Work, Energy And PowerClass 11 Physics NCERT Solutions

The sign of work done is determined by the angle $\theta$ between the force vector $\mathbf{F}$ and the displacement vector $\mathbf{d}$, according to the formula $W = Fd \cos \theta$.

(a) Positive: The man applies an upward force to lift the bucket, and the displacement of the bucket is also upward. The angle $\theta$ between the force and displacement is $0^\circ$. Since $\cos 0^\circ = 1$, the work done is positive.

(b) Negative: The gravitational force acts downward on the bucket, while the displacement is upward. The angle $\theta$ between the force and displacement is $180^\circ$. Since $\cos 180^\circ = -1$, the work done by gravity is negative.

(c) Negative: Frictional force always opposes the motion. As the body slides down the inclined plane, the frictional force acts up the plane. The angle $\theta$ between the frictional force and the displacement is $180^\circ$. Therefore, the work done by friction is negative.

(d) Positive: The body is moving with uniform velocity, which means the net force is zero. The applied force must be equal in magnitude and opposite in direction to the frictional force. The applied force is in the direction of motion (displacement). The angle $\theta$ between the applied force and displacement is $0^\circ$. Therefore, the work done by the applied force is positive.

(e) Negative: The resistive force of air always opposes the motion of the pendulum's bob. The angle $\theta$ between the resistive force and the displacement of the bob is always $180^\circ$. Therefore, the work done by the air resistance is negative, which causes the pendulum's mechanical energy to decrease and eventually brings it to rest.

Given: Mass of the body, $m = 2 \text{ kg}$ Initial velocity, $u = 0 \text{ m/s}$ Applied horizontal force, $F_{app} = 7 \text{ N}$ Coefficient of kinetic friction, $\mu_k = 0.1$ Time, $t = 10 \text{ s}$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

Calculations: First, we calculate the forces and the acceleration of the body. The force of kinetic friction is given by: $f_k = \mu_k N = \mu_k (mg) = 0.1 \times 2 \text{ kg} \times 9.8 \text{ m/s}^2 = 1.96 \text{ N}$ The net force on the body is: $F_{net} = F_{app} - f_k = 7 \text{ N} - 1.96 \text{ N} = 5.04 \text{ N}$ From Newton's second law, the acceleration of the body is: $a = \frac{F_{net}}{m} = \frac{5.04 \text{ N}}{2 \text{ kg}} = 2.52 \text{ m/s}^2$ Now, we find the distance traversed by the body in 10 s using the equation of motion: $s = ut + \frac{1}{2}at^2 = 0 \times 10 + \frac{1}{2} \times 2.52 \times (10)^2 = 1.26 \times 100 = 126 \text{ m}$

(a) Work done by the applied force in 10 s: $W_{app} = F_{app} \times s = 7 \text{ N} \times 126 \text{ m} = 882 \text{ J}$

(b) Work done by friction in 10 s: The frictional force opposes the motion, so the angle is $180^\circ$. $W_f = f_k \times s \times \cos(180^\circ) = 1.96 \text{ N} \times 126 \text{ m} \times (-1) = -246.96 \text{ J}$

(c) Work done by the net force on the body in 10 s: $W_{net} = F_{net} \times s = 5.04 \text{ N} \times 126 \text{ m} = 635.04 \text{ J}$ Alternatively, $W_{net} = W_{app} + W_f = 882 \text{ J} - 246.96 \text{ J} = 635.04 \text{ J}$.

(d) Change in kinetic energy of the body in 10 s: First, find the final velocity after 10 s: $v = u + at = 0 + 2.52 \times 10 = 25.2 \text{ m/s}$ The change in kinetic energy is: $\Delta K = K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2} \times 2 \times (25.2)^2 - 0 = 635.04 \text{ J}$

Interpretation: The result from part (c) and part (d) are the same. This verifies the Work-Energy Theorem, which states that the work done by the net force on a body is equal to the change in its kinetic energy ($W_{net} = \Delta K$). The work done by the applied force is used to overcome friction and to increase the kinetic energy of the body.

