Work, Energy and Power

Introduction and the Scalar Product

  • Everyday vs. Physical Definitions:

    • In common usage, activities such as ploughing a field, carrying bricks, studying for an exam, or painting a landscape are described as "work".

    • In physics, work has a precise definition requiring both an applied force and a displacement along or against that force.

    • Stamina or the physical capacity to labor for long hours corresponds to the definition of energy.

    • Power in daily language conveys strength or speed (e.g., a powerful punch in boxing); in physics, it specifically measures the time rate of doing work.

  • Definition of the Scalar (Dot) Product:

    • For two vectors A\mathbf{A} and B\mathbf{B} with an angle θ\theta (0≤θ≤π0 \le \theta \le \pi) between them, the scalar product is defined as:         A⋅B=ABcos⁡(θ)\mathbf{A} \cdot \mathbf{B} = A B \cos(\theta)

    • Since AA, BB, and cos⁡(θ)\cos(\theta) are scalar quantities, the dot product has magnitude but no direction.

  • Geometric Interpretation:

    • Bcos⁡(θ)B \cos(\theta) represents the geometric projection of B\mathbf{B} onto A\mathbf{A}.

    • Acos⁡(θ)A \cos(\theta) represents the geometric projection of A\mathbf{A} onto B\mathbf{B}.

    • The scalar product equals the magnitude of A\mathbf{A} multiplied by the component of B\mathbf{B} along A\mathbf{A}, or vice versa:         A⋅B=A(Bcos⁡(θ))=B(Acos⁡(θ))\mathbf{A} \cdot \mathbf{B} = A (B \cos(\theta)) = B (A \cos(\theta))

    

Scalar product of two vectors and geometric projections

  • Algebraic Properties of the Scalar Product:

    • Commutative Law:         A⋅B=B⋅A\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}

    • Distributive Law:         A⋅(B+C)=A⋅B+A⋅C\mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = \mathbf{A} \cdot \mathbf{B} + \mathbf{A} \cdot \mathbf{C}

    • Scalar Multiplication:         A⋅(λB)=λ(A⋅B)\mathbf{A} \cdot (\lambda \mathbf{B}) = \lambda (\mathbf{A} \cdot \mathbf{B})

    • Orthonormal Unit Vectors:         i^⋅i^=j^⋅j^=k^⋅k^=1\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{k}} = 1         i^⋅j^=j^⋅k^=k^⋅i^=0\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{k}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{i}} = 0

    • Cartesian Component Form:         For A=Axi^+Ayj^+Azk^\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}} and B=Bxi^+Byj^+Bzk^\mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}} + B_z \hat{\mathbf{k}}:         A⋅B=AxBx+AyBy+AzBz\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z

    • Self Dot Product:         A⋅A=Ax2+Ay2+Az2=A2\mathbf{A} \cdot \mathbf{A} = A_x^2 + A_y^2 + A_z^2 = A^2

    • Perpendicularity Condition:         A⋅B=0if and only if A⊥B(θ=90∘)\mathbf{A} \cdot \mathbf{B} = 0 \quad \text{if and only if } \mathbf{A} \perp \mathbf{B} \quad (\theta = 90^\circ)

  • Worked Example 5.1 (Angle and Projection):

    • Given force F=3i^+4j^−5k^ units\mathbf{F} = 3\hat{\mathbf{i}} + 4\hat{\mathbf{j}} - 5\hat{\mathbf{k}}\,\text{units} and displacement d=5i^+4j^+3k^ units\mathbf{d} = 5\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 3\hat{\mathbf{k}}\,\text{units}:         F⋅d=(3)(5)+(4)(4)+(−5)(3)=15+16−15=16 units\mathbf{F} \cdot \mathbf{d} = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16\,\text{units}         F2=F⋅F=32+42+(−5)2=50⇒F=50 unitsF^2 = \mathbf{F} \cdot \mathbf{F} = 3^2 + 4^2 + (-5)^2 = 50 \Rightarrow F = \sqrt{50}\,\text{units}         d2=d⋅d=52+42+32=50⇒d=50 unitsd^2 = \mathbf{d} \cdot \mathbf{d} = 5^2 + 4^2 + 3^2 = 50 \Rightarrow d = \sqrt{50}\,\text{units}         cos⁡(θ)=F⋅dFd=165050=1650=0.32⇒θ=cos⁡−1(0.32)\cos(\theta) = \frac{\mathbf{F} \cdot \mathbf{d}}{F d} = \frac{16}{\sqrt{50} \sqrt{50}} = \frac{16}{50} = 0.32 \Rightarrow \theta = \cos^{-1}(0.32)

