WORK, ENERGY AND POWER-Exercise

Q1. The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Solution

The sign of work done depends on the angle between force and displacement vectors, given by \( W = \vec{F} \cdot \vec{d} = F d \cos\theta \). When force and displacement align (\( \theta = 0^\circ \)), work is positive; when opposite (\( \theta = 180^\circ \)), work is negative.

• (a) Work done by a man lifting a bucket out of a well: The man applies an upward force through the rope, matching the bucket's upward displacement. Thus, \( \theta = 0^\circ \), making the work positive.

• (b) Work done by gravitational force in the above case: Gravity pulls the bucket downward while displacement remains upward, so \( \theta = 180^\circ \) and \( \cos\theta = -1 \). The work done by gravity is negative.

• (c) Work done by friction on a body sliding down an inclined plane: Friction acts up the plane, opposing the downward motion and displacement. With \( \theta = 180^\circ \), friction always does negative work.

• (d) Work done by applied force on a body moving with uniform velocity on rough horizontal plane: Uniform velocity means applied force balances friction, acting in the direction of motion. Therefore, \( \theta = 0^\circ \) and work is positive.

• (e) Work done by resistive force of air on a vibrating pendulum: Air resistance opposes the pendulum's oscillatory motion, making force antiparallel to displacement throughout. The work is negative, dissipating energy to bring it to rest.

Q2. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s,
and interpret your results.

Solution

Given: Mass \( m = 2 \) kg, initial velocity \( u = 0 \), applied force \( F = 7 \) N, coefficient of kinetic friction \( \mu_k = 0.1 \), time \( t = 10 \) s, \( g = 9.8 \) m/s².

(a) Work done by applied force: $$\begin{aligned} f_k &= \mu_k mg \\ &= 0.1 \times 2 \times 9.8 \\ &= 1.96\, \text{N} \\\\ F_{net} &= F - f_k \\ &= 7 - 1.96 \\ &= 5.04\, \text{N} \\\\ a &= \frac{F_{net}}{m} \\&= \frac{5.04}{2} \\&= 2.52\, \text{m/s}^2 \\\\ v &= u + at \\&= 0 + 2.52 \times 10 \\&= 25.2\, \text{m/s} \\\\ s &= \frac{v^2 - u^2}{2a} \\&= \frac{25.2^2}{2 \times 2.52} \&= 126\, \text{m} \\\\ W_F &= F \cdot s \\&= 7 \times 126 \\&= 882\, \text{J} \end{aligned}$$

(b) Work done by friction: $$\begin{aligned} W_f &= -f_k \cdot s \\ &= -\mu_k mg \times 126 \\ &= -0.1 \times 2 \times 9.8 \times 126 \\ &= -246.96\, \text{J} \end{aligned}$$

(c) Work done by net force: $$\begin{aligned} W_{net} &= F_{net} \cdot s \\ &= 5.04 \times 126 \\ &= 635\, \text{J} \end{aligned}$$

(d) Change in kinetic energy: $$\begin{aligned} \Delta K &= \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \\ &= \frac{1}{2} \times 2 \times (25.2)^2 - 0 \\ &= 635\, \text{J} \end{aligned}$$

Work-energy theorem confirmed: \( W_{net} = \Delta K = 635\, \text{J} \). Applied force supplies 882 J while friction dissipates 247 J as heat.

Q4. The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Solution

Simple Harmonic Motion: Turning Points

In simple harmonic motion, particle turns back at turning points where total energy equals potential energy: \( E = V(x) \). Beyond these points, kinetic energy becomes negative, which is impossible classically.

Given: \( V(x) = \frac{kx^2}{2} \), \( k = 0.5 \) N/m, total energy \( E = 1 \) J.

At turning points: $$\begin{aligned} E &= V(x) \\ 1 &= \frac{kx^2}{2} \\ 1 &= \frac{0.5 \cdot x^2}{2} \\ 1 &= 0.25 x^2 \\ x^2 &= \frac{1}{0.25} \\ x^2 &= 4 \\ x &= \pm \sqrt{4} \\ x &= \pm 2\, \text{m} \end{aligned}$$

Particle reaches maximum displacement at \( x = \pm 2 \) m where all energy converts to potential energy, velocity becomes zero, and motion reverses direction.

