Work Energy And Power – NEET UG PhysicsPractice Questions & PYQs

The total energy radiated by a device is determined by the product of its power output and the duration of its operation. Since power is given in kilowatts, it must be converted into watts, while time must be converted from hours into seconds to ensure the final result is in joules. Power is $100 \times 10^{3}$ W, and the time interval is 3600 seconds. Calculating the total energy using the formula $E = P \times t$ leads to $E = 100 \times 10^{3} \times 3600$, which simplifies to $36 \times 10^{7}$ J.

Since the lift is moving at a constant velocity, the net force on it must be zero, meaning the upward force supplied by the motor has to exactly balance the downward forces of weight and friction. The gravitational weight of the lift is calculated using $W = mg = 2000 \times 10 = 20000 \text{ N}$. Given that friction opposes motion with 3000 N, the total force required by the motor is $F = 20000 + 3000 = 23000 \text{ N}$. Because power is the product of force and velocity, applying the formula $P = F \cdot v$ with the provided speed of 1.5 ms$^{-1}$ results in $P = 23000 \times 1.5 = 34500 \text{ W}$.

Linear momentum captures the quantity of motion an object possesses, defined as the product of its mass and velocity through the relationship $p = mv$. To find the distance traveled, first convert the mass of 5 g to 0.005 kg, then rearrange the momentum formula to solve for velocity using $v = \frac{p}{m}$. Substituting the given values provides a velocity of 60 m/s. Because distance is calculated as the product of constant velocity and time, multiplying 60 m/s by 5 s results in a total distance of 300 m.

Since the particle falls from rest at height $S$ under the influence of gravity, the principle of conservation of mechanical energy applies. At any height $y$ above the ground, the potential energy is $U = mgy$, and the given condition specifies that the kinetic energy is $K = 3mgy$. Because the total mechanical energy is conserved, the sum of kinetic and potential energy at height $y$ must equal the initial potential energy at height $S$, which is expressed as $K + U = mgS$. Substituting the kinetic energy condition into this equation results in $3mgy + mgy = mgS$, which simplifies to $4mgy = mgS$, yielding a height of $y = \frac{S}{4}$. The speed at this position relates to the kinetic energy through the expression $\frac{1}{2}mv^2 = 3mgy$. Substituting the calculated height $y = \frac{S}{4}$ into this relationship gives $\frac{1}{2}mv^2 = 3mg \left( \frac{S}{4} \right)$, which simplifies to $v^2 = \frac{3gS}{2}$ and indicates that the speed is $v = \sqrt{\frac{3gS}{2}}$.

To achieve static equilibrium for a rigid body, the net torque acting around the pivot point must be zero, meaning the sum of all clockwise torques must exactly balance the sum of all counter-clockwise torques. For a uniform rod, the weight of the rod acts at its center of mass, which is located at the geometric midpoint.

Given the wedge is at the 40 cm mark, the 2 kg mass at the 20 cm mark creates a counter-clockwise torque with a lever arm of $40 - 20 = 20$ cm or 0.2 m. This torque is calculated as $\tau = 2 \times 10 \times 0.2 = 4$ Nm. Conversely, the weight of the 0.5 kg rod acts at the 100 cm mark, providing a clockwise torque with a lever arm of $100 - 40 = 60$ cm or 0.6 m, resulting in $\tau = 0.5 \times 10 \times 0.6 = 3$ Nm. The unknown mass $m$ at the 160 cm mark also produces a clockwise torque with a lever arm of $160 - 40 = 120$ cm or 1.2 m, calculated as $\tau = m \times 10 \times 1.2 = 12m$ Nm. Equating the counter-clockwise torque to the sum of the clockwise torques results in the following relation:

$4 = 3 + 12m$

This simplifies to $1 = 12m$, which yields a mass of $m = \frac{1}{12}$ kg.

The total power supplied to the turbine is derived from the rate at which gravitational potential energy is delivered, calculated using the formula $P = \frac{dm}{dt}gh$. By substituting the mass flow rate of 15 kg/s, gravity of 10 m/s$^2$, and height of 60 m, the input power becomes $15 \times 10 \times 60 = 9000$ W. Since frictional losses account for 10% of this energy, the turbine retains 90% of the total input power, resulting in an output of $9000 \times 0.9 = 8100$ W, which is equivalent to 8.1 kW.

When a force changes depending on the position of an object, the total work done is determined by integrating the force function over the displacement interval. Because the force varies linearly with the position $y$, the work is calculated by finding the definite integral from the starting position of 0 m to the final position of 1 m.

$W = \int_{0}^{1} (20 + 10 y) \, dy$

Evaluating this integral term by term using the power rule results in the expression $20y + 5y^2$ calculated between the boundaries of 0 and 1. Substituting 1 for the variable $y$ gives a sum of 20 and 5, while substituting 0 results in 0, leading to a total work of 25 J.

