Unit 5: Work, Energy, and Power
Comprehensive theoretical derivations, vector line diagrams, mathematical formulations, conservative dynamics, elastic and inelastic collision mechanics, and solved practice numericals tailored for National Examinations Board (NEB) Nepal Grade XI Physics.
Work Done by a Constant Force and a Variable Force
1. Definition of Work in Physics
In common parlance, any physical or mental effort is termed "work". However, in physics, work is defined rigorously: Work is said to be done by a force when the point of application of the force undergoes a displacement along or opposite to the line of action of the force.
If a constant force $\vec{F}$ acts on a body and causes a displacement $\vec{d}$ making an angle $\theta$ with the direction of the force, the work done $W$ is defined as the scalar (dot) product of the force vector and displacement vector:
Here, $F = |\vec{F}|$ is the magnitude of force, $d = |\vec{d}|$ is the magnitude of displacement, and $\theta$ is the angle between $\vec{F}$ and $\vec{d}$. Alternatively, work is the product of the component of force in the direction of displacement ($(F\cos\theta)$) and the magnitude of displacement ($d$).
Dimensions & Units of Work
- Dimensional Formula: $[\text{Work}] = [\text{Force}] \times [\text{Displacement}] = [M^1 L^1 T^{-2}] \times [L^1] = \mathbf{[M^1 L^2 T^{-2}]}$.
- SI Unit: The SI unit of work is the Joule ($\text{J}$). One Joule is defined as the work done when a force of $1\text{ Newton}$ displaces an object through $1\text{ meter}$ in the direction of the force ($1\text{ J} = 1\text{ N} \cdot 1\text{ m} = 1\text{ kg}\cdot\text{m}^2\text{s}^{-2}$).
- CGS Unit: The CGS unit of work is the Erg ($1\text{ erg} = 1\text{ dyne} \cdot 1\text{ cm}$).
- Conversion Relation: $$1\text{ Joule} = 1\text{ N} \times 1\text{ m} = (10^5\text{ dynes}) \times (10^2\text{ cm}) = \mathbf{10^7\text{ ergs}}$$
Sign Convention & Physical Nature of Work
Positive Work ($\theta < 90^\circ$)
Done when force has a component in direction of motion ($\cos\theta > 0$). Energy is transferred to the body.
Example: Work done by gravity on a freely falling body ($\theta = 0^\circ$).
Negative Work ($90^\circ < \theta \le 180^\circ$)
Done when force opposes motion ($\cos\theta < 0$). Energy is removed from the body.
Example: Work done by friction on a sliding block ($\theta = 180^\circ$).
Zero Work ($\theta = 90^\circ$ or $d=0$)
No displacement or force acts perpendicular to motion ($\cos 90^\circ = 0$).
Example: Work done by centripetal force in circular motion ($\theta = 90^\circ$).
2. Work Done by a Variable Force
In practical scenario, forces are rarely constant. When a variable force $F(x)$ acts on a particle moving along the x-axis, its magnitude changes with position $x$. We cannot use $W = F d \cos\theta$ directly.
To formulate the work done, we divide the total displacement from initial position $x_1$ to final position $x_2$ into a very large number of infinitesimal displacements $dx$. During a tiny displacement $dx$, the force $F(x)$ can be treated as virtually constant.
The small elementary work done $dW$ for displacement $dx$ is given by: $$dW = F(x) \, dx$$
Summing up the elementary work over the entire interval from $x_1$ to $x_2$ by taking the definite integral as $dx \to 0$:
3. Work Done in Stretching/Compressing an Ideal Spring
Consider a block of mass $m$ attached to an ideal spring of spring constant (stiffness constant) $k$ resting on a frictionless horizontal plane.
According to Hooke's Law, when the spring is extended or compressed by a displacement $x$ from its natural unstretched length ($x=0$), it exerts a restoring force $F_{\text{restoring}}$ given by: $$F_{\text{restoring}} = -k x$$
To stretch the spring slowly without acceleration, an external applied force $F_{\text{ext}}$ equal and opposite to restoring force must be applied: $$F_{\text{ext}} = +k x$$
The infinitesimal work done $dW$ by external force in stretching the spring through further distance $dx$ is: $$dW = F_{\text{ext}} \, dx = (k x) \, dx$$
Therefore, the total work done $W$ in stretching the spring from its relaxed length ($x = 0$) to a final elongation $x$ is:
Power and Machine Efficiency
1. Definition and Mathematical Formulation
Power is defined as the time rate at which work is done or the rate at which energy is transferred/transformed by a force. While work measures the total effect, power measures how fast or slowly that work is executed.