The total energy $E$ of a particle is the sum of its kinetic energy $K$ and potential energy $V(x)$, i.e., $E = K + V(x)$. Since kinetic energy $K = \frac{1}{2}mv^2$ must be non-negative ($K \ge 0$), the particle can only be found in regions where its potential energy is less than or equal to its total energy, i.e., $V(x) \le E$.

Given: Potential energy function, $V(x) = \frac{1}{2} kx^2$ Force constant, $k = 0.5 \text{ N/m}$ Total energy of the particle, $E = 1 \text{ J}$

To Show: The particle must 'turn back' at $x = \pm 2 \text{ m}$.

Concept: The total energy $E$ of a particle is the sum of its kinetic energy ($K$) and potential energy ($V(x)$). $E = K + V(x)$ A particle 'turns back' at the points where its velocity becomes zero, which means its kinetic energy $K$ becomes zero. At these turning points, the total energy is equal to the potential energy. $E = V(x)$

Calculation: We set the total energy equal to the potential energy to find the positions of the turning points. $E = \frac{1}{2} kx^2$ Substituting the given values: $1 \text{ J} = \frac{1}{2} \times (0.5 \text{ N/m}) \times x^2$ $1 = 0.25 \times x^2$ $x^2 = \frac{1}{0.25} = 4$ $x = \pm \sqrt{4}$ $x = \pm 2 \text{ m}$

Conclusion: At $x = +2 \text{ m}$ and $x = -2 \text{ m}$, the potential energy of the particle is equal to its total energy (1 J). This means its kinetic energy is zero at these points. The particle momentarily stops and reverses its direction of motion. Therefore, a particle of total energy 1 J moving under this potential must 'turn back' when it reaches $x = \pm 2 \text{ m}$.

(a) The heat energy required for the burning of the rocket's casing is obtained at the expense of the rocket's mechanical energy (kinetic and potential energy). The work done by the force of air friction on the rocket is negative. According to the work-energy theorem modified for non-conservative forces, this negative work leads to a decrease in the total mechanical energy of the rocket. This lost mechanical energy is converted into heat, which causes the casing to burn up. So, the energy comes from the rocket itself.

(b) The gravitational force is a conservative force. A key property of a conservative force is that the work done by it over any closed path is zero. Since a comet's orbit is a closed path, the net work done by the sun's gravitational force on the comet over one complete orbit is zero. Although the work is non-zero over parts of the orbit (negative when moving away from the sun, positive when moving towards it), the total work over the entire elliptical loop cancels out to zero.

(c) The total energy of the satellite is the sum of its kinetic energy ($K$) and potential energy ($V$), $E = K + V$. The potential energy is negative ($V = -GMm/r$) and kinetic energy is positive ($K = GMm/2r$). The total energy is $E = -GMm/2r$. As the satellite loses energy due to atmospheric resistance, its total energy $E$ decreases, meaning it becomes more negative. For $E$ to become more negative, the orbital radius $r$ must decrease. As $r$ decreases, the satellite moves closer to the Earth. The kinetic energy is given by $K = GMm/2r$. Since $r$ decreases, the kinetic energy $K$ increases, and therefore its speed increases. The loss in potential energy is greater than the loss in total energy, and the difference is converted into kinetic energy.

(d) The work done is greater in case (ii).

(a) When a conservative force does positive work on a body, the potential energy of the body decreases. The work done by a conservative force is equal to the negative of the change in potential energy: $W_c = -\Delta V = -(V_f - V_i)$. If $W_c$ is positive, then $V_i - V_f > 0$, which means $V_f < V_i$. Thus, the potential energy decreases.

(b) Work done by a body against friction always results in a loss of its kinetic energy. Friction is a dissipative force that converts mechanical energy (usually kinetic energy) into heat. For a body moving on a horizontal surface, only kinetic energy is involved. If it is on an incline, both may change, but the loss is fundamentally from the mechanical energy, which often manifests as a reduction in kinetic energy or a smaller increase in kinetic energy than would otherwise be expected.