    • The projection of F\mathbf{F} onto d\mathbf{d} is:         Fcos⁡(θ)=1650 unitsF \cos(\theta) = \frac{16}{\sqrt{50}}\,\text{units}

Work and Kinetic Energy: The Work-Energy Theorem

  • Derivation in One Dimension (Constant Acceleration):

    • The standard kinematic relation for rectilinear motion with constant acceleration aa is:         v2−u2=2asv^2 - u^2 = 2 a s

    • Multiplying both sides by 12m\frac{1}{2} m yields:         12mv2−12mu2=mas\frac{1}{2} m v^2 - \frac{1}{2} m u^2 = m a s

    • Applying Newton's Second Law (F=maF = m a):         12mv2−12mu2=Fs\frac{1}{2} m v^2 - \frac{1}{2} m u^2 = F s

  • Vector Generalization in Three Dimensions:

    • For displacement vector d\mathbf{d} and constant acceleration vector a\mathbf{a}:         v2−u2=2a⋅dv^2 - u^2 = 2 \mathbf{a} \cdot \mathbf{d}

    • Multiplying by 12m\frac{1}{2} m gives:         12mv2−12mu2=ma⋅d=F⋅d\frac{1}{2} m v^2 - \frac{1}{2} m u^2 = m \mathbf{a} \cdot \mathbf{d} = \mathbf{F} \cdot \mathbf{d}

  • Definitions:

    • Kinetic Energy (KK): Half the mass times the square of the speed:         K=12mv2K = \frac{1}{2} m v^2

    • Work (WW): The dot product of the applied force and displacement:         W=F⋅dW = \mathbf{F} \cdot \mathbf{d}

  • Statement of the Work-Energy (WE) Theorem:

    • Kf−Ki=WK_f - K_i = W

    • Theorem Statement: The change in kinetic energy of a particle is equal to the net work done on it by the resultant force.

  • Worked Example 5.2 (Falling Raindrop):

    • A raindrop of mass m=1.00 g=10−3 kgm = 1.00\,\text{g} = 10^{-3}\,\text{kg} falls from rest (u=0 m s−1u = 0\,\text{m\,s}^{-1}) from height h=1.00 km=103 mh = 1.00\,\text{km} = 10^3\,\text{m} and hits the ground with speed v=50.0 m s−1v = 50.0\,\text{m\,s}^{-1}.

    • (a) Work done by gravity (WgW_g):         ΔK=12mv2−0=12(10−3 kg)(50.0 m s−1)2=1.25 J\Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} (10^{-3}\,\text{kg}) (50.0\,\text{m\,s}^{-1})^2 = 1.25\,\text{J}         Taking g=10 m s−2g = 10\,\text{m\,s}^{-2}:         Wg=mgh=(10−3 kg)(10 m s−2)(103 m)=10.0 JW_g = m g h = (10^{-3}\,\text{kg}) (10\,\text{m\,s}^{-2}) (10^3\,\text{m}) = 10.0\,\text{J}

    • (b) Work done by resistive air force (WrW_r):         Using the Work-Energy Theorem (ΔK=Wg+Wr\Delta K = W_g + W_r):         Wr=ΔK−Wg=1.25 J−10.0 J=−8.75 JW_r = \Delta K - W_g = 1.25\,\text{J} - 10.0\,\text{J} = -8.75\,\text{J}

Work: Definition, Conditions, and Units

  • Mathematical Definition:

    • Work done by a force F\mathbf{F} over displacement d\mathbf{d} is:         W=(Fcos⁡(θ))d=F⋅dW = (F \cos(\theta)) d = \mathbf{F} \cdot \mathbf{d}

    

An object undergoing displacement under a force

  • Conditions for Zero Work (W=0W = 0):

    1. Zero Displacement (d=0d = 0): Pushing against a fixed wall does no work on the wall despite muscular fatigue. A weightlifter holding a 150 kg150\,\text{kg} barbell stationary overhead for 30 s30\,\text{s} does zero work on the barbell.