Average Power ($P_{\text{avg}}$)
Total work done divided by total time taken:
$$P_{\text{avg}} = \frac{\Delta W}{\Delta t}$$
Instantaneous Power ($P$)
Power at a specific instant of time $t$:
$$P = \lim_{\Delta t \to 0} \frac{\Delta W}{\Delta t} = \frac{dW}{dt}$$
2. Relation Between Power, Force, and Velocity
Suppose a force $\vec{F}$ produces an infinitesimal displacement $d\vec{r}$ of a particle in time $dt$. The elementary work done is $dW = \vec{F} \cdot d\vec{r}$.
Dividing both sides by $dt$: $$P = \frac{dW}{dt} = \frac{\vec{F} \cdot d\vec{r}}{dt} = \vec{F} \cdot \left(\frac{d\vec{r}}{dt}\right)$$
Since $\frac{d\vec{r}}{dt} = \vec{v}$ (instantaneous velocity vector):
Thus, instantaneous power equals the scalar dot product of force vector $\vec{F}$ and velocity vector $\vec{v}$.
3. Units of Power
- SI Unit: The SI unit of power is the Watt ($\text{W}$), named in honor of James Watt. One Watt is defined as 1 Joule of work done per second ($1\text{ W} = 1\text{ J/s} = 1\text{ kg}\cdot\text{m}^2\cdot\text{s}^{-3}$).
- Larger Practical Units:
- $1\text{ Kilowatt (kW)} = 10^3\text{ W} = 1000\text{ W}$
- $1\text{ Megawatt (MW)} = 10^6\text{ W}$
- $1\text{ Horsepower (HP)} = \mathbf{746\text{ Watts}}$ (Imperial unit widely used for motor ratings).
- Commercial Energy Unit ($\text{kWh}$): Kilowatt-hour is a practical unit of electrical energy (not power). $1\text{ kWh} = (1000\text{ W}) \times (3600\text{ s}) = \mathbf{3.6 \times 10^6\text{ Joules}}$.
4. Efficiency of Machines ($\eta$)
In practical mechanical systems, due to frictional and thermal dissipation, output power is always less than input power. The efficiency ($\eta$) is defined as the ratio of useful output power to total input power expressed as a percentage:
NEB Board Application Tip: Water Pump Efficiency
If a water motor pump of efficiency $\eta$ lifts mass $m$ of water to height $h$ in time $t$:
$$P_{\text{out}} = \frac{mgh}{t} \implies P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{mgh}{\eta \cdot t}$$
Work-Energy Theorem; Kinetic and Potential Energy
1. Kinetic Energy ($E_k$) and its Calculus Derivation
Kinetic energy is the energy possessed by a body by virtue of its motion.
Derivation: Consider a body of mass $m$ initially at rest ($u=0$) on a smooth horizontal surface. A variable force $F$ acts on it, displacing it through distance $x$ and imparting a velocity $v$.
According to Newton's second law, $F = m a = m \frac{dv}{dt}$. Using the chain rule for acceleration: $$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$$
The elementary work $dW$ done by force $F$ for displacement $dx$ is: $$dW = F \, dx = \left(m v \frac{dv}{dx}\right) dx = m v \, dv$$
Integrating both sides from initial velocity $0$ to final velocity $v$:
This total work done in setting the body into motion is stored as its Kinetic Energy: $E_k = \frac{1}{2} m v^2$.