Q5. Answer the following : (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ?
(d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?

Solution

Work-Energy Concepts: Conceptual Analysis

(a) Rocket casing burns due to friction: Heat energy comes from the rocket's kinetic energy. Frictional force acts opposite to rocket's motion, doing negative work on rocket which converts kinetic energy to thermal energy. Atmosphere provides resistance but doesn't supply energy - rocket loses KE to heat both casing and atmosphere.

(b) Zero work by gravity over complete comet orbit: Gravitational force is always perpendicular to instantaneous velocity in closed elliptical orbit (conservative force, closed path). Over complete orbit, displacement vector returns to starting point, so $$\begin{aligned} \int \vec{F_g} \cdot d\vec{r} &= 0 \end{aligned}$$ Total work done is zero despite non-perpendicular force at individual points.

(c) Satellite speed increases despite energy loss: Atmospheric drag does negative work, reducing total mechanical energy. As satellite spirals inward, gravitational potential energy decreases more significantly ($$V = -\frac{GM}{r}$$). Conservation of angular momentum increases tangential speed: $$v \propto \frac{1}{\sqrt{r}}$$ Closer orbit requires higher speed despite overall energy loss.

(d) Work comparison - carrying vs pulley system:

Fig (i) carrying: $$ \begin{aligned} W &= F_g \times 0 \\ &= 15 \times 9.8 \times 0 \\ &= 0\, \text{J} \end{aligned} $$

Force vertical, displacement horizontal

Fig (ii) pulley: $$ \begin{aligned} W &= T \times 2\, \text{m} \\ T &= 15 \times 9.8 \\&= 147\, \text{N} \\\\ W &= 147 \times 2 \\&= 294\, \text{J} \end{aligned}$$ Work done greater in pulley case (ii).

Key principle: Work depends on force component along displacement direction, not just force magnitude.

Q6. Underline the correct alternative :
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Solution

Multiple Choice: Work-Energy-Momentum Concepts

(a) Conservative force doing positive work: decreases When conservative force does positive work, potential energy decreases by equal amount (work-energy theorem: $$W_{cons} = -\Delta U$$). Positive work means force moves particle toward lower potential.

(b) Work against friction: kinetic Friction converts kinetic energy to thermal energy irreversibly. Body loses KE to overcome frictional resistance regardless of potential energy changes.

(c) Rate of change of total momentum: external force Newton's second law for system: $$\frac{d\vec{P}}{dt} = \vec{F}_{ext}$$ Internal forces cancel by action-reaction pairs, only external forces affect total momentum.

(d) Inelastic collision conservation: total linear momentum Momentum conserves (no external forces), but kinetic energy lost to deformation/sound/heat. Total energy conserved but redistributed.

Correct answers: (a) decreases, (b) kinetic, (c) external force, (d) total linear momentum.

Q7. State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Solution

True/False: Conservation Laws Analysis

(a) False: In elastic collision, total momentum and total kinetic energy of system are conserved, not individual quantities of each body. Individual bodies exchange momentum and energy: $$\begin{aligned} m_1u_1 + m_2u_2 &= m_1v_1 + m_2v_2 \\ \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 &= \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \end{aligned}$$

(b) False: Total energy conserved only when no non-conservative forces do work. External friction, air resistance convert mechanical energy to thermal energy: $$W_{non-cons} = \Delta E_{mech}$$ Isolated systems conserve total energy including all forms.

(c) False: Zero work over closed loop only for conservative forces (gravity, spring). Frictional force does negative work: $$\int \vec{F}_{fr} \cdot d\vec{r} < 0$$ Non-conservative forces create energy dissipation.

(d) True: Inelastic collisions convert kinetic energy to other forms (deformation, sound, heat). By definition: $$\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 < \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2$$ Momentum conserved, KE decreases.

Answers: (a) False, (b) False, (c) False, (d) True.