(c) The rate of change of total momentum of a many-particle system is proportional to the external force on the system. This is Newton's second law for a system of particles. The internal forces occur in equal and opposite pairs (Newton's third law), so their vector sum is zero and they cannot change the total momentum of the system.

(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total linear momentum and total energy of the system of two bodies. Total linear momentum is conserved in all collisions if there are no external forces. Total kinetic energy is not conserved in an inelastic collision (it decreases). However, the total energy of the system (including heat, sound, etc.) is always conserved according to the law of conservation of energy.

(a) False. Reason: In an elastic collision, the total momentum and total kinetic energy of the system of two bodies are conserved. The momentum and kinetic energy of the individual bodies almost always change. Momentum and energy are transferred between the colliding bodies.

(b) False. Reason: The total energy of a system is conserved only when the external forces acting on the system do no work. If there is work done by external forces (like an external push or pull) or dissipative internal forces (like friction), the total mechanical energy of the system is not conserved. The law of conservation of energy states that energy can neither be created nor destroyed, but it can be transferred into or out of a system.

(c) False. Reason: The work done in the motion of a body over a closed loop is zero only for conservative forces (e.g., gravitational force, electrostatic force, spring force). For non-conservative forces, such as friction or air resistance, the work done over a closed loop is always negative and not zero.

(d) True. Reason: By definition, an inelastic collision is one in which the total kinetic energy of the system is not conserved. During the collision, some of the initial kinetic energy is converted into other forms of energy, such as heat, sound, or potential energy of deformation. Therefore, the final kinetic energy is always less than the initial kinetic energy.

(a) No. During the short time of collision, the total kinetic energy is not conserved. When the balls are in contact, they deform. The initial kinetic energy is temporarily converted into potential energy of deformation. In a perfectly elastic collision, this stored potential energy is fully converted back into kinetic energy as the balls separate, so the final kinetic energy is equal to the initial kinetic energy. But at the instant of maximum deformation, the kinetic energy is at a minimum.

(b) Yes. The total linear momentum of the system is conserved during the short time of collision. The forces involved in the collision are internal forces (action-reaction pair). As long as there are no significant external forces (like friction) acting on the system during the brief collision time, the total linear momentum remains constant throughout the process.

(c) For an inelastic collision:

(d) The collision is elastic. A force that depends only on the separation distance (position) is a conservative force. The potential energy function described corresponds to such a force. When the forces involved in a collision are conservative, the total mechanical energy (kinetic + potential) is conserved. This means that any kinetic energy converted to potential energy during the collision is fully recovered as kinetic energy after the collision. This is the definition of an elastic collision.

Given: Initial velocity, $u = 0$ Constant acceleration, $a$

Formula: Power is given by $P = \mathbf{F} \cdot \mathbf{v}$. For one-dimensional motion, $P = Fv$. From Newton's second law, Force $F = ma$. Since $m$ and $a$ are constant, $F$ is constant. The velocity of the body at time $t$ is given by the first equation of motion: $v = u + at$. Since $u=0$, we have $v = at$.

Derivation: Substitute the expressions for $F$ and $v$ into the power formula: $P = (ma)(at)$ $P = ma^2 t$ Since mass $m$ and acceleration $a$ are constants, $ma^2$ is a constant. Therefore, the power $P$ is directly proportional to time $t$. $P \propto t$

Final Answer: The correct option is (ii) $t$.

Given: Constant power, $P = \text{constant}$.