2. Rigorous Proof of Work-Energy Theorem (WET)
Statement: The net work done by all resultant forces acting on a body is equal to the change in its kinetic energy. $$W_{\text{net}} = \Delta E_k = E_{k,f} - E_{k,i} = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$$
A. Proof for Constant Force:
From third equation of motion under constant acceleration $a$: $$v^2 - u^2 = 2 a s \implies a s = \frac{v^2 - u^2}{2}$$
Multiplying both sides by mass $m$: $$F \cdot s = (m a) s = m (a s) = m \left( \frac{v^2 - u^2}{2} \right) = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$$ $$\mathbf{W = \Delta E_k} \quad \text{(Proved)}$$
B. Proof for Variable Force:
Work done by variable force $F(x)$ from $x_1$ to $x_2$: $$W = \int_{x_1}^{x_2} F(x) \, dx = \int_{x_1}^{x_2} m \left( \frac{dv}{dt} \right) dx = \int_{x_1}^{x_2} m \left( v \frac{dv}{dx} \right) dx = \int_{u}^{v} m v \, dv$$ $$W = m \left[ \frac{v^2}{2} \right]_{u}^{v} = \mathbf{\frac{1}{2} m v^2 - \frac{1}{2} m u^2 = \Delta E_k} \quad \text{(Proved)}$$
3. Potential Energy ($U$) & Reference Levels
Potential Energy is defined as the energy stored in a system by virtue of its position, configuration, or state of strain in a conservative field.
- Gravitational Potential Energy ($U = mgh$): Work done against gravitational force in lifting mass $m$ slowly to height $h$: $W = F \cdot h = (mg) h = \mathbf{mgh}$.
- Elastic Potential Energy ($U = \frac{1}{2}kx^2$): Work done in compressing or stretching a spring by distance $x$.
- Zero Reference Level: Potential energy is relative. By convention, ground surface is taken as $U=0$ for gravity, and natural length $x=0$ is taken as $U=0$ for springs.
4. Relation Between Kinetic Energy ($E_k$) and Linear Momentum ($p$)
Linear momentum is $p = m v \implies v = \frac{p}{m}$. Substituting into kinetic energy formula: $$E_k = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{p}{m}\right)^2 = \mathbf{\frac{p^2}{2m}}$$ Rearranging for linear momentum $p$: $$\mathbf{p = \sqrt{2m E_k}}$$
Case I: Same Momentum ($p_1 = p_2$)
$E_k = \frac{p^2}{2m} \implies E_k \propto \frac{1}{m}$.
The lighter mass body possesses GREATER kinetic energy!
Case II: Same Kinetic Energy ($E_{k1} = E_{k2}$)
$p = \sqrt{2m E_k} \implies p \propto \sqrt{m}$.
The heavier mass body possesses GREATER linear momentum!
Conservation of Energy
1. Principle of Conservation of Mechanical Energy
Law of Conservation of Total Energy: Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant.
Conservation of Mechanical Energy: In an isolated system where only conservative forces do work, the total mechanical energy $E$ (sum of kinetic energy $K$ and potential energy $U$) remains constant at all points: $$E = K + U = \text{Constant} \implies K_A + U_A = K_B + U_B$$
2. Rigorous Mathematical Verification for a Freely Falling Body
Consider a body of mass $m$ initially held at rest at point A at height $h$ above the ground. Let it fall freely under gravity. We calculate total energy at three points: A (top), B (intermediate after falling distance $x$), and C (just before striking ground).
Point A (Top, height $y = h$):
At highest position, initial velocity $u = 0$.
$$K_A = \frac{1}{2}m(0)^2 = 0, \quad U_A = mgh \implies \mathbf{E_A = K_A + U_A = mgh} \quad \text{--- (1)}$$
Point B (Intermediate, after falling distance $x$, height $y = h - x$):
Velocity $v_B$ at point B: $v_B^2 = u^2 + 2gx = 0 + 2gx = 2gx$.
$$K_B = \frac{1}{2}m v_B^2 = \frac{1}{2}m(2gx) = mgx$$ $$U_B = mg(h - x)$$ $$\mathbf{E_B = K_B + U_B = mgx + mg(h - x) = mgh} \quad \text{--- (2)}$$
Point C (Bottom, just before striking ground $y = 0$):
Velocity $v_C$ at point C: $v_C^2 = u^2 + 2gh = 2gh$.
$$K_C = \frac{1}{2}m v_C^2 = \frac{1}{2}m(2gh) = mgh, \quad U_C = 0$$ $$\mathbf{E_C = K_C + U_C = mgh + 0 = mgh} \quad \text{--- (3)}$$
Since $E_A = E_B = E_C = mgh$, total mechanical energy is conserved throughout the motion under gravity.