Q8. Answer carefully, with reasons :
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ?
(c) What are the answers to (a) and (b) for an inelastic collision ?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Solution

Collision Dynamics: During Contact Analysis

(a) Elastic collision - KE during contact: No During collision, balls deform elastically, storing kinetic energy as elastic potential energy in compressed material. Total mechanical energy conserved, but kinetic energy decreases temporarily: $$KE_{initial} > KE_{during} + PE_{elastic}$$

(b) Elastic collision - momentum during contact: Yes No external forces act horizontally during short collision time. Internal forces between balls form action-reaction pairs: $$\vec{F}_{12} = -\vec{F}_{21}$$ Total momentum remains constant throughout contact.

(c) Inelastic collision: $$\begin{aligned} \text{KE during contact:} & \quad \text{No (deformation energy + heat)} \\ \text{Momentum during contact:} & \quad \text{Yes (no external forces)} \end{aligned}$$ KE lost permanently to plastic deformation and thermal energy.

(d) Potential energy depends only on separation: Elastic Conservative force field (depends only on position/distance). During collision, elastic potential energy stored during compression, fully recovered during separation: $$W_{cons} = 0$$ over closed deformation path. No dissipation occurs.

Key distinction: Momentum conserved always (no external forces), KE conservation depends on energy dissipation mechanism during contact.

Q9. A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) \(t^{1/2}\) (ii) \(t\) (iii) \(t^{3/2}\) (iv) \(t^2\)

Solution

Power in Constant Acceleration Motion

Body starts from rest (\(u = 0\)) with constant acceleration \(a\). Power \(P = \vec{F} \cdot \vec{v}\).

Solution: $$\begin{aligned} P &= F \cdot v \\ F &= ma \\\\ v &= u + at \\&= 0 + at \\&= at \\\\ P &= ma \cdot (at) \\ P &= ma^2 t \\ P &\propto t \end{aligned}$$

Correct answer: (ii) \(t\)

Q10. A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time \(t\) is proportional to
(i) \(t^{1/2}\) (ii) \(t\) (iii) \(t^{3/2}\) (iv) \(t^2\)

Solution

Displacement under Constant Power

Body moves unidirectionally with constant power \(P\). Find displacement \(s \propto t^n\).

Solution: $$\begin{aligned} P &= F \cdot v = \text{constant} \\ F &= m \frac{dv}{dt} \\ P &= m v \frac{dv}{dt} \\ P \, dt &= m v \, dv \\ \int\limits_0^t P \, dt &= m \int\limits_0^v v \, dv \\ P t &= \frac{1}{2} m v^2 \\ v^2 &\propto t \\ v &\propto t^{1/2} \\ v &= \frac{ds}{dt} \propto t^{1/2} \\\\ ds &\propto t^{1/2} \, dt \\ s &\propto \int t^{1/2} \, dt \propto t^{3/2} \end{aligned}$$

Correct answer: (iii) \(t^{3/2}\)

Q11. A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by \[\vec{F}=-\hat{i}+2\hat{j}+3\hat{k} \mathrm{N}\] where \(\hat{i},\ \hat{j},\ \hat{k}\) are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ?

Solution

Work Done Along Z-axis

Body moves 4 m along z-axis under force \(\vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k}\) N. Work = \(\vec{F} \cdot \vec{dr}\).

Solution: $$\begin{aligned} \vec{F} &= -\hat{i} + 2\hat{j} + 3\hat{k} \\ d\vec{r} &= 4\hat{k} \\ W &= \vec{F} \cdot d\vec{r} \\ &= (-\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (4\hat{k}) \\ &= -4(\hat{i}\cdot\hat{k}) + 8(\hat{j}\cdot\hat{k}) + 12(\hat{k}\cdot\hat{k}) \\ &= 0 + 0 + 12(1) \\ W &= 12\, \text{J} \end{aligned}$$

Only z-component \(F_z = 3\) N contributes to work since displacement is purely along z-axis.

Q12. An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = \(9.11×10^{-31}\) kg, proton mass = \(1.67×10^{–27}\) kg, 1 eV = \(1.60 ×10^{–19}\) J).