In an elastic head-on collision where a moving object impacts a stationary one, the fractional kinetic energy lost by the projectile is determined by the specific mass relationship:

$\frac{\Delta KE}{KE} = \frac{4 m_1 m_2}{(m_1 + m_2)^2}$

Substituting the mass of the colliding body $m_1 = 4m$ and the target body $m_2 = 2m$ into this formula gives the expression $\frac{4(4m)(2m)}{(4m + 2m)^2}$. Simplifying the terms, the numerator becomes $32m^2$ and the denominator evaluates to $(6m)^2$, or $36m^2$. Consequently, the ratio of these values simplifies to $\frac{8}{9}$.

Understanding the motion of an object in a vertical circle requires analyzing the balance of forces acting upon the mass at different positions. At the lowest point of the path, the tension in the wire must not only provide the necessary centripetal force to keep the object moving in a circle but also support the weight of the mass itself. This leads to the relationship $T = mg + \mu^2$, which demonstrates that the tension reaches its peak value here. Since the wire is under the greatest amount of tension at this point, it is where the structural integrity of the wire is most likely to fail.

Since the initial system is at rest, the vector sum of the final momenta of the three fragments must be zero to satisfy the principle of conservation of linear momentum. The two fragments of mass $m$ move at speed $v$ in mutually perpendicular directions, creating momentum vectors $mv\hat{i}$ and $mv\hat{j}$. For the system to maintain a net momentum of zero, the remaining fragment of mass $3m$ must have a momentum vector equal to $-(mv\hat{i} + mv\hat{j})$, which possesses a magnitude of $\sqrt{2}mv$.

Calculating the speed of this third fragment involves dividing its momentum magnitude by its mass, which results in a speed of $\frac{\sqrt{2}v}{3}$. Summing the kinetic energies of all three fragments accounts for the total energy released during the breakup:
$\frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}(3m) \left( \frac{\sqrt{2}v}{3} \right)^2 = \frac{4}{3}mv^2$

Since the track is frictionless, the total mechanical energy remains conserved throughout the motion. To successfully complete a vertical loop, the object must maintain a minimum velocity at the lowest point of the circle equal to $\sqrt{5gR}$, where $R$ represents the radius of the track. Setting the initial gravitational potential energy at height $h$ equal to the kinetic energy at the bottom of the loop yields the following relationship:

$mgh = \frac{1}{2}mv^2$

Substituting the minimum velocity condition $v^2 = 5gR$ into this energy balance simplifies to $h = \frac{5}{2}R$. Because the diameter $D$ of the circular path is equal to $2R$, we substitute $R = \frac{D}{2}$ into the height expression to arrive at the final result $h = \frac{5}{4}D$.

The law of conservation of linear momentum dictates that the total momentum of the two-block system remains unchanged during the collision. Given that the lighter block of mass $m$ moves with initial velocity $v$ and the heavier block of mass $4m$ is initially at rest, the total initial momentum is $mv$. Since the lighter block stops after the impact, the entire momentum is transferred to the heavier block, resulting in the relationship $mv = 4mv'$, which shows the heavier block moves away with a velocity of $v' = v/4$.

The coefficient of restitution is defined as the ratio of the relative velocity of separation to the relative velocity of approach. Before the collision, the blocks approach each other with a relative velocity of $v - 0 = v$. Following the collision, the lighter block is stationary, so the relative velocity of separation is simply the final velocity of the heavier block, which is $v/4$. Dividing the separation velocity by the approach velocity gives $e = \frac{v/4}{v}$, which simplifies to $0.25$.

The work done by gravity is found using the potential energy formula $W_g = mgh$, which with a mass of 0.001 kg, acceleration of 10 m/s², and a fall of 1000 m results in 10 J. By the work-energy theorem, the sum of this gravitational work and the work of air resistance must equal the final kinetic energy:
$W_g + W_{\text{res}} = \frac{1}{2}mv^2$
Substituting the values, $10 + W_{\text{res}} = \frac{1}{2} \times 0.001 \times 50^2$ leads to $10 + W_{\text{res}} = 1.25$ J, which means the resistive force performs -8.75 J of work.

Power is defined as the scalar product of the force and velocity vectors, so we must first derive the velocity from the provided force. Since the mass is $1 \text{ kg}$, Newton's second law gives an acceleration $\vec{a} = \vec{F}/m = (2t \hat{i} + 3t^2 \hat{j}) \text{ m/s}^2$. By integrating acceleration with respect to time, assuming the body starts from rest, we obtain the velocity $\vec{v} = \int \vec{a} \, dt = (t^2 \hat{i} + t^3 \hat{j}) \text{ m/s}$.