Derivation: Power is the rate of doing work, $P = \frac{dW}{dt}$. Since power is constant, the work done in time $t$ is $W = Pt$. According to the work-energy theorem, the work done on the body is equal to the change in its kinetic energy. Assuming the body starts from rest: $W = \Delta K = \frac{1}{2}mv^2 - 0$ Equating the two expressions for work: $Pt = \frac{1}{2}mv^2$ Solving for velocity $v$: $v^2 = \frac{2Pt}{m}$ $v = \sqrt{\frac{2P}{m}} t^{1/2}$ Since $P$ and $m$ are constants, $v$ is proportional to $t^{1/2}$. $v \propto t^{1/2}$ Now, velocity is the rate of change of displacement, $v = \frac{ds}{dt}$. $\frac{ds}{dt} = C t^{1/2}$ where $C = \sqrt{\frac{2P}{m}}$ is a constant. To find the displacement $s$, we integrate with respect to time $t$: $ds = C t^{1/2} dt$ $\int ds = \int C t^{1/2} dt$ $s = C \frac{t^{3/2}}{3/2} + C'$ Assuming the displacement is zero at $t=0$, the integration constant $C'=0$. $s = \frac{2C}{3} t^{3/2}$ Since $\frac{2C}{3}$ is a constant, the displacement $s$ is proportional to $t^{3/2}$. $s \propto t^{3/2}$

Final Answer: The correct option is (iii) $t^{3/2}$.

Given: Constant force, $\mathbf{F} = (-\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}) \text{ N}$ Distance moved along the z-axis = 4 m.

To Find: The work done by the force $\mathbf{F}$.

Formula: Work done is given by the scalar product (dot product) of the force vector and the displacement vector: $W = \mathbf{F} \cdot \mathbf{d}$.

Calculation: The body moves a distance of 4 m along the z-axis. So, the displacement vector is: $\mathbf{d} = (0\hat{\mathbf{i}} + 0\hat{\mathbf{j}} + 4\hat{\mathbf{k}}) \text{ m}$ Now, we calculate the work done: $W = \mathbf{F} \cdot \mathbf{d} = (-\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}) \cdot (0\hat{\mathbf{i}} + 0\hat{\mathbf{j}} + 4\hat{\mathbf{k}})$ Using the property of dot product for orthogonal unit vectors ($\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = 0$, etc.) and parallel unit vectors ($\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = 1$, etc.): $W = (-1)(0) + (2)(0) + (3)(4)$ $W = 0 + 0 + 12$ $W = 12 \text{ J}$

Final Answer: The work done by the force is $12 \text{ J}$.

Given: Kinetic energy of electron, $K_e = 10 \text{ keV} = 10 \times 10^3 \text{ eV} = 10^4 \times 1.60 \times 10^{-19} \text{ J} = 1.60 \times 10^{-15} \text{ J}$ Kinetic energy of proton, $K_p = 100 \text{ keV} = 100 \times 10^3 \text{ eV} = 10^5 \times 1.60 \times 10^{-19} \text{ J} = 1.60 \times 10^{-14} \text{ J}$ Mass of electron, $m_e = 9.11 \times 10^{-31} \text{ kg}$ Mass of proton, $m_p = 1.67 \times 10^{-27} \text{ kg}$

To Find: Which particle is faster and the ratio of their speeds ($v_e / v_p$).

Formula: Kinetic energy is given by $K = \frac{1}{2}mv^2$. Therefore, speed is $v = \sqrt{\frac{2K}{m}}$.

Calculation: Speed of the electron: $v_e = \sqrt{\frac{2K_e}{m_e}} = \sqrt{\frac{2 \times 1.60 \times 10^{-15}}{9.11 \times 10^{-31}}} = \sqrt{0.351 \times 10^{16}} = 0.592 \times 10^8 \text{ m/s}$ $v_e = 5.92 \times 10^7 \text{ m/s}$

Speed of the proton: $v_p = \sqrt{\frac{2K_p}{m_p}} = \sqrt{\frac{2 \times 1.60 \times 10^{-14}}{1.67 \times 10^{-27}}} = \sqrt{1.916 \times 10^{13}} = \sqrt{19.16 \times 10^{12}} = 4.38 \times 10^6 \text{ m/s}$

Comparing the speeds, $5.92 \times 10^7 \text{ m/s} > 4.38 \times 10^6 \text{ m/s}$. Thus, the electron is faster.