Solution

Electron vs Proton Speed Comparison

Given: \( m_e = 9.11 \times 10^{-31} \) kg, \( m_p = 1.67 \times 10^{-27} \) kg, \( 1 \) eV \( = 1.60 \times 10^{-19} \) J.

Electron speed: $$\begin{aligned} KE_e &= 10\, \text{keV} \\ &= 10^4 \times 1.60 \times 10^{-19} \\ &= 1.60 \times 10^{-15}\, \text{J} \\\\ \frac{1}{2} m_e v_e^2 &= 1.60 \times 10^{-15} \\ v_e^2 &= \frac{2 \times 1.60 \times 10^{-15}}{9.11 \times 10^{-31}} \\ &= 3.51 \times 10^{15} \\ v_e &= \sqrt{3.51 \times 10^{15}} \\ &= 1.87 \times 10^7\, \text{m/s} \end{aligned}$$

Proton speed: $$\begin{aligned} KE_p &= 100\, \text{keV} \\ &= 10^5 \times 1.60 \times 10^{-19} \\ &= 1.60 \times 10^{-14}\, \text{J} \\\\ \frac{1}{2} m_p v_p^2 &= 1.60 \times 10^{-14} \\ v_p^2 &= \frac{2 \times 1.60 \times 10^{-14}}{1.67 \times 10^{-27}} \\ &= 1.91 \times 10^{13} \\ v_p &= \sqrt{1.91 \times 10^{13}} \\ &= 1.38 \times 10^6\, \text{m/s} \end{aligned}$$

Ratio: $$\begin{aligned} \frac{v_e}{v_p} &= \frac{1.87 \times 10^7}{1.38 \times 10^6} \\ \frac{v_e}{v_p} &= 13.55 \end{aligned}$$

Electron is faster by factor of 13.55 due to much smaller mass despite lower kinetic energy.

Q13. A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 \(\mathrm{m\ s^{–1}}\) ?

Solution

Work Done on Raindrop Journey

Radius \( r = 2 \) mm = \( 2 \times 10^{-3} \) m,
height \( h = 500 \) m, terminal speed \( v = 10 \) m/s,
density of water \( \rho = 10^3 \) kg/m³.

Mass of raindrop: $$\begin{aligned} m &= \rho \cdot \frac{4}{3} \pi r^3 \\ &= 10^3 \cdot \frac{4}{3} \cdot \frac{22}{7} \cdot (2 \times 10^{-3})^3 \\ &= 10^3 \cdot \frac{4}{3} \cdot \frac{22}{7} \cdot 8 \times 10^{-9} \\ &= 3.35 \times 10^{-5}\, \text{kg} \end{aligned}$$

Work by gravity - first half (500 m to 250 m): $$\begin{aligned} W_{g1} &= mg \cdot \Delta h \\ &= 3.35 \times 10^{-5} \times 9.8 \times 250 \\ &= 0.082\, \text{J} \end{aligned}$$

Work by gravity - second half (250 m to 0 m): $$\begin{aligned} W_{g2} &= mg \cdot 250 \\ &= 3.35 \times 10^{-5} \times 9.8 \times 250 \\ &= 0.082\, \text{J} \end{aligned}$$

Total work by gravity: $$\begin{aligned} W_{g,total} &= mgh = 3.35 \times 10^{-5} \times 9.8 \times 500 \\ &= 0.164\, \text{J} \end{aligned}$$

Work by resistive force (entire journey): $$\begin{aligned} KE_{final} &= \frac{1}{2} m v^2 \\ &= \frac{1}{2} \times 3.35 \times 10^{-5} \times 10^2 \\ &= 0.001675\, \text{J} \\\\ W_R &= W_{g,total} - KE_{final} \\ &= 0.164 - 0.001675 \\ &= 0.162\, \text{J} \end{aligned}$$

Gravity does equal work (0.082 J) in both halves. Resistive force does -0.162 J (dissipates most gravitational work as heat).

Q14. A molecule in a gas container hits a horizontal wall with speed \(\mathrm{200\ m\ s^{–1}}\) and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?

