate at which an engine imparts kinetic energy to the water corresponds to the power delivered to the jet, which depends on how much mass flows through the system each second and the square of the water's velocity. Since the jet has a linear mass density of $m$ and travels at a speed of $v$, the total mass passing a specific point per second is the product of these values, resulting in a mass flow rate of $mv$. By substituting this mass flow rate into the standard kinetic energy formula $\frac{1}{2}Mv^2$, the rate of energy transfer becomes $\frac{1}{2}(mv)v^2$, which simplifies to $\frac{1}{2}mv^3$.

The law of conservation of linear momentum dictates that since the rock was initially at rest, the vector sum of the momenta of its resulting pieces must equal zero. This implies that the momentum of the third piece must be equal in magnitude and opposite in direction to the resultant of the first two pieces.

We first determine the individual momenta of the two parts moving at right angles, calculating $p_1 = 1 \times 12 = 12$ kg m s$^{-1}$ and $p_2 = 2 \times 8 = 16$ kg m s$^{-1}$. Because these momenta are perpendicular, their combined resultant magnitude is found using the Pythagorean theorem as $p = \sqrt{12^2 + 16^2} = 20$ kg m s$^{-1}$.

To conserve momentum, the third piece must also possess a momentum such that $p = 20$ kg m s$^{-1}$. Given this piece moves with a velocity of 4 m s$^{-1}$, we find its mass by calculating $m = 20 / 4 = 5$ kg.

Conservation of energy governs this scenario because the gravitational potential energy lost as the mass falls a distance $x$ must be equivalent to the elastic potential energy stored in the stretched spring at the bottom of its motion. Setting the energy lost by the mass, $Mgx$, equal to the elastic energy gained by the spring, $\frac{1}{2}kx^2$, provides the necessary balance for the system at the point of maximum extension. Solving this equation by dividing both sides by $x$ allows us to determine the final displacement as

$x = \frac{2Mg}{k}$

Energy conservation principles dictate that, in the absence of non-conservative forces, the initial kinetic energy of a projectile should completely convert into gravitational potential energy at its maximum height. We start by calculating the launch energy using the formula $K = \frac{1}{2}mv^2$, which gives us a total of 200 J. At the actual height of 18 m, the object possesses a potential energy calculated by $U = mgh$, which equals 180 J. Any discrepancy between the initial energy available and the potential energy attained is due to the work done against air friction, calculated as the difference between these two values, yielding a total energy loss of 20 J.

The conservation of linear momentum requires that the backward momentum of the gun must match the forward momentum of the shell, leading to the recoil velocity relation $v' = \frac{m}{M}v$. Because the explosive energy manifests as the total kinetic energy of the system, we can express this relationship as:
$K = \frac{1}{2}mv^2 + \frac{1}{2}Mv'^2 = \frac{1}{2}mv^2\left(1 + \frac{m}{M}\right)$
Using the shell mass of 0.2 kg, the gun mass of 4 kg, and the total energy of 1050 J, we substitute these into the equation to get:
$\frac{1}{2}(0.2)v^2\left(1 + \frac{0.2}{4}\right) = 1050$
This expression simplifies to $0.1(1.05)v^2 = 1050$, which reduces to $0.105v^2 = 1050$. By dividing, we find $v^2 = 10000$, which results in an initial shell velocity of 100 m s^-1.

The power available from falling water is determined by the rate of potential energy conversion, represented by the formula $P = \frac{m}{t}gh$. Given the flow rate of 15 kg/s from a height of 60 m with an acceleration due to gravity of 10 m/s^2, the theoretical power generated is 15 times 10 times 60, totaling 9000 Watts. Since 10% of this energy is dissipated by frictional forces, the turbine operates at 90% efficiency. Calculating 90% of the initial power yields 8100 Watts, which corresponds to 8.1 kW.

The total work performed on the falling mass is equivalent to the sum of the work done by the gravitational force and the work done by the spring force. During the entire descent, gravity acts downward over a total displacement of $(h+d)$, generating a positive work value of $mg(h+d)$. Meanwhile, the spring applies a restoring force opposite to the direction of the ball's displacement during the compression phase, resulting in work defined by $-\frac{1}{2}kd^2$. By adding these two quantities together, the overall net work is calculated as $mg(h+d) - \frac{1}{2}kd^2$.

The total work performed while moving an object up an inclined plane is divided between increasing its gravitational potential energy and overcoming frictional resistance. The gain in potential energy is calculated using $PE = mgh$, which results in $2 \times 10 \times 10 = 200$ joules. Since the total energy input is 300 joules, the remainder represents the work spent fighting friction, calculated as $300 - 200 = 100$ joules.

The potential energy stored in a stretched spring is proportional to the square of its extension, expressed by the relationship $U = \frac{1}{2}kx^2$ where $k$ is the spring constant. Because the energy relates to the square of the displacement, changing the extension from 2 cm to 8 cm effectively multiplies the initial stretch by a factor of 4. Squaring this factor of 4 yields 16, resulting in a final potential energy of 16U.

The work done on an object by a constant force is determined by the product of the force and the total displacement experienced by the body. Calculating the force requires finding the acceleration by taking the second derivative of the displacement function $s = \frac{1}{3} t^{2}$. Differentiating once with respect to time yields the velocity $v = \frac{ds}{dt} = \frac{2}{3}t$, and differentiating a second time yields the constant acceleration $a = \frac{d^{2}s}{dt^{2}} = \frac{2}{3} \text{ m/s}^{2}$.