    2. Zero Force (F=0F = 0): A block moving at constant velocity on a smooth frictionless surface has zero net horizontal force, so zero work is done on it.

    3. Perpendicular Force and Displacement (θ=90∘\theta = 90^\circ):

      • Gravity does zero work on a body moving horizontally.

      • Earth's gravity does zero work on the Moon assuming a perfectly circular orbit (force is radially inward, displacement is tangential).

  • Sign of Work:

    • Positive Work: 0∘≤θ<90∘0^\circ \le \theta < 90^\circ (e.g., force aligned with motion).

    • Negative Work: 90∘<θ≤180∘90^\circ < \theta \le 180^\circ (e.g., retarding friction acting opposite to displacement, θ=180∘\theta = 180^\circ).

  • Dimensions and Units:

    • Dimensions: [ML2T−2][M L^2 T^{-2}]

    • SI Unit: Joule (J\text{J}), named after James Prescott Joule.

    • Unit Conversion Factors:

      • 1 erg=10−7 J1\,\text{erg} = 10^{-7}\,\text{J}

      • 1 electron volt (eV)=1.60×10−19 J1\,\text{electron volt (eV)} = 1.60 \times 10^{-19}\,\text{J}

      • 1 calorie (cal)=4.186 J1\,\text{calorie (cal)} = 4.186\,\text{J}

      • 1 kilowatt hour (kWh)=3.6×106 J1\,\text{kilowatt hour (kWh)} = 3.6 \times 10^6\,\text{J}

  • Worked Example 5.3 (Skidding Cyclist):

    • A cyclist comes to a skidding stop in d=10 md = 10\,\text{m} under a road frictional force F=200 NF = 200\,\text{N} opposing motion (θ=180∘\theta = 180^\circ).

    • (a) Work done by the road on the cycle:         Wr=Fdcos⁡(180∘)=(200 N)(10 m)(−1)=−2000 JW_r = F d \cos(180^\circ) = (200\,\text{N}) (10\,\text{m}) (-1) = -2000\,\text{J}

    • (b) Work done by the cycle on the road:         By Newton's Third Law, the cycle exerts an equal 200 N200\,\text{N} force on the road. However, the road undergoes no displacement (d=0d = 0). Thus, work done on the road is 0 J0\,\text{J}.

    • Principle: Work done by body B on body A is not necessarily equal and opposite to work done by body A on body B.

Kinetic Energy

  • Definition and Scalar Nature:

    • The kinetic energy KK of a mass mm with velocity v\mathbf{v} is:         K=12mv2=12m(v⋅v)K = \frac{1}{2} m v^2 = \frac{1}{2} m (\mathbf{v} \cdot \mathbf{v})

    • It is a non-negative scalar quantity measuring the capacity of an object to perform work by virtue of its motion.

  • Typical Kinetic Energies:

    • Running athlete (70 kg70\,\text{kg}, 10 m s−110\,\text{m\,s}^{-1}): 3.5×103 J3.5 \times 10^3\,\text{J}

    • Bullet (50 g50\,\text{g}, 200 m s−1200\,\text{m\,s}^{-1}): 103 J10^3\,\text{J}

    • Stone dropped from 10 m10\,\text{m} (1 kg1\,\text{kg}, 14 m s−114\,\text{m\,s}^{-1}): 102 J10^2\,\text{J}

    • Raindrop at terminal speed (3.5×10−5 kg3.5 \times 10^{-5}\,\text{kg}, 9 m s−19\,\text{m\,s}^{-1}): 1.4×10−3 J1.4 \times 10^{-3}\,\text{J}

    • Air molecule (≈10−26 kg\approx 10^{-26}\,\text{kg}, 500 m s−1500\,\text{m\,s}^{-1}): ≈10−21 J\approx 10^{-21}\,\text{J}

  • Worked Example 5.4 (Bullet penetrating plywood):

    • Bullet mass m=50.0 g=0.05 kgm = 50.0\,\text{g} = 0.05\,\text{kg}, initial speed vi=200 m s−1v_i = 200\,\text{m\,s}^{-1}, plywood thickness 2.00 cm2.00\,\text{cm}. Bullet emerges with 10%10\% of its initial kinetic energy.