Solution

Molecular Collision Analysis

Molecule hits horizontal wall at speed \(v\) with 30° angle to normal, rebounds with same speed. Analyze momentum conservation and collision type.

Momentum analysis:

Component parallel to wall (x-direction):

Component normal to wall (y-direction):

Kinetic energy analysis: $$\begin{aligned} KE_{initial} &= \frac{1}{2} m v^2 \\ KE_{final} &= \frac{1}{2} m v^2 \\ \Delta KE &= 0 \end{aligned}$$ Speed unchanged means kinetic energy conserved.

Momentum not conserved for molecule alone (external impulse from wall). Collision is elastic (KE conserved).

Q15. A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?

Solution

Pump Power Consumption Calculation

Tank volume = 30 m³, time = 15 min = 900 s, height = 40 m, pump efficiency = 30%.

Mass of water: $$\begin{aligned} m &= \rho V \\ &= 10^3 \times 30 \\ &= 3 \times 10^4\, \text{kg} \end{aligned}$$

Useful work (potential energy gain): $$\begin{aligned} W_{useful} &= mgh \\ &= 3 \times 10^4 \times 9.8 \times 40 \\ &= 1.176 \times 10^7\, \text{J} \end{aligned}$$

Total energy consumed by pump: $$\begin{aligned} \eta &= \frac{W_{useful}}{W_{consumed}} = 0.30 \\ W_{consumed} &= \frac{W_{useful}}{\eta} \\ &= \frac{1.176 \times 10^7}{0.30} \\ &= 3.92 \times 10^7\, \text{J} \end{aligned}$$

Power consumption: $$\begin{aligned} P &= \frac{W_{consumed}}{t} \\ &= \frac{3.92 \times 10^7}{900} \\ &= 4.36 \times 10^4\, \text{W} \\ &= 43.6\, \text{kW} \end{aligned}$$

Pump consumes 43.6 kW electrical power to deliver 13.1 kW useful hydraulic power.

Q16. Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision ?

Solution

Elastic Collision: Three Identical Balls

Incoming ball mass m, speed V hits two stationary identical balls at rest. Elastic collision conserves both momentum and kinetic energy.

Conservation laws:

Momentum:

Kinetic energy

Physical solution: Incoming ball stops (v₁ = 0), two target balls move forward with speed V/2 each (head-on symmetric collision). $$\begin{aligned} v_2 &= v_3 = \frac{V}{2} \\\\ V &= 0 + \frac{V}{2} + \frac{V}{2} \quad \checkmark \\\\ V^2 &= 0 + \left(\frac{V}{2}\right)^2 + \left(\frac{V}{2}\right)^2 \\\\ &= \frac{V^2}{2} + \frac{V^2}{2} \quad \checkmark \end{aligned}$$

Correct result: Incoming ball stops, two target balls move together with equal speed V/2 (symmetric forward motion).

Q17. The bob A of a pendulum released from \(30^\circ\) to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Solution

Pendulum Collision: Height After Elastic Impact

Bob A released from 30° swings down hitting stationary bob B (same mass). Elastic collision on table level. Find height bob A rises after collision.

Initial speed of A at bottom: $$\begin{aligned} h &= L(1 - \cos 30^\circ) \\ &= L\left(1 - \frac{\sqrt{3}}{2}\right) \\\\ v_A &= \sqrt{2gh} \\ &= \sqrt{2gL\left(1 - \cos 30^\circ\right)} \end{aligned}$$ Bob B initially at rest: \(v_B = 0\).

Elastic collision (equal masses, head-on): $$\begin{aligned} v_A' &= \frac{m_A v_A + m_B v_B}{m_A + m_B} - v_B \\ &= 0 \\\\ v_B' &= \frac{m_A v_A + m_B v_B}{m_A + m_B} + v_A \\ &= v_A \end{aligned}$$ After collision: A stops (\(v_A' = 0\)), B moves with speed \(v_B' = v_A\).

