Work Energy And Power – NEET UG PhysicsPractice Questions & PYQs

The potential energy $E$ stored in a spring deformed by a force $F$ is given by the formula $E = \frac{F^2}{2K}$, where $K$ represents the spring constant. Because the force applied to both springs is identical, the stored energy is inversely proportional to the spring constant. Given that spring $A$ has a spring constant $K_{A} = 2 K_{B}$, we can set up the ratio of the energies as $\frac{E_{A}}{E_{B}} = \frac{K_{B}}{K_{A}}$. Substituting the given relationship into this ratio yields $\frac{E_{A}}{E_{B}} = \frac{K_{B}}{2 K_{B}} = \frac{1}{2}$, which confirms that $E_{B} = 2 E_{A}$.

The motion of a child on a swing is a classic application of the law of conservation of mechanical energy, where the total energy remains constant throughout the trajectory. Because there is no energy loss, the maximum speed is achieved at the lowest point of the arc, where the gravitational potential energy lost during the descent is entirely converted into kinetic energy.

The velocity is determined by equating the decrease in potential energy to the increase in kinetic energy. The vertical displacement for the child is the difference between the maximum height of 2 m and the minimum height of 0.75 m, which results in a height drop of 1.25 m. By applying the energy conservation principle:

$mg(h_{\text{max}} - h_{\text{min}}) = \frac{1}{2}mv^2$

Canceling the mass from both sides and solving for velocity yields the relationship:

$v = \sqrt{2g(1.25)}$

Using an acceleration due to gravity of 10 m/s^2, the velocity becomes the square root of 25, resulting in 5 m/s.

At the highest point of a projectile's trajectory, the vertical component of velocity becomes zero, leaving the particle with only its constant horizontal velocity. Since the particle is projected with an initial velocity $v$ at an angle of $45^{\circ}$, the horizontal velocity at the peak is defined by $v \cos 45^{\circ}$, which simplifies to $\frac{v}{\sqrt{2}}$. Given the initial kinetic energy is $K = \frac{1}{2}mv^2$, the kinetic energy at the highest point is calculated by substituting this peak velocity into the kinetic energy formula $\frac{1}{2}m v_x^2$, resulting in $\frac{1}{2}m \left( \frac{v}{\sqrt{2}} \right)^2$. This expression further simplifies to $\frac{1}{4}mv^2$, which is exactly $\frac{K}{2}$.

When a force varies with position, the work done is calculated by determining the definite integral of the force function over the given displacement, which is expressed as $W = \int_{0}^{5} (7 - 2x + 3x^{2}) dx$. The antiderivative of the force function is $7x - x^{2} + x^{3}$, which represents the accumulation of work as the body moves from the starting point. Evaluating this antiderivative at the limit of 5, while noting that the value at 0 is 0, yields a final result of 135 J.

When an object moves at a constant speed along a circular path, its velocity continuously changes direction, which confirms that both centripetal acceleration and a centripetal force are acting on the body. This centripetal force is directed toward the center of the circle, making it perpendicular to the object's instantaneous displacement at every point along the path. Because the scalar product of force and displacement involves the cosine of the angle between them, which is 90 degrees in this case, the work done is determined by $W = \vec{F} \cdot d\vec{s} = 0$, meaning no work is done on the body.

Kinetic energy relates to linear momentum through the expression $K = \frac{p^2}{2m}$, which implies that for objects with equal kinetic energies, the momentum of each object must be proportional to the square root of its mass. To determine the ratio of the momenta for two objects with masses 1 g and 9 g, we use the relationship $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$. Substituting the given mass values yields $\frac{p_1}{p_2} = \sqrt{\frac{1}{9}}$, simplifying the ratio of the magnitudes of their linear momenta to 1:3.

Work done by a constant force is determined by calculating the scalar dot product of the force vector and the displacement vector. Because the dot product of orthogonal unit vectors is zero, the only non-zero contribution comes from the component of the force acting in the same direction as the displacement. Substituting the given values into the formula $W = \vec{F} \cdot \vec{d}$, we find the expression $W = (-2 \hat{i} + 15 \hat{j} + 6 \hat{k}) \cdot (10 \hat{j})$. Since only the $\hat{j}$ components interact to produce a non-zero product, the calculation yields 150 J.

Instantaneous power $P$ is determined by calculating the scalar dot product of the force vector $\vec{F}$ and the velocity vector $\vec{v}$, expressed by the formula $P = \vec{F} \cdot \vec{v}$. This process requires multiplying the corresponding components of the vectors along the $\hat{i}$, $\hat{j}$, and $\hat{k}$ directions and then summing the resulting values.

$P = (60)(2) + (15)(-4) + (-3)(5)$

Performing the multiplication for these components yields 120, -60, and -15. Adding these values together, $120 - 60 - 15$, results in a final power of 45 watts.

The linear momentum $p$ of a body is related to its kinetic energy $E$ and mass $m$ by the expression $p = \sqrt{2mE}$. Because both bodies share the same kinetic energy, the momentum of each body is directly proportional to the square root of its mass, represented as $p \propto \sqrt{m}$. By comparing the two masses $m$ and $4m$, we find the ratio of their linear momenta using the equation:
$\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{m}{4m}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
This confirms the ratio is $1: 2$.

The fundamental principle relating work to motion is that work done on an object can be determined by the product of the force applied and the displacement, where the force itself is the product of mass and acceleration. Calculating the total work done on the particle requires evaluating its kinematics at the end of the specified interval using the position function $x = 3t - 4t^{2} + t^{3}$.

First, convert the mass from 3 g to $3 \times 10^{-3}$ kg. To obtain the acceleration function, differentiate the position equation twice with respect to time, giving $a = \frac{d^{2}x}{dt^{2}} = -8 + 6t$. At the 4 second mark, the acceleration evaluates to $a = -8 + 6(4) = 16$ m/s$^2$ and the position evaluates to $x = 3(4) - 4(4)^{2} + (4)^{3} = 12$ m. Multiplying these values according to the relation $W = m \cdot a \cdot x$ yields a total work of 576 mJ.

The fundamental principle is that power represents the rate of doing work, where the work accomplished involves lifting water against gravity as defined by $P = \frac{mgh}{t}$. Converting our parameters to standard units, the pump provides 2000 Watts of power over 60 seconds. Rearranging the relationship to solve for mass, $m = \frac{2000 \times 60}{10 \times 10}$ results in a value of 1200 kilograms. Since one kilogram of water equals one litre, the pump raises 1200 litres.

The principle of conservation of linear momentum dictates that the total momentum of the system remains constant during the explosion, as no external force acts on the mass during this brief event. First, we find the vertical velocity of the 1 kg object just before the explosion using the kinematic formula $v = u - gt$. Plugging in the known values, we get $v = 100 - 10 \times 5 = 50$ m/s.

The total initial momentum is therefore $1 \times 50 = 50$ kg m/s. After the explosion, one fragment of 0.4 kg moves downward at 25 m/s, which corresponds to a velocity of -25 m/s, while the 0.6 kg fragment moves at an unknown velocity $v'$. By conserving momentum, we have $1 \times 50 = 0.4 \times (-25) + 0.6 \times v'$. This simplifies to $50 = -10 + 0.6 \times v'$, or $60 = 0.6 \times v'$, which confirms that $v' = 100$ m/s.

The work-energy theorem establishes that the work done by a constant force on an object equals the change in its kinetic energy. Because the object begins from rest, its final kinetic energy is exactly the product of the force and the distance $d$. Since this calculation involves only the force and displacement without any term involving mass, the kinetic energy is independent of the mass $m$, which is expressed as being directly proportional to $m^{0}$.