Conservative and Non-Conservative Forces
1. Comparison & Distinguishing Properties
Forces in nature are fundamentally divided into two categories based on path dependency of work done:
| Property | Conservative Force | Non-Conservative Force |
|---|---|---|
| Path Dependency | Work done depends ONLY on initial and final positions, independent of path. | Work done depends on the actual path taken between two points. |
| Closed Loop Work | Work done over a closed round-trip path is ZERO ($\oint \vec{F} \cdot d\vec{r} = 0$). | Work done over a closed loop is NON-ZERO ($\oint \vec{F} \cdot d\vec{r} \neq 0$). |
| Energy Conservation | Total mechanical energy ($K + U$) is completely conserved. | Mechanical energy is dissipated into heat, sound, or deformation. |
| Potential Energy Relation | Potential energy $U$ can be uniquely defined ($F = -\frac{dU}{dx}$). | No potential energy function exists for non-conservative forces. |
| Physical Examples | Gravitational force, Electrostatic force, Elastic spring force. | Friction force, Viscous drag force, Air resistance. |
2. Proof: Gravitational Force is a Conservative Force
To prove gravitational force is conservative, consider raising a mass $m$ to height $h$ along three different paths:
- Vertical Path 1 (Direct Lift): $$W_1 = F \cdot h = mg \cdot h = mgh$$
- Inclined Plane Path 2 (Angle $\theta$, length $L = h / \sin\theta$): Component of force needed against gravity $= mg \sin\theta$. $$W_2 = (mg \sin\theta) \cdot L = (mg \sin\theta) \cdot \left(\frac{h}{\sin\theta}\right) = mgh$$
- Staircase / Stepwise Path 3: Dividing path into horizontal steps $dx_i$ and vertical steps $dy_i$. Work along horizontal steps $= \sum mg \cdot dx_i \cos 90^\circ = 0$. Work along vertical steps $= \sum mg \cdot dy_i = mg \sum dy_i = mgh$.
Since $W_1 = W_2 = W_3 = mgh$, work done against gravity is strictly path-independent. Hence, Gravitational Force is Conservative!
Elastic and Inelastic Collisions
1. Definition and Classification of Collisions
A collision is an intense physical interaction between two or more bodies occurring in a relatively short time interval, resulting in mutual exchange of momentum and energy.
- Elastic Collision: A collision in which both total linear momentum and total kinetic energy are conserved. No mechanical energy is lost as heat, sound, or permanent deformation.
- Inelastic Collision: A collision in which total linear momentum is conserved, but kinetic energy is NOT conserved.
- Perfectly Inelastic Collision: An extreme inelastic collision where the colliding bodies stick together after impact and move with a single common velocity. Maximum possible kinetic energy is lost.
2. One-Dimensional (Head-On) Elastic Collision
Consider two bodies $A$ and $B$ of masses $m_1$ and $m_2$ moving along a straight line with initial velocities $u_1$ and $u_2$ ($u_1 > u_2$). Let $v_1$ and $v_2$ be their final velocities after collision.
Step 1: Apply Conservation of Linear Momentum
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ Rearranging terms of $m_1$ and $m_2$: $$m_1 (u_1 - v_1) = m_2 (v_2 - u_2) \quad \text{--- (Equation 1)}$$
Step 2: Apply Conservation of Kinetic Energy
$$\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$$ $$m_1 (u_1^2 - v_1^2) = m_2 (v_2^2 - u_2^2)$$ $$m_1 (u_1 - v_1)(u_1 + v_1) = m_2 (v_2 - u_2)(v_2 + u_2) \quad \text{--- (Equation 2)}$$
Step 3: Derive Velocity Relation
Dividing Equation (2) by Equation (1): $$\frac{m_1 (u_1 - v_1)(u_1 + v_1)}{m_1 (u_1 - v_1)} = \frac{m_2 (v_2 - u_2)(v_2 + u_2)}{m_2 (v_2 - u_2)}$$ $$u_1 + v_1 = v_2 + u_2$$ $$\mathbf{u_1 - u_2 = v_2 - v_1} \quad \text{--- (Equation 3)}$$
Step 4: Express Final Velocity Formulas
From Equation (3), $v_2 = u_1 - u_2 + v_1$. Substituting $v_2$ into Equation (1):
$$m_1 (u_1 - v_1) = m_2 (u_1 - u_2 + v_1 - u_2) = m_2 (u_1 - 2u_2 + v_1)$$
$$m_1 u_1 - m_1 v_1 = m_2 u_1 - 2 m_2 u_2 + m_2 v_1$$
$$(m_1 + m_2) v_1 = (m_1 - m_2) u_1 + 2 m_2 u_2$$
Similarly, substituting $v_1 = v_2 - u_1 + u_2$ into Equation (1) yields:
3. Analysis of 4 Critical Special Cases
Case 1: Equal Masses ($m_1 = m_2 = m$)
$v_1 = 0 \cdot u_1 + \frac{2m}{2m} u_2 = u_2$
$v_2 = \frac{2m}{2m} u_1 + 0 \cdot u_2 = u_1$
Result: Bodies swap/exchange their velocities completely!