    • Initial kinetic energy:         Ki=12mvi2=12(0.05 kg)(200 m s−1)2=1000 JK_i = \frac{1}{2} m v_i^2 = \frac{1}{2} (0.05\,\text{kg}) (200\,\text{m\,s}^{-1})^2 = 1000\,\text{J}

    • Final kinetic energy:         Kf=0.10×1000 J=100 JK_f = 0.10 \times 1000\,\text{J} = 100\,\text{J}

    • Emergent speed vfv_f:         12mvf2=100 J⇒vf=2×100 J0.05 kg=4000≈63.2 m s−1\frac{1}{2} m v_f^2 = 100\,\text{J} \Rightarrow v_f = \sqrt{\frac{2 \times 100\,\text{J}}{0.05\,\text{kg}}} = \sqrt{4000} \approx 63.2\,\text{m\,s}^{-1}

    • The bullet's speed is reduced by approximately 68%68\% (not 90%90\%).

Work Done by a Variable Force

  • Mathematical Integration:

    • When force F(x)F(x) varies along displacement xx, divide the displacement into infinitesimal intervals Δx\Delta x.

    • Work for a small interval is:         ΔW=F(x)Δx\Delta W = F(x) \Delta x

    • Summing over all intervals from initial position xix_i to final position xfx_f:         W≈∑xixfF(x)ΔxW \approx \sum_{x_i}^{x_f} F(x) \Delta x

    • Taking the limit Δx→0\Delta x \to 0 transforms the sum into a definite integral:         W=lim⁡Δx→0∑xixfF(x)Δx=∫xixfF(x) dxW = \lim_{\Delta x \to 0} \sum_{x_i}^{x_f} F(x) \Delta x = \int_{x_i}^{x_f} F(x)\,dx

    

Shaded area representing work done by a varying force

  • Geometrical Relation: Work done by a variable force equals the area under the force-displacement curve between xix_i and xfx_f

  • Worked Example 5.5 (Woman pushing a trunk):

    • Applied force F=100 NF = 100\,\text{N} for x=0x = 0 to 10 m10\,\text{m}, then decreases linearly to 50 N50\,\text{N} at x=20 mx = 20\,\text{m}. Constant frictional force f=−50 Nf = -50\,\text{N}.

    

Plot of applied force F and opposing frictional force f versus displacement
*   Work done by the woman (WFW_F):

        WF=Area(ABCD)+Area(CEID)W_F = \text{Area}(ABCD) + \text{Area}(CEID)         WF=(100 N×10 m)+12(100 N+50 N)×(20 m−10 m)W_F = (100\,\text{N} \times 10\,\text{m}) + \frac{1}{2} (100\,\text{N} + 50\,\text{N}) \times (20\,\text{m} - 10\,\text{m})         WF=1000 J+750 J=1750 JW_F = 1000\,\text{J} + 750\,\text{J} = 1750\,\text{J} * Work done by friction (WfW_f):         Wf=Area(AGHI)=(−50 N)×20 m=−1000 JW_f = \text{Area}(AGHI) = (-50\,\text{N}) \times 20\,\text{m} = -1000\,\text{J}

The Work-Energy Theorem for a Variable Force

  • Proof in One Dimension:

    • Rate of change of kinetic energy with time:         dKdt=ddt(12mv2)=mvdvdt=mav\frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) = m v \frac{dv}{dt} = m a v

    • By Newton's Second Law (F=maF = m a) and velocity definition (v=dxdtv = \frac{dx}{dt}):         dKdt=Fv=Fdxdt\frac{dK}{dt} = F v = F \frac{dx}{dt}

    • Canceling dtdt yields dK=F dxdK = F\,dx.

