Work Energy And Power – NEET UG PhysicsPractice Questions & PYQs

Since the speed $v$ of the particle remains constant in uniform circular motion, the kinetic energy does not change at all. Momentum is a vector quantity, meaning that when the particle reaches a diametrically opposite point, its velocity direction is reversed while maintaining the same magnitude. The change in momentum is calculated by subtracting the initial momentum vector from the final one, which is given by $Mv - (-Mv) = 2Mv$.

The work done by a constant force is defined by the formula $W = F s \cos \theta$, where $\theta$ is the angle between the force vector and the direction of displacement. Substituting the given values of 25 J for work, 5 N for force, and 10 m for distance into this expression results in:

$25 = 5 \times 10 \cos \theta$

Solving the equation $25 = 50 \cos \theta$ yields $\cos \theta = 0.5$, which indicates that the angle between the force and the direction of motion is $\theta = 60^{\circ}$.

Stable equilibrium for a physical system occurs at the configuration where the net force, defined as the negative derivative of potential energy, vanishes. Differentiating $U(x)= \frac{a}{x^{12}}-\frac{b}{x^{6}}$ with respect to $x$ yields the following expression:

$\frac{dU}{dx}= -\frac{12a}{x^{13}} + \frac{6b}{x^{7}}$

Setting this derivative to zero allows us to equate the terms as $\frac{6b}{x^{7}} = \frac{12a}{x^{13}}$. Multiplying both sides by $x^{13}$ simplifies the expression to $6b x^6 = 12a$, which reduces to $x^6 = \frac{2a}{b}$ and confirms the equilibrium distance as $x=\left(\frac{2 a}{b}\right)^{1 / 6}$.

In a perfectly inelastic collision, momentum is conserved, but a portion of the system's kinetic energy is inevitably dissipated, typically as thermal or sound energy. First, we convert the speed of the metal ball to SI units, noting that 36 km/h is equivalent to 10 m/s. According to the law of conservation of momentum, $m_1 u_1 = (m_1 + m_2) v$. Substituting the given mass and velocity values, $2 \times 10 = (2 + 3) \times v$, which simplifies to $20 = 5v$, resulting in a final velocity of 4 m/s for the combined mass.

We determine the kinetic energy before and after the collision to find the total loss. The initial kinetic energy is $K_i = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2 \times 10^2 = 100$ J. After the collision, both balls move together with a total mass of 5 kg at 4 m/s, yielding a final kinetic energy of $K_f = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 5 \times 4^2 = 40$ J. Subtracting the final kinetic energy from the initial value, the energy lost during the collision is $100 - 40 = 60$ J.

An explosion during flight is driven entirely by internal forces, and because the process occurs almost instantaneously, the impulse from external forces like gravity is negligible. According to the principle of conservation of linear momentum, the vector sum of the momenta of the four fragments must remain equal to the original momentum of the shell, expressed as $\vec{P}_{initial} = \vec{P}_{final}$. Although the chemical energy released during the blast increases the total kinetic energy of the system, this conversion of energy does not affect the total momentum, which remains invariant throughout the event.

Calculating the work done by a force that varies with position requires integration over the entire path of motion. In this scenario, the frictional force at any point $x$ is equal to the normal force $Mg$ multiplied by the variable coefficient $\mu_k(x) = \mu_0 - \alpha x$. To find the total work done by friction, we compute the definite integral of this force function with respect to $x$ from the starting position $x=0$ to the final position $x=L$.

Before performing the integration, define the constant $\alpha$ using the boundary condition that the coefficient of friction reaches zero at the end of the journey. Because $\mu_0 - \alpha L = 0$, it follows that $\alpha = \frac{\mu_0}{L}$. Integrating the force function results in the expression $W = \int_0^L (\mu_0 - \alpha x)Mg \, dx$, which evaluates to $Mg \left[ \mu_0 x - \frac{\alpha x^2}{2} \right]_0^L$. Substituting the limits of integration gives $Mg (\mu_0 L - \frac{\alpha L^2}{2})$. By substituting $\alpha = \frac{\mu_0}{L}$ into this result, the expression simplifies to $Mg (\mu_0 L - \frac{\mu_0 L^2}{2L})$, which further reduces to $\frac{1}{2} \mu_0 M g L$. Equating this calculated value to the given form $n \mu_0 M g L$ confirms that $n$ must be $\frac{1}{2}$.