Calculating the instantaneous power requires the dot product of the force vector and the velocity vector:
$P = \vec{F} \cdot \vec{v} = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j})$
Evaluating the scalar product components results in:
$P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 \text{ W}$

When two objects of equal mass undergo a perfectly elastic collision, they satisfy the requirements of both conservation of momentum and conservation of kinetic energy, a condition that results in the two objects simply interchanging their initial velocities. Ball A begins with a velocity of $v_A = 0.5 \text{ m/s}$ and ball B begins with a velocity of $v_B = -0.3 \text{ m/s}$. Upon impact, the final velocity of A becomes $v_A' = -0.3 \text{ m/s}$ and the final velocity of B becomes $v_B' = 0.5 \text{ m/s}$. Since the question requests the final velocities of B and A respectively, the final values are $v_B' = 0.5 \text{ m/s}$ and $v_A' = -0.3 \text{ m/s}$.

To ensure a body completes a vertical loop, it must possess enough energy at the lowest point to reach the highest point without losing contact with the track. At the very top of the loop, the normal force acting on the body must be at least zero, requiring the centripetal force to be provided entirely by gravity, which corresponds to a velocity of $\sqrt{gR}$. Conservation of energy dictates that the kinetic energy at the bottom must account for both the potential energy gained at the top and the minimum kinetic energy required at that peak, resulting in a necessary minimum velocity of $\sqrt{5gR}$ at the lowest point.

The work done by a constant force is defined as the scalar dot product of the force vector and the displacement vector. We determine the displacement by subtracting the initial position coordinates from the final position coordinates, which yields the displacement vector $\vec{S} = (4\hat{j} + 3\hat{k}) - (-2\hat{i} + 5\hat{j}) = 2\hat{i} - \hat{j} + 3\hat{k}$.

Once the displacement is established, we calculate the total work by taking the dot product of the applied force $\vec{F} = 4\hat{i} + 3\hat{j}$ with this displacement vector. Applying the rules of dot products where corresponding directional components are multiplied, we compute the work using $\vec{F} \cdot \vec{S} = (4\hat{i} + 3\hat{j}) \cdot (2\hat{i} - \hat{j} + 3\hat{k})$. This results in 8 minus 3, which equals 5 J of work done.

The interaction between the bullet and the wooden block is governed by the conservation of linear momentum during the collision, while the subsequent swing of the block is determined by the conservation of mechanical energy. The velocity of the block immediately after the impact is found by equating the gain in gravitational potential energy to the kinetic energy, using the formula $\frac{1}{2}MV^2 = Mgh$. With a height rise of 0.1 m, this velocity is $V = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.1} = \sqrt{2} \approx 1.4$ m/s. Applying the conservation of linear momentum involves equating the initial momentum of the bullet, $0.01 \times 400$, to the sum of the final momenta of the block and the bullet, $2 \times V + 0.01 \times v$, which yields a final bullet velocity of $v = 120$ m/s.

The work-energy theorem states that the work done by the net force acting along the direction of motion equals the change in kinetic energy. Since the particle starts from rest, the work performed by the tangential force accounts for the final kinetic energy of $8 \times 10^{-4}$ J. This tangential force is defined as the product of mass and tangential acceleration, $F_t = m a_t$, and the total displacement covered over two revolutions is $s = 2(2\pi r) = 4\pi r$.

Substituting the given values into the work equation $W = F_t s$, we set $(10 \times 10^{-3}) (a_t) (4\pi \times 6.4 \times 10^{-2}) = 8 \times 10^{-4}$. By isolating $a_t$ in this equation, we calculate $a_t = \frac{8 \times 10^{-4}}{4\pi \times 6.4 \times 10^{-4}}$, which simplifies to $a_t = 0.1 \text{ m/s}^2$.

The motion of the ball is governed by the conservation of mechanical energy, which links the velocity just before impact to the velocity immediately after rebounding. By analyzing the kinetic energy before and after the collision, we can account for the specific energy loss that dictates how high the ball returns.

To reach a height of $h = 20 \text{ m}$ after the bounce, the ball must leave the ground with a velocity $v_f = \sqrt{2gh}$. The impact velocity just before the collision, however, is determined by the initial throw $v_0$ and the acceleration due to gravity, expressed as $v_i = \sqrt{v_0^2 + 2gh}$. Since the collision removes 50 percent of the kinetic energy, the relationship between these velocities is defined by $v_f^2 = \frac{1}{2} v_i^2$, which rearranges to $2v_f^2 = v_i^2$. Substituting $2gh$ for $v_f^2$ and $v_0^2 + 2gh$ for $v_i^2$, the equation becomes $4gh = v_0^2 + 2gh$, simplifying to $v_0^2 = 2gh$. With $g = 10 \text{ m s}^{-2}$ and $h = 20 \text{ m}$, we find $v_0^2 = 2 \times 10 \times 20 = 400$, resulting in an initial velocity of $20 \text{ m s}^{-1}$.