Case 2: Equal Masses ($m_1 = m_2$), Target at Rest ($u_2 = 0$)
$v_1 = 0$, $\quad v_2 = u_1$
Result: Incident body stops completely; target moves off with speed $u_1$.
Case 3: Heavy Target at Rest ($m_2 \gg m_1, u_2 = 0$)
$v_1 \approx \left(\frac{-m_2}{m_2}\right) u_1 = -u_1$
$v_2 \approx 0$
Result: Light body rebounds with same speed; heavy target remains stationary.
Case 4: Heavy Incident Body ($m_1 \gg m_2, u_2 = 0$)
$v_1 \approx u_1$
$v_2 \approx 2 u_1$
Result: Heavy incident body continues unaffected; light target kicked forward at $2u_1$.
4. Perfectly Inelastic Collision & Loss of Kinetic Energy
In a 1D perfectly inelastic collision, two bodies stick together after impact and move with common velocity $v$.
By Conservation of Linear Momentum: $$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \implies \mathbf{v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}}$$
Mathematical Proof of Kinetic Energy Loss ($\Delta E_k$):
Initial KE: $E_{k,i} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2$
Final KE: $E_{k,f} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (m_1 + m_2) \left[ \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \right]^2 = \frac{(m_1 u_1 + m_2 u_2)^2}{2(m_1 + m_2)}$
Loss of Kinetic Energy $\Delta E_k = E_{k,i} - E_{k,f}$:
Since $m_1, m_2 > 0$ and $(u_1 - u_2)^2 > 0$, $\Delta E_k$ is strictly positive, proving kinetic energy is always lost in an inelastic collision.
5. Coefficient of Restitution ($e$)
The coefficient of restitution ($e$) is defined as the ratio of relative velocity of separation to relative velocity of approach:
- For Perfectly Elastic Collision: $\mathbf{e = 1}$.
- For Inelastic Collision: $\mathbf{0 < e < 1}$.
- For Perfectly Inelastic Collision: $\mathbf{e = 0}$.
Unit 5 Key Formula Summary Sheet
| Physical Quantity / Theorem | Formula | SI Units |
|---|---|---|
| Work Done (Constant Force) | $W = \vec{F} \cdot \vec{d} = F d \cos\theta$ | Joule ($\text{J}$) |
| Work Done (Variable Force) | $W = \int_{x_1}^{x_2} F(x) \, dx$ | Joule ($\text{J}$) |
| Spring Elastic Potential Work | $W = \frac{1}{2} k x^2$ | Joule ($\text{J}$) |
| Instantaneous Power | $P = \frac{dW}{dt} = \vec{F} \cdot \vec{v} = F v \cos\theta$ | Watt ($\text{W}$) |
| Work-Energy Theorem | $W_{\text{net}} = \Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ | Joule ($\text{J}$) |
| Momentum & KE Relation | $p = \sqrt{2m E_k} \quad \text{and} \quad E_k = \frac{p^2}{2m}$ | $\text{kg}\cdot\text{m/s}$, $\text{J}$ |
| 1D Elastic Final Velocities |
$v_1 = \left(\frac{m_1-m_2}{m_1+m_2}\right)u_1 + \left(\frac{2m_2}{m_1+m_2}\right)u_2$ $v_2 = \left(\frac{2m_1}{m_1+m_2}\right)u_1 + \left(\frac{m_2-m_1}{m_1+m_2}\right)u_2$ |
$\text{m/s}$ |
| Inelastic Collision KE Loss | $\Delta E_k = \frac{m_1 m_2}{2(m_1 + m_2)} (u_1 - u_2)^2$ | Joule ($\text{J}$) |
| Coefficient of Restitution | $e = \frac{v_2 - v_1}{u_1 - u_2}$ | Dimensionless |
10 Essential NEB Conceptual Questions & Answers
Answer:
No, kinetic energy can never be negative. Kinetic energy is given by $E_k = \frac{1}{2} m v^2$. Since mass $m$ is a scalar positive quantity ($m > 0$) and velocity squared $v^2$ is always non-negative ($v^2 \ge 0$), kinetic energy must be $\ge 0$. It can be zero (at rest) or positive, but never negative.