When two objects of identical mass undergo a one-dimensional elastic collision, they satisfy the conservation of momentum and kinetic energy by effectively swapping their initial velocity vectors, a condition that can be described as $v_{1f} = v_{2i}$ and $v_{2f} = v_{1i}$. Since the first mass begins with a velocity of 3 m/s and the second mass moves at -5 m/s, these two bodies simply exchange their momentum values during the interaction. Consequently, the first object departs with a velocity of -5 m/s, while the second object proceeds with a velocity of 3 m/s.

The coefficient of restitution represents the degree to which kinetic energy is preserved during an impact, defined as the ratio of the relative velocity of separation after a collision to the relative velocity of approach before it. In a perfectly elastic collision, there is no loss of kinetic energy, which implies that the bodies rebound with the same relative speed that they possessed prior to the interaction. Consequently, this state of perfect conservation dictates that $e=1$.

Power measures how quickly energy is transferred or work is completed. When a crane lifts an object, it performs work equal to the change in gravitational potential energy, which is defined by the formula $P = \frac{mgh}{t}$.

Calculating the work requires multiplying the 1000 kg mass by the 9.8 m/s^2 acceleration due to gravity and the 20 m height, resulting in a total of 196000 J of energy. Dividing this work by the 10 s interval shows that the crane operates at 19600 W. Converting this value into kilowatts by dividing by 1000 results in 19.6 kW.

The relationship between kinetic energy $K$, momentum $p$, and mass $m$ is expressed by the equation:
$K = \frac{p^2}{2m}$
When two bodies share the same linear momentum, kinetic energy is inversely proportional to mass, so the ratio of kinetic energies is the inverse of the mass ratio. Since the kinetic energies exist in a 4:1 ratio, the relationship $\frac{m_2}{m_1} = \frac{4}{1}$ applies. Therefore, the ratio of the first mass to the second mass is 1:4.

The law of conservation of linear momentum states that the total initial momentum of a system must equal the total final momentum after a collision when no external forces act on the bodies. The first object has a momentum of $m \times 3$, while the second object, having a mass of $2m$ and being at rest, contributes zero momentum, resulting in a total system momentum of $3m$. Because the bodies stick together to form a combined mass of $3m$, the state after the collision is represented by the equation:
$3m = (3m)v$
Solving for the velocity $v$ reveals that the combined mass moves at 1 km/hour.

The Work-Energy Theorem dictates that the work performed by resistive forces corresponds to the decrease in an object's kinetic energy. Because air resistance acts against the motion of the bullet, the energy lost represents the work done in overcoming that resistance.

Converting the mass of 10 g into the standard SI unit gives 0.01 kg. The magnitude of energy loss is calculated using the formula $W = \frac{1}{2} m \left( v_i^2 - v_f^2 \right)$, where the initial velocity is 1000 m/s and the final velocity is 500 m/s. Substituting these values into the calculation $W = \frac{1}{2} \times 0.01 \times \left( 1000^2 - 500^2 \right)$ leads to $W = 0.005 \times \left( 1,000,000 - 250,000 \right)$, which equals 3750 joule.

When the string goes slack at point $P$, the tension vanishes, meaning the centripetal force required to maintain circular motion is provided solely by the component of gravity directed toward the center. This condition is expressed as $mg \sin \theta = \frac{mv^2}{l}$, which simplifies to $gl = \frac{v^2}{\sin \theta}$.

Simultaneously, the conservation of mechanical energy dictates that the total energy at the start, $\frac{1}{2}mv_0^2$, must equal the sum of the kinetic energy at point $P$ and the gravitational potential energy at that height. The height of point $P$ relative to the lowest point is $l + l \sin \theta$, leading to the energy equation $\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mg(l + l \sin \theta)$. Dividing by $m$ and multiplying the entire equation by 2 yields $v_0^2 = v^2 + 2gl(1 + \sin \theta)$.