Ratio of their speeds: $\frac{v_e}{v_p} = \frac{5.92 \times 10^7}{4.38 \times 10^6} \approx 13.5$

Alternatively, we can find the ratio directly: $\frac{v_e}{v_p} = \frac{\sqrt{2K_e/m_e}}{\sqrt{2K_p/m_p}} = \sqrt{\frac{K_e}{K_p} \times \frac{m_p}{m_e}}$ $\frac{v_e}{v_p} = \sqrt{\frac{10 \text{ keV}}{100 \text{ keV}} \times \frac{1.67 \times 10^{-27} \text{ kg}}{9.11 \times 10^{-31} \text{ kg}}} = \sqrt{0.1 \times 1833} = \sqrt{183.3} \approx 13.54$

Final Answer: The electron is faster than the proton. The ratio of their speeds, $v_e / v_p$, is approximately $13.54$.

Given: Radius of raindrop, $r = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$ Total height, $H = 500 \text{ m}$ Final speed on reaching the ground, $v_f = 10 \text{ m/s}$ Density of water, $\rho = 1000 \text{ kg/m}^3$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

Calculation of Mass: First, we calculate the mass of the raindrop. Volume of the drop, $V = \frac{4}{3}\pi r^3 = \frac{4}{3} \pi (2 \times 10^{-3})^3 = \frac{32}{3} \pi \times 10^{-9} \text{ m}^3$ Mass of the drop, $m = \rho V = 1000 \times \frac{32}{3} \pi \times 10^{-9} = \frac{32}{3} \pi \times 10^{-6} \text{ kg} \approx 3.35 \times 10^{-5} \text{ kg}$

Work done by gravitational force: The gravitational force is $mg$ and it acts downwards. The displacement is also downwards. The height for each half of the journey is $h = H/2 = 500/2 = 250 \text{ m}$. Work done by gravity is independent of the path or speed, it only depends on the vertical displacement. Work done in the first half of the journey: $W_{g1} = mgh = (3.35 \times 10^{-5}) \times 9.8 \times 250 = 0.082 \text{ J}$ Work done in the second half of the journey: $W_{g2} = mgh = (3.35 \times 10^{-5}) \times 9.8 \times 250 = 0.082 \text{ J}$

Work done by the resistive force: We use the Work-Energy Theorem for the entire journey from $H = 500 \text{ m}$ to the ground. $W_{net} = \Delta K = K_f - K_i$ The net work done is the sum of the work done by gravity ($W_g$) and the work done by the resistive force ($W_r$). The drop starts from rest, so its initial kinetic energy $K_i = 0$. $W_g + W_r = \frac{1}{2}mv_f^2 - 0$ Total work done by gravity over the entire journey is: $W_g = W_{g1} + W_{g2} = mgH = (3.35 \times 10^{-5}) \times 9.8 \times 500 = 0.164 \text{ J}$ Final kinetic energy is: $K_f = \frac{1}{2}mv_f^2 = \frac{1}{2} \times (3.35 \times 10^{-5}) \times (10)^2 = 1.675 \times 10^{-3} \text{ J}$ Now, substitute these values into the work-energy equation: $0.164 + W_r = 1.675 \times 10^{-3}$ $W_r = 0.001675 - 0.164 = -0.1623 \text{ J}$

Final Answer: Work done by gravitational force in the first half: $0.082 \text{ J}$. Work done by gravitational force in the second half: $0.082 \text{ J}$. Work done by the resistive force in the entire journey: $-0.1623 \text{ J}$.

Is momentum conserved? No, the momentum of the molecule is not conserved.