Answer:
The lighter body has greater kinetic energy. Kinetic energy in terms of momentum is $E_k = \frac{p^2}{2m}$. Since momentum $p$ is constant for both bodies, $E_k \propto \frac{1}{m}$. Therefore, the body with smaller mass $m$ will possess a larger kinetic energy.
Answer:
Centripetal force acts towards the center of the circular path, whereas instantaneous displacement is directed tangentially along the circle. Thus, the angle between centripetal force vector $\vec{F}_c$ and displacement vector $d\vec{r}$ is $\theta = 90^\circ$. Work done $W = F d \cos(90^\circ) = 0$.
Answer:
No. Relation between momentum and kinetic energy is $p = \sqrt{2m E_k}$. If $E_k = 0$, velocity $v=0$, so $p=0$. Conversely if $p=0$, then $v=0$, so $E_k = 0$. Hence, a body cannot have momentum without kinetic energy or vice versa.
Answer:
The "lost" kinetic energy is converted into non-mechanical forms of energy, such as thermal energy (heat generation during impact), acoustic energy (sound produced during clashing), and strain energy stored as permanent mechanical deformation of the colliding bodies.
Answer:
Usually friction opposes relative motion ($\theta = 180^\circ$), doing negative work. However, static friction can do positive work! For example, when a car accelerates, static friction exerted by the road on the tires propels the car forward in the direction of motion ($\theta = 0^\circ$), doing positive work on the car.
Answer:
Yes. Absolute potential energy depends on the chosen zero reference height $y=0$. However, the change in potential energy ($\Delta U = U_2 - U_1$) is independent of reference level choosing.
Answer:
Work done in stretching a spring is $W = \frac{1}{2} k x^2$. For equal extension $x$, $W \propto k$. Since $k_A > k_B$, Spring A requires more work.
Answer:
The collision between ball and floor is inelastic ($e < 1$). Part of the kinetic energy is lost as heat and sound during deformation. With reduced kinetic energy, the rebound velocity $v < u$, so it reaches a lower maximum height ($h' < h$).
Answer:
No, total mechanical energy ($K+U$) is not conserved because non-conservative deformation forces dissipate kinetic energy into heat/sound. However, the total energy of the universe remains conserved.
8 Fully Solved Standard NEB Numericals
Numerical 1: Work done in stretching a spring
A spring with force constant $k = 300\text{ N/m}$ is stretched from initial elongation $x_1 = 2\text{ cm}$ to final elongation $x_2 = 8\text{ cm}$. Calculate the work done on the spring.
Given: $k = 300\text{ N/m}$, $x_1 = 0.02\text{ m}$, $x_2 = 0.08\text{ m}$.
Formula: $W = \frac{1}{2} k x_2^2 - \frac{1}{2} k x_1^2 = \frac{1}{2} k (x_2^2 - x_1^2)$
Calculation: $W = \frac{1}{2} \times 300 \times [(0.08)^2 - (0.02)^2]$
$W = 150 \times [0.0064 - 0.0004] = 150 \times 0.0060 = \mathbf{0.9\text{ Joules}}$
Numerical 2: Water pump power & motor efficiency
An electric pump lifts $1200\text{ kg}$ of water to a overhead tank at height $h = 20\text{ m}$ in $1\text{ minute}$. If efficiency of pump motor is $80\%$, find input electric power required in kW. ($g = 9.8\text{ m/s}^2$).
Given: $m = 1200\text{ kg}$, $h = 20\text{ m}$, $t = 60\text{ s}$, $\eta = 0.80$.
Useful Power Output: $P_{\text{out}} = \frac{mgh}{t} = \frac{1200 \times 9.8 \times 20}{60} = 3920\text{ W}$
Input Electrical Power: $P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{3920}{0.80} = 4900\text{ W} = \mathbf{4.9\text{ kW}}$
Numerical 3: Work-Energy theorem stopping distance
A car of mass $1000\text{ kg}$ travelling at $72\text{ km/h}$ is brought to rest over a distance of $40\text{ m}$ by applying brakes. Calculate average braking force using Work-Energy Theorem.