Substituting the expression $gl = \frac{v^2}{\sin \theta}$ into this energy equation results in $v_0^2 = v^2 + 2 \left( \frac{v^2}{\sin \theta} \right) (1 + \sin \theta)$. Distributing the terms reveals $v_0^2 = v^2 + \frac{2v^2}{\sin \theta} + 2v^2$, which simplifies further to $v_0^2 = v^2 \left( 3 + \frac{2}{\sin \theta} \right)$. Rewriting the term in the parentheses as $\frac{3 \sin \theta + 2}{\sin \theta}$ allows us to isolate the ratio $\frac{v^2}{v_0^2} = \frac{\sin \theta}{3 \sin \theta + 2}$, which gives the final result of $\left( \frac{\sin \theta}{2 + 3 \sin \theta} \right)^{\frac{1}{2}}$.

According to the work-energy theorem, the work done by a braking force to bring a vehicle to a stop is equal to the magnitude of its initial kinetic energy, which is represented by the formula $F \cdot S = KE$ where $F$ is the braking force, $S$ is the stopping distance, and $KE$ is the kinetic energy. By rearranging this relationship to solve for force, we find that the ratio of the forces applied to car $A$ and car $B$ is given by $\frac{F_A}{F_B} = \frac{KE_A}{KE_B} \cdot \frac{S_B}{S_A}$ Substituting the provided values into this expression results \in the calculation $\frac{F_A}{F_B} = \frac{100}{225} \cdot \frac{1500}{1000}$ By simplifying the terms within this equation, we arrive at a final ratio of $\frac{2}{3}$.

Energy conservation principles indicate that an object's potential energy at a starting height $h_1$ is fully converted into kinetic energy $K_1$ just before it strikes the ground. With an initial height of 10 m, we express the energy as $K_1 = mgh_1$. Because the object loses half of its kinetic energy upon impact, the rebound kinetic energy becomes $K_2 = \frac{1}{2}K_1$. As the object rises to a new height $h_2$, this remaining kinetic energy is transformed back into potential energy, leading to the relation:

$mgh_2 = \frac{1}{2}mgh_1$

By canceling the common terms $mg$ on both sides, we simplify the equation to $h_2 = \frac{1}{2}h_1$. Given that the initial height is 10 m, the object reaches a rebound height of 5 m.

Instantaneous power is determined by the product of the applied force and the instantaneous velocity of a particle, expressed by the relation $P = F \cdot v$. To find this velocity, we take the derivative of the displacement with respect to time, which means differentiating the expression $x = 2t - 1$ to obtain a velocity of 2 m/s.

With the velocity known, multiplying it by the force of 5 N provides the instantaneous power. Calculating the final value using the equation $P = F \cdot v$ yields 10 W.

The work-energy theorem establishes that the net work done on an object equals the change in its kinetic energy, represented by $W = K_f - K_i$. Calculating the work done by the force involves determining the dot product of the force vector and the displacement vector, which gives $W = (-2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i})$, simplifying to $W = -6$ J. Substituting this work and the initial kinetic energy of 10 J into the energy relationship yields the equation $-6 = K_f - 10$. Solving for the final state shows that the kinetic energy of the object at the end of the displacement is $K_f = 4$ J.

The elastic potential energy stored in a spring is defined by $U = \frac{1}{2} k x^2$, demonstrating that energy is directly proportional to the square of the displacement $x$. Increasing the stretch from 2 cm to 8 cm increases the displacement by a factor of 4. Because the potential energy is proportional to the square of this displacement, the new energy becomes $4^2 = 16$ times the initial value, which is $16 U$.

In a perfectly elastic collision between two objects, the efficiency of kinetic energy transfer is governed by the ratio of their masses. This transfer is optimized and becomes total when the mass of the striking bullet is identical to the mass of the target block, represented by $M=m$. Under this specific condition, the striking object imparts all of its kinetic energy to the block and comes to a complete rest, representing the most effective energy exchange possible.

Instantaneous power is defined as the scalar product of the force vector and the velocity vector applied to an object. Using the relation $P = \vec{F} \cdot \vec{v}$, we substitute the provided vectors to obtain $(10\hat{i} + 10\hat{j} + 20\hat{k}) \cdot (5\hat{i} - 3\hat{j} + 6\hat{k})$. By multiplying the corresponding components along the coordinate axes, we perform the calculation $50 - 30 + 120$. Summing these values gives a final result of 140 W.