    • Integrating from initial state (xi,Ki)(x_i, K_i) to final state (xf,Kf)(x_f, K_f):         ∫KiKfdK=∫xixfF(x) dx\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F(x)\,dx         Kf−Ki=∫xixfF(x) dx=WK_f - K_i = \int_{x_i}^{x_f} F(x)\,dx = W

  • Dynamical Significance:

    • Newton's Second Law is an instantaneous differential vector equation.

    • The Work-Energy Theorem is an integrated scalar form over time and space intervals; explicit time dependence and direction information are integrated out.

  • Worked Example 5.6 (Block crossing a rough patch):

    • Mass m=1 kgm = 1\,\text{kg}, initial speed vi=2 m s−1v_i = 2\,\text{m\,s}^{-1}. Retarding force Fr=−kxF_r = -\frac{k}{x} over 0.10 m<x<2.01 m0.10\,\text{m} < x < 2.01\,\text{m} where k=0.5 Jk = 0.5\,\text{J}.

    • Ki=12mvi2=12(1 kg)(2 m s−1)2=2 JK_i = \frac{1}{2} m v_i^2 = \frac{1}{2} (1\,\text{kg}) (2\,\text{m\,s}^{-1})^2 = 2\,\text{J}

    • Kf−Ki=∫0.12.01(−kx)dx=−kln⁡(x)∣0.12.01=−kln⁡(2.010.1)K_f - K_i = \int_{0.1}^{2.01} \left( -\frac{k}{x} \right) dx = -k \ln(x) \Big|_{0.1}^{2.01} = -k \ln \left( \frac{2.01}{0.1} \right)

    • Kf=2−0.5ln⁡(20.1)≈2−0.5(3.0)=2−1.5=0.5 JK_f = 2 - 0.5 \ln(20.1) \approx 2 - 0.5 (3.0) = 2 - 1.5 = 0.5\,\text{J}

    • Final speed vfv_f:         vf=2Kfm=2×0.51=1 m s−1v_f = \sqrt{\frac{2 K_f}{m}} = \sqrt{\frac{2 \times 0.5}{1}} = 1\,\text{m\,s}^{-1}

Concept and Properties of Potential Energy

  • Definition:

    • Potential energy V(x)V(x) is stored energy possessed by a body due to its position or configuration.

    • Examples include a stretched bowstring, fault lines in the Earth's crust, or a raised object in a gravitational field.

  • Gravitational Potential Energy:

    • For heights h≪REh \ll R_E (where Earth's radius RER_E variations in gg are negligible), taking upward as positive:         V(h)=mghV(h) = m g h

    • The conservative force is related to potential energy by the negative derivative:         F(x)=−dV(x)dxF(x) = -\frac{dV(x)}{dx}

    • In integral form:         V(xf)−V(xi)=−∫xixfF(x) dxV(x_f) - V(x_i) = -\int_{x_i}^{x_f} F(x)\,dx         ΔV=−F(x)Δx=−W\Delta V = -F(x) \Delta x = -W

  • Conservative vs. Non-Conservative Forces:

    • Conservative Force:

      1. Derivable from a scalar potential energy function: F(x)=−dV(x)dxF(x) = -\frac{dV(x)}{dx}.

      2. Work done depends solely on initial and final positions {xi,xf}\{x_i, x_f\}, independent of path taken.

      3. Work done over any arbitrary closed path is zero:             ∮F⋅dr=0\oint \mathbf{F} \cdot d\mathbf{r} = 0

      • Examples: Gravitational force, electrostatic force, spring force.

    • Non-Conservative Force:

      • Work done depends on path taken and speed.

      • Examples: Friction, air resistance, viscosity.

    • Dimensions and Units: Dimensions [ML2T−2][M L^2 T^{-2}], Unit: Joule (J\text{J}).