Reason: Momentum is a vector quantity. Let the initial velocity be $\mathbf{v}_i$ and the final velocity be $\mathbf{v}_f$. Let the normal to the wall be along the y-axis and the wall be along the x-axis. The initial velocity vector can be written as: $\mathbf{v}_i = v \sin(30^\circ) \hat{\mathbf{i}} - v \cos(30^\circ) \hat{\mathbf{j}}$ After rebounding with the same speed $v$, the angle with the normal is also $30^\circ$. The final velocity vector is: $\mathbf{v}_f = v \sin(30^\circ) \hat{\mathbf{i}} + v \cos(30^\circ) \hat{\mathbf{j}}$ Initial momentum: $\mathbf{p}_i = m\mathbf{v}_i = m(v \sin(30^\circ) \hat{\mathbf{i}} - v \cos(30^\circ) \hat{\mathbf{j}})$ Final momentum: $\mathbf{p}_f = m\mathbf{v}_f = m(v \sin(30^\circ) \hat{\mathbf{i}} + v \cos(30^\circ) \hat{\mathbf{j}})$ Since $\mathbf{p}_i \neq \mathbf{p}_f$, the momentum of the molecule is not conserved. The y-component of the momentum has changed direction. This change is due to the force exerted by the wall on the molecule. However, the momentum of the entire system (molecule + wall) is conserved.

Is the collision elastic or inelastic? The collision is elastic.

Reason: An elastic collision is one in which the total kinetic energy is conserved. Initial kinetic energy: $K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}m(200)^2$ Final kinetic energy: $K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}m(200)^2$ Since the molecule rebounds with the same speed, its kinetic energy remains unchanged ($K_i = K_f$). Therefore, the collision is elastic.

Given: Volume of the tank, $V = 30 \text{ m}^3$ Time to fill the tank, $t = 15 \text{ min} = 15 \times 60 = 900 \text{ s}$ Height of the tank, $h = 40 \text{ m}$ Efficiency of the pump, $\eta = 30\% = 0.30$ Density of water, $\rho = 1000 \text{ kg/m}^3$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

To Find: The electric power consumed by the pump ($P_{in}$).

Final Answer: The electric power consumed by the pump is approximately $43.6 \text{ kW}$.

Let the mass of each ball bearing be $m$. The first ball (ball 1) has initial velocity $V$. The other two balls (ball 2 and ball 3) are initially at rest.

Applying Principles of Elastic Collisions: In an elastic head-on collision between two identical masses, where one is at rest, the moving mass comes to rest and the stationary mass moves off with the velocity of the first mass. This is a direct exchange of velocities.

Let's analyze the sequence of collisions:

Final State: After this rapid sequence of collisions:

This outcome conserves both momentum and kinetic energy:

Conclusion: The only possible result shown in Fig. 5.14 that matches this analysis is the one where the first two balls are stationary and the third ball moves off with speed $V$.

Final Answer: The correct scenario is the one depicted in the third case of Fig 5.14, where the striking ball and the middle ball come to rest, and the last ball moves on with the initial speed $V$.

Key Principle: For a one-dimensional (head-on) elastic collision between two objects of equal mass, where one is initially at rest, the velocities of the two objects are exchanged.

Therefore, after the collision, bob A will transfer all of its momentum and kinetic energy to bob B. Bob A will come to a complete rest at the point of collision, and bob B will move forward with the velocity $v$ that bob A had.

Conclusion: Since bob A comes to rest immediately after the collision, it will not swing up on the other side. Its velocity becomes zero, so it cannot rise.

Final Answer: Bob A will not rise after the collision. It will come to a stop at the lowest point of its swing.

Given: Length of the pendulum, $L = 1.5 \text{ m}$ Initial position: horizontal. This means the initial height $h$ above the lowermost point is equal to the length of the pendulum, $h = L = 1.5 \text{ m}$. Energy dissipated against air resistance = $5\%$ of initial energy.

To Find: The speed of the bob at the lowermost point, $v$.

Formula: We will use the principle of conservation of energy, modified to account for the energy lost. Initial Energy ($E_i$) = Final Energy ($E_f$) + Energy Lost

Final Answer: The speed with which the bob arrives at the lowermost point is approximately $5.3 \text{ m/s}$.