Given: $m = 1000\text{ kg}$, $u = 72 \times \frac{5}{18} = 20\text{ m/s}$, $v = 0$, $s = 40\text{ m}$.
Change in KE: $\Delta E_k = 0 - \frac{1}{2} m u^2 = -\frac{1}{2} (1000) (20)^2 = -200,000\text{ J}$
Work-Energy Theorem: $W = -F_{\text{brake}} \cdot s = \Delta E_k$
$-F_{\text{brake}} \times 40 = -200,000 \implies F_{\text{brake}} = \mathbf{5000\text{ N}}$
Numerical 4: 1D Elastic Collision Velocities
A sphere of mass $4\text{ kg}$ moving at $6\text{ m/s}$ collides head-on elastically with a stationary sphere of mass $2\text{ kg}$. Find final velocities $v_1$ and $v_2$.
Given: $m_1 = 4\text{ kg}$, $u_1 = 6\text{ m/s}$, $m_2 = 2\text{ kg}$, $u_2 = 0$.
$v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) u_1 = \left(\frac{4 - 2}{4 + 2}\right) \times 6 = \frac{2}{6} \times 6 = \mathbf{2\text{ m/s}}$
$v_2 = \left(\frac{2 m_1}{m_1 + m_2}\right) u_1 = \left(\frac{2 \times 4}{4 + 2}\right) \times 6 = \frac{8}{6} \times 6 = \mathbf{8\text{ m/s}}$
Numerical 5: Inelastic ballistic pendulum bullet loss of KE
A bullet of mass $20\text{ g}$ moving at $300\text{ m/s}$ embeds into a wooden block of mass $980\text{ g}$ at rest. Calculate common final velocity and loss of KE.
$m_1 = 0.02\text{ kg}, u_1 = 300\text{ m/s}, m_2 = 0.98\text{ kg}, u_2 = 0$.
Common velocity $v = \frac{m_1 u_1}{m_1 + m_2} = \frac{0.02 \times 300}{1.0} = \mathbf{6\text{ m/s}}$.
$\Delta E_k = \frac{m_1 m_2}{2(m_1 + m_2)} u_1^2 = \frac{0.02 \times 0.98}{2(1.0)} (300)^2 = \mathbf{882\text{ Joules}}$.
Numerical 6: Variable force work integral
A force $F(x) = (3x^2 + 2x + 1)\text{ N}$ acts on a particle from $x = 1\text{ m}$ to $x = 3\text{ m}$. Find total work done.
$W = \int_{1}^{3} (3x^2 + 2x + 1) \, dx = \left[ x^3 + x^2 + x \right]_{1}^{3}$
$W = (3^3 + 3^2 + 3) - (1^3 + 1^2 + 1) = (27 + 9 + 3) - (3) = 39 - 3 = \mathbf{36\text{ Joules}}$.
Numerical 7: Simple Pendulum energy speed at lowest point
A pendulum bob of mass $0.1\text{ kg}$ is suspended by string of length $L = 1\text{ m}$. It is released from an angle of $60^\circ$ with vertical. Calculate speed at mean position ($g=9.8\text{ m/s}^2$).
Height raised $h = L(1 - \cos 60^\circ) = 1(1 - 0.5) = 0.5\text{ m}$.
Conservation of Energy: $\frac{1}{2} m v^2 = mgh \implies v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.5} = \sqrt{9.8} \approx \mathbf{3.13\text{ m/s}}$.
Numerical 8: Freely falling body energy verification
A body of mass $2\text{ kg}$ is dropped from height $20\text{ m}$. Verify total mechanical energy at midpoint ($10\text{ m}$). ($g=9.8\text{ m/s}^2$).
At top ($20\text{ m}$): $E = mgh = 2 \times 9.8 \times 20 = 392\text{ J}$.
At midpoint ($y=10\text{ m}$): $v^2 = 2g(10) = 196$. $K = \frac{1}{2}(2)(196) = 196\text{ J}$. $U = 2 \times 9.8 \times 10 = 196\text{ J}$.
Total $E = K + U = 196 + 196 = \mathbf{392\text{ J}}$ (Verified equal!).