Conservation of Mechanical Energy

  • Mechanical Energy Definition: Total mechanical energy EE is the sum of kinetic energy KK and potential energy V(x)V(x).     E=K+V(x)E = K + V(x)

  • Mathematical Derivation:

    • From Work-Energy theorem: ΔK=F(x)Δx\Delta K = F(x) \Delta x

    • For a conservative force: ΔV=−F(x)Δx\Delta V = -F(x) \Delta x

    • Adding equations: ΔK+ΔV=0⇒Δ(K+V)=0\Delta K + \Delta V = 0 \Rightarrow \Delta (K + V) = 0

    • Ki+V(xi)=Kf+V(xf)=E=constantK_i + V(x_i) = K_f + V(x_f) = E = \text{constant}

  • Statement of Conservation Principle:     The total mechanical energy of a system is conserved if the forces doing work on it are conservative.

  • Case Study: Object Dropped from Height HH:

    

Ball of mass m dropped from height H
*   At top position HH (v=0v = 0):

        EH=mgHE_H = m g H * At intermediate height hh:         Eh=mgh+12mvh2E_h = m g h + \frac{1}{2} m v_h^2         Equating Eh=EH⇒mgH=mgh+12mvh2⇒vh=2g(H−h)E_h = E_H \Rightarrow m g H = m g h + \frac{1}{2} m v_h^2 \Rightarrow v_h = \sqrt{2 g (H - h)} * At ground level (h=0h = 0):         E0=12mvf2E_0 = \frac{1}{2} m v_f^2         Equating E0=EH⇒mgH=12mvf2⇒vf=2gHE_0 = E_H \Rightarrow m g H = \frac{1}{2} m v_f^2 \Rightarrow v_f = \sqrt{2 g H}

  • Worked Example 5.7 (Vertical Circular Motion of a Pendulum Bob):

    

Bob attached to a light string undergoing vertical circular motion
*   Bob of mass mm on string length LL. Velocity v0v_0 imparted at lowest point A. String slackens at highest point C (TC=0T_C = 0).
*   *(i) Expression for initial speed v0v_0:*
    *   Reference potential energy VA=0V_A = 0 at lowest point A. Energy at A:

            EA=12mv02E_A = \frac{1}{2} m v_0^2 * At highest point C (height 2L2L), potential energy VC=2mgLV_C = 2 m g L. Newton's second law at C:             TC+mg=mvC2LT_C + m g = \frac{m v_C^2}{L} * Setting TC=0⇒mg=mvC2L⇒vC=gLT_C = 0 \Rightarrow m g = \frac{m v_C^2}{L} \Rightarrow v_C = \sqrt{g L} * Total mechanical energy at C:             EC=2mgL+12mvC2=2mgL+12m(gL)=52mgLE_C = 2 m g L + \frac{1}{2} m v_C^2 = 2 m g L + \frac{1}{2} m (g L) = \frac{5}{2} m g L * Equating EA=ECE_A = E_C:             12mv02=52mgL⇒v0=5gL\frac{1}{2} m v_0^2 = \frac{5}{2} m g L \Rightarrow v_0 = \sqrt{5 g L} * (ii) Speeds at points B and C: * Speed at C: vC=gLv_C = \sqrt{g L} * At mid-height point B (height LL): VB=mgLV_B = m g L. Energy equation:             EB=mgL+12mvB2=52mgL⇒12mvB2=32mgL⇒vB=3gLE_B = m g L + \frac{1}{2} m v_B^2 = \frac{5}{2} m g L \Rightarrow \frac{1}{2} m v_B^2 = \frac{3}{2} m g L \Rightarrow v_B = \sqrt{3 g L} * (iii) Ratio of kinetic energies at B and C: * KB=12mvB2=32mgLK_B = \frac{1}{2} m v_B^2 = \frac{3}{2} m g L * KC=12mvC2=12mgLK_C = \frac{1}{2} m v_C^2 = \frac{1}{2} m g L * KBKC=32mgL12mgL=3\frac{K_B}{K_C} = \frac{\frac{3}{2} m g L}{\frac{1}{2} m g L} = 3 * Trajectory after C: If string is cut at C, bob executes parabolic projectile motion starting with horizontal velocity vCv_C to the left. If string remains intact, bob continues on circular path.

Potential Energy of a Spring

  • Hooke's Law:

    • An ideal massless spring exerts a restoring force proportional to displacement xx from equilibrium:         Fs=−kxF_s = -k x

    • kk is the spring constant (stiffness) in N m−1\text{N\,m}^{-1}.