Given: Mass of trolley, $M = 300 \text{ kg}$ Mass of sandbag, $m_{sand} = 25 \text{ kg}$ Initial speed, $v_i = 27 \text{ km/h} = 27 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 7.5 \text{ m/s}$ The track is frictionless.

To Find: The final speed of the trolley ($v_f$) after the sandbag is empty.

Concept: This problem is about the conservation of linear momentum. The system consists of the trolley and the sand. Since the track is frictionless, there are no external horizontal forces acting on the system. Therefore, the total horizontal momentum of the system must be conserved.

Analysis: The sand leaks out of a hole in the floor. When the sand leaves the trolley, it has the same horizontal velocity as the trolley. Since the leaking sand continues to move with the same horizontal velocity as the trolley, there is no change in the horizontal momentum of the leaking sand.

Let's consider the system of (trolley + remaining sand). The sand that has leaked out is no longer part of this system, but it leaves with the system's velocity. Let $M(t)$ be the mass of the trolley + sand at time $t$, and $v(t)$ be its velocity. The momentum of the system at time $t$ is $P(t) = M(t)v(t)$. In a small time interval $dt$, a mass $dM$ leaks out. The mass of the system becomes $M(t) - dM$, and its velocity becomes $v(t) + dv$. The leaked mass $dM$ has velocity $v(t)$.

By conservation of momentum: Initial momentum = Final momentum $M(t)v(t) = (M(t) - dM)(v(t) + dv) + dM \cdot v(t)$ $Mv = Mv + Mdv - v dM - dM dv + v dM$ Neglecting the very small term $dM dv$, we get: $0 = Mdv$ Since the mass $M$ is not zero, this implies $dv = 0$. The change in velocity is zero.

Conclusion: The velocity of the trolley does not change. The forces involved in the leaking process are purely vertical (gravity pulling the sand down). There is no horizontal force to change the horizontal velocity of the trolley.

Final Answer: The speed of the trolley remains unchanged. Its final speed will be $27 \text{ km/h}$ or $7.5 \text{ m/s}$.

Given: Mass of the body, $m = 0.5 \text{ kg}$ Velocity function, $v = ax^{3/2}$ Value of constant, $a = 5 \text{ m}^{-1/2} \text{ s}^{-1}$ Initial position, $x_i = 0 \text{ m}$ Final position, $x_f = 2 \text{ m}$

To Find: The work done by the net force, $W_{net}$.

Formula: According to the Work-Energy Theorem, the work done by the net force on a body is equal to the change in its kinetic energy. $W_{net} = \Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Final Answer: The work done by the net force during its displacement from $x=0$ to $x=2 \text{ m}$ is $50 \text{ J}$.

Given: Area swept by blades, $A = 30 \text{ m}^2$ Wind velocity, $v = 36 \text{ km/h} = 36 \times \frac{1000}{3600} = 10 \text{ m/s}$ Density of air, $\rho = 1.2 \text{ kg/m}^3$ Efficiency of conversion, $\eta = 25\% = 0.25$

(a) Mass of the air passing through in time $t$: In time $t$, the wind travels a distance $d = v \times t$. The volume of air passing through the area $A$ in time $t$ is: $\text{Volume} = \text{Area} \times \text{distance} = A \times (vt)$ The mass of this air is: $m = \text{Volume} \times \text{Density} = (Avt) \rho$ $m = \rho A v t$

(b) Kinetic energy of the air: The kinetic energy of this mass of air is: $K = \frac{1}{2}mv^2 = \frac{1}{2}(\rho A v t)v^2$ $K = \frac{1}{2} \rho A v^3 t$