Height bob A rises: $$\begin{aligned} v_A' &= 0 \\ \text{KE after collision} &= 0 \\ h' &= 0 \end{aligned}$$

Bob A does not rise (remains at table level, h' = 0 m). All energy transfers to bob B.

Q18. The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ?

Solution

Pendulum Speed with Air Resistance

Pendulum length L = 1.5 m, released from horizontal position. Dissipates 5% initial energy to air resistance.

Initial potential energy: $$\begin{aligned} h &= L = 1.5\, \text{m} \\ PE_{initial} &= mgh \\ &= mg \times 1.5 \end{aligned}$$

Energy available at bottom: $$\begin{aligned} \text{Energy dissipated} &= 5\% \text{ of } PE_{initial} \\ KE_{bottom} &= 95\% \text{ of } PE_{initial} \\ \frac{1}{2}mv^2 &= 0.95 \times mgh \\ \frac{1}{2}v^2 &= 0.95 \times 9.8 \times 1.5 \end{aligned}$$

Speed calculation: $$\begin{aligned} v^2 &= 2 \times 0.95 \times 9.8 \times 1.5 \\ v^2 &= 1.9 \times 9.8 \times 1.5 \\ v^2 &= 27.93 \\ v &= \sqrt{27.93} \\ v &= 5.28\, \text{m/s} \end{aligned}$$

Bob arrives at lowermost point with speed 5.28 m/s (compared to 5.42 m/s without resistance).

Q19. A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty ?

Solution

Trolley Speed: Sand Leakage Analysis

Trolley mass 300 kg + sandbag 25 kg = 325 kg total, initial speed 27 km/h = 7.5 m/s, frictionless track. Sand leaks at 0.05 kg/s. No external horizontal forces act on system.

Conservation of momentum: $$\begin{aligned} \text{Initial momentum} &= (300 + 25) \times 7.5 \\ &= 325 \times 7.5 \\ &= 2437.5\, \text{kg m/s} \end{aligned}$$

Final condition (sandbag empty): $$\begin{aligned} \text{Final mass of trolley} &= 300\, \text{kg} \\ \text{Sand leaked} &= 25\, \text{kg} \\ \text{Time to empty} &= \frac{25}{0.05} = 500\, \text{s} \end{aligned}$$

Sand leaves with trolley's velocity: $$\begin{aligned} \text{No external force} &\implies \text{momentum conserved} \\ m_{\text{final}} v_{\text{final}} &= m_{\text{initial}} v_{\text{initial}} \\ 300 \times v_f &= 325 \times 7.5 \\ v_f &= \frac{325 \times 7.5}{300} \\ v_f &= \frac{2437.5}{300} \\ v_f &= 8.125\, \text{m/s} = 29.25\, \text{km/h} \end{aligned}$$

Trolley speed increases to 8.125 m/s (29.25 km/h) as lighter mass carries same momentum.

Q20. A body of mass 0.5 kg travels in a straight line with velocity \(\mathrm{v =a x^{3/2}}\) where \(\mathrm{a = 5\ m–^{1/2}\ s^{–1}}\). What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

Solution

Work Done: Variable Velocity Motion

Mass \( m = 0.5 \) kg,
velocity \( v = a x^{3/2} \), \( a = 5 \) m\(^{-1/2}\) s\(^{-1}\).
Find work from \( x = 0 \) to \( x = 2 \) m.

Work-energy theorem: \( W_{net} = \Delta KE \).

Initial and final speeds: $$\begin{aligned} v(x=0) &= 5 \cdot 0^{3/2} = 0\, \text{m/s} \\ v(x=2) &= 5 \cdot (2)^{3/2} \\ &= 5 \cdot (2\sqrt{2}) \\ &= 10\sqrt{2}\, \text{m/s} \end{aligned}$$

Change in kinetic energy: $$\begin{aligned} KE_i &= \frac{1}{2} m (0)^2 = 0 \\ KE_f &= \frac{1}{2} \cdot 0.5 \cdot (10\sqrt{2})^2 \\ &= 0.25 \cdot 100 \cdot 2 \\ &= 0.25 \cdot 200 \\ &= 50\, \text{J} \\ W_{net} &= \Delta KE = 50\, \text{J} \end{aligned}$$

Net force does 50 J work, increasing kinetic energy from 0 to 50 J.