With the mass given as $3 \text{ kg}$, Newton's second law allows us to calculate the force as $F = ma = 3 \times \frac{2}{3} = 2 \text{ N}$. The displacement at $t = 2 \text{ s}$ is found by substituting the time into the original equation, which gives $s = \frac{1}{3} \times (2)^{2} = \frac{4}{3} \text{ m}$. Multiplying the force by this displacement results in the work done, $W = Fs = 2 \times \frac{4}{3} = \frac{8}{3} \text{ J}$.

When a force varies as an object moves, the total work done is equal to the area enclosed by the force-displacement graph over the specified interval. To find this value, we calculate the geometric area of the shape formed between the curve and the horizontal axis from the starting position ($x=0$) to the final position ($x=6$).

The shape formed is a trapezium with parallel sides of length 6 and 3, and a perpendicular height of 3. By applying the geometric formula ($Area = \frac{1}{2} (a + b) h$) where $a$ and $b$ are the parallel sides, the calculation becomes ($Area = \frac{1}{2} (6 + 3) \times 3$). Solving this arithmetic yields a total work done of 13.5 J.

When a bomb explodes while at rest, the law of conservation of linear momentum dictates that the net momentum of the system must remain zero. This means the momentum carried by the 18 kg fragment must be exactly balanced by the momentum of the 12 kg fragment, as shown by the equation $(18 \times 6) + (12 \times v) = 0$. Solving this reveals that the second piece moves at a velocity of $v = -9$ m/s, where the negative sign simply indicates it travels in the opposite direction to the first fragment. Because kinetic energy is determined by the formula $K = \frac{1}{2}mv^2$, the direction of motion does not affect the energy, allowing us to calculate the value using the expression $\frac{1}{2} \times 12 \times (-9)^2$, which results in 486 J.

The principle of conservation of mechanical energy governs this interaction, as there is no friction or external force to dissipate energy. When the moving mass strikes the spring, its kinetic energy is entirely converted into elastic potential energy at the point where the spring reaches its maximum compression and the mass momentarily comes to a stop.

The initial kinetic energy, $\frac{1}{2} m v^2$, must equal the potential energy, $\frac{1}{2} k x^2$, stored in the spring. Canceling the $\frac{1}{2}$ from both sides leads to the relationship $m v^2 = k x^2$, which allows for solving the displacement as $x = \sqrt{\frac{m v^2}{k}}$. Substituting the mass of 0.5 kg, the velocity of 1.5 m/s, and the force constant of 50 N/m into the formula yields $x = \sqrt{\frac{0.5 \times 1.5^2}{50}}$. This simplifies to the square root of 0.0225, which means the maximum compression of the spring is 0.15 m.

When an object is dropped from rest, it gains kinetic energy exactly equal to the potential energy lost during the fall, which is represented by the formula:
$KE = mgs$
Because acceleration due to gravity and the distance fallen are identical for both objects regardless of their mass, the kinetic energy of each ball is directly proportional to its mass. For the $2 \,\text{kg}$ and $4 \,\text{kg}$ balls, the ratio of their kinetic energies is determined by the ratio of their masses:
$\frac{KE_1}{KE_2} = \frac{m_1}{m_2} = \frac{2}{4}$
This calculation simplifies to a final ratio of $1:2$.

The relationship between kinetic energy $E$, momentum $p$, and mass $m$ is expressed by the formula $E = \frac{p^2}{2m}$. When two particles share the same momentum, the kinetic energy becomes inversely proportional to their respective masses. This implies that for a constant momentum, an object with a larger mass will necessarily have less kinetic energy. Since the problem specifies that $m_1 > m_2$, it follows logically that $E_1 < E_2$.

The elastic potential energy stored in a spring is described by the formula $U = \frac{1}{2}kx^2$, which indicates that the energy is directly proportional to the square of the displacement. When the spring is stretched by 2 cm, the potential energy is U; stretching it instead by 10 cm increases the displacement by a factor of 5. Because the potential energy relies on the square of the displacement, this 5-fold increase in stretch leads to a 25-fold increase in energy, resulting in a total stored potential energy of 25 U.

The law of conservation of linear momentum dictates that because the particle is initially stationary, the magnitudes of the momenta of the two resulting fragments must be equal, represented by $m_{1} v_{1} = m_{2} v_{2}$. The ratio of their kinetic energies is defined as $\frac{E_{1}}{E_{2}} = \frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} v_{2}^{2}}$, which simplifies to $\frac{m_{1}}{m_{2}} \left( \frac{v_{1}}{v_{2}} \right)^{2}$. By substituting the relationship $\frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}}$ derived from the conservation of momentum into this energy ratio, the expression becomes $\frac{m_{1}}{m_{2}} \left( \frac{m_{2}}{m_{1}} \right)^{2}$, yielding a final result of $\frac{m_{2}}{m_{1}}$.

The relationship between kinetic energy $K$ and momentum $p$ for a body of mass $m$ is expressed by the equation
$K = \frac{p^{2}}{2m}$
which indicates that momentum is directly proportional to the square root of kinetic energy, or $p \propto \sqrt{K}$. When the kinetic energy increases by $300\%$, the final value becomes the initial energy plus three times that initial amount, resulting in a total of $4K$. Applying the square root relationship, the final momentum $p_{new}$ corresponds to $\sqrt{4K}$, which simplifies to $2\sqrt{K}$, meaning the final momentum is twice the initial momentum. The percentage change is determined by the difference between the final and initial momentum divided by the initial value, calculated as $\frac{2p - p}{p} \times 100\%$, which results in a $100\%$ increase.