(c) Electrical power produced: Power is the rate at which energy is transferred. The power of the wind is the kinetic energy of the air passing through the area per unit time. $P_{wind} = \frac{K}{t} = \frac{\frac{1}{2} \rho A v^3 t}{t} = \frac{1}{2} \rho A v^3$ The windmill converts $25\%$ of this power into electrical power. $P_{electrical} = \eta \times P_{wind} = 0.25 \times \frac{1}{2} \rho A v^3$ Now, we substitute the given values: $P_{electrical} = 0.25 \times \frac{1}{2} \times (1.2 \text{ kg/m}^3) \times (30 \text{ m}^2) \times (10 \text{ m/s})^3$ $P_{electrical} = 0.25 \times 0.5 \times 1.2 \times 30 \times 1000$ $P_{electrical} = 0.125 \times 36 \times 1000$ $P_{electrical} = 4.5 \times 1000 = 4500 \text{ W}$ This is equal to $4.5 \text{ kW}$.

Final Answer: (a) The mass of the air is $m = \rho A v t$. (b) The kinetic energy of the air is $K = \frac{1}{2} \rho A v^3 t$. (c) The electrical power produced is $4500 \text{ W}$ or $4.5 \text{ kW}$.

Given: Mass lifted, $m = 10 \text{ kg}$ Number of lifts, $n = 1000$ Height of each lift, $h = 0.5 \text{ m}$ Energy supplied by fat, $E_{fat} = 3.8 \times 10^7 \text{ J/kg}$ Efficiency of conversion, $\eta = 20\% = 0.20$ Acceleration due to gravity, $g \approx 9.8 \text{ m/s}^2$

(a) Work done against the gravitational force: The work done in one lift against gravity is $W_1 = mgh$. The total work done in 1000 lifts is: $W_{total} = n \times W_1 = n \times mgh$ $W_{total} = 1000 \times 10 \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.5 \text{ m}$ $W_{total} = 1000 \times 49 = 49000 \text{ J}$ This is the total mechanical work done.

(b) Fat used up: The total work done ($W_{total}$) is the useful mechanical energy output. The body is not perfectly efficient; it must consume more energy from fat to produce this mechanical work. The energy consumed from fat ($E_{consumed}$) is related to the mechanical work by the efficiency: $\eta = \frac{\text{Mechanical Energy Output}}{\text{Energy Consumed}} = \frac{W_{total}}{E_{consumed}}$ Therefore, the total energy that must be supplied by burning fat is: $E_{consumed} = \frac{W_{total}}{\eta} = \frac{49000 \text{ J}}{0.20} = 245000 \text{ J}$ Now, we can find the mass of fat that must be burned to supply this amount of energy: $\text{Mass of fat} = \frac{\text{Total energy consumed}}{\text{Energy per kg of fat}}$ $\text{Mass of fat} = \frac{245000 \text{ J}}{3.8 \times 10^7 \text{ J/kg}} \approx 0.00645 \text{ kg}$ Converting to grams: $0.00645 \text{ kg} \times 1000 \text{ g/kg} = 6.45 \text{ g}$

Final Answer: (a) She does $49000 \text{ J}$ of work against the gravitational force. (b) The dieter will use up approximately $6.45 \times 10^{-3} \text{ kg}$ (or $6.45 \text{ g}$) of fat.

Given: Power required by the family, $P_{req} = 8 \text{ kW} = 8000 \text{ W}$ Incident solar power per unit area, $I = 200 \text{ W/m}^2$ Conversion efficiency, $\eta = 20\% = 0.20$

(b) Comparison with the roof of a typical house: A typical house might have a roof area ranging from about $100 \text{ m}^2$ to $250 \text{ m}^2$. For example, a house that is 15 m long and 10 m wide would have a roof area of $150 \text{ m}^2$ (assuming a flat roof for simplicity).

The calculated area of $200 \text{ m}^2$ is quite large but is comparable to the roof area of a large single-family house. It is certainly a feasible area to cover with solar panels for a household committed to solar energy.

Final Answer: (a) An area of $200 \text{ m}^2$ is needed to supply 8 kW of power. (b) This area is comparable to the roof area of a typical large house, making it a plausible but significant installation.