Q21. The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ?
(b) What is the kinetic energy of the air ?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that \(\mathrm{A = 30\ m^2,\ v = 36\ km/h}\) and the density of air is \(\mathrm{1.2\ kg\ m^{–3}}\). What is the electrical power produced ?

Solution

Windmill Power Generation Analysis

Windmill blades sweep area A, wind speed v perpendicular to circle, density ρ = 1.2 kg/m³.

(a) Mass of air passing through in time t: $$\begin{aligned} \text{Volume swept} &= A \cdot v \cdot t \\ m &= \rho \cdot \text{Volume} \\ m &= \rho A v t \end{aligned}$$

(b) Kinetic energy of air: $$\begin{aligned} KE &= \frac{1}{2} m v^2 \\ &= \frac{1}{2} (\rho A v t) v^2 \\ KE &= \frac{1}{2} \rho A v^3 t \end{aligned}$$

(c) Electrical power (A = 30 m², v = 36 km/h = 10 m/s, 25% efficiency): $$\begin{aligned} P_{wind} &= \frac{1}{2} \rho A v^3 \\ &= \frac{1}{2} \times 1.2 \times 30 \times (10)^3 \\ &= 0.6 \times 30 \times 1000 \\ &= 18000\, \text{W} \\\\ P_{electrical} &= 0.25 \times 18000 \\ &= 4500\, \text{W} = 4.5\, \text{kW} \end{aligned}$$

Windmill produces 4.5 kW electrical power from 18 kW wind kinetic power.

Q22. A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.
(a) How much work does she do against the gravitational force ?
(b) Fat supplies \(\mathrm{3.8 × 10^{7J}}\) of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Solution

Dieting Work and Fat Loss Calculation

Mass m = 10 kg, height h = 0.5 m, 1000 lifts. Potential energy lost when lowering dissipated as heat.

(a) Work against gravity: $$\begin{aligned} W_{single\ lift} &= mgh \\ &= 10 \times 9.8 \times 0.5 \\ &= 49\, \text{J} \\\\ W_{total} &= 49 \times 1000 \\ &= 49000\, \text{J} \end{aligned}$$

(b) Fat consumption: $$\begin{aligned} \text{Energy per kg fat} &= 3.8 \times 10^7\, \text{J/kg} \\ \text{Efficiency} &= 20\% = 0.20 \\ \text{Useful energy per kg} &= 3.8 \times 10^7 \times 0.20 \\ &= 7.6 \times 10^6\, \text{J/kg} \\\\ \text{Mass of fat} &= \frac{49000}{7.6 \times 10^6} \\ &= 6.45 \times 10^{-3}\, \text{kg} = 6.45\, \text{g} \end{aligned}$$

Dieter does 49 kJ work, loses 6.45 g fat (considering 20% mechanical efficiency).

Q23. A family uses 8 kW of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a typical house.

Solution

Solar Panel Area Calculation

Family power need = 8 kW. Solar insolation = 200 W/m², conversion efficiency = 20%.

(a) Required solar panel area: $$\begin{aligned} P_{electrical} &= 8\, \text{kW} = 8000\, \text{W} \\\\ \eta &= 20\% = 0.20 \\\\ P_{solar\ required} &= \frac{P_{electrical}}{\eta} \\\\ &= \frac{8000}{0.20} \\\\ &= 40000\, \text{W} = 40\, \text{kW} \end{aligned}$$

Area calculation: $$\begin{aligned} \text{Solar flux} &= 200\, \text{W/m}^2 \\\\ A &= \frac{P_{solar\ required}}{\text{solar flux}} \\\\ &= \frac{40000}{200} \\\\ A &= 200\, \text{m}^2 \end{aligned}$$

(b) Comparison: Typical house roof area ≈ 150-250 m². Required 200 m² matches typical roof size, making solar feasible for 8 kW system.

Answer: 200 m² (comparable to typical house roof).