    

Illustration of spring force for extended and compressed spring

  • Work Done by Spring Force:

    • For extension xmx_m from equilibrium (x=0x = 0):         Ws=∫0xm(−kx) dx=−12kxm2W_s = \int_0^{x_m} (-k x)\,dx = -\frac{1}{2} k x_m^2

    • Work done by external pulling force:         Wext=+12kxm2W_{ext} = +\frac{1}{2} k x_m^2

    • For displacement from xix_i to xfx_f:         Ws=∫xixf(−kx) dx=12kxi2−12kxf2W_s = \int_{x_i}^{x_f} (-k x)\,dx = \frac{1}{2} k x_i^2 - \frac{1}{2} k x_f^2

    • For a closed cyclic process (xf=xix_f = x_i):         Wcyclic=0W_{cyclic} = 0

  • Spring Potential Energy Function:

    • Setting V(0)=0V(0) = 0 at equilibrium:         V(x)=12kx2V(x) = \frac{1}{2} k x^2

    • Note that −dVdx=−ddx(12kx2)=−kx=Fs-\frac{dV}{dx} = -\frac{d}{dx} \left( \frac{1}{2} k x^2 \right) = -k x = F_s

  • Energy Conservation and Maximum Speed:

    • For a block released from initial extension xmx_m:         12mv2+12kx2=12kxm2=constant\frac{1}{2} m v^2 + \frac{1}{2} k x^2 = \frac{1}{2} k x_m^2 = \text{constant}

    • At equilibrium (x=0x = 0), speed reaches its maximum vmv_m:         12mvm2=12kxm2⇒vm=xmkm\frac{1}{2} m v_m^2 = \frac{1}{2} k x_m^2 \Rightarrow v_m = x_m \sqrt{\frac{k}{m}}

    

Parabolic potential energy V and kinetic energy K plots

  • Worked Example 5.8 (Frictionless Car-Spring Collision):

    • Car mass m=1000 kgm = 1000\,\text{kg}, speed v=18.0 km/h=5.0 m s−1v = 18.0\,\text{km/h} = 5.0\,\text{m\,s}^{-1}. Spring constant k=5.25×103 N m−1k = 5.25 \times 10^3\,\text{N\,m}^{-1}.

    • Initial kinetic energy:         K=12mv2=12(1000 kg)(5.0 m s−1)2=1.25×104 JK = \frac{1}{2} m v^2 = \frac{1}{2} (1000\,\text{kg}) (5.0\,\text{m\,s}^{-1})^2 = 1.25 \times 10^4\,\text{J}

    • At maximum compression xmx_m, kinetic energy converts completely into potential energy:         12kxm2=1.25×104 J⇒12(5.25×103)xm2=1.25×104\frac{1}{2} k x_m^2 = 1.25 \times 10^4\,\text{J} \Rightarrow \frac{1}{2} (5.25 \times 10^3) x_m^2 = 1.25 \times 10^4         xm=2×1.25×1045.25×103=4.76=2.00 mx_m = \sqrt{\frac{2 \times 1.25 \times 10^4}{5.25 \times 10^3}} = \sqrt{4.76} = 2.00\,\text{m}

  • Worked Example 5.9 (Car-Spring Collision with Friction):

    • Coefficient of kinetic friction μ=0.5\mu = 0.5.

    

Forces acting on car during spring collision with friction
*   Frictional force magnitude: f=μmg=(0.5)(1000 kg)(10 m s−2)=5×103 Nf = \mu m g = (0.5) (1000\,\text{kg}) (10\,\text{m\,s}^{-2}) = 5 \times 10^3\,\text{N}
*   Work done by net retarding force over displacement xmx_m:

        Wnet=−12kxm2−μmgxmW_{net} = -\frac{1}{2} k x_m^2 - \mu m g x_m * By Work-Energy theorem (ΔK=Wnet\Delta K = W_{net}):         0−12mv2=−12kxm2−μmgxm⇒12kxm2+μmgxm−12mv2=00 - \frac{1}{2} m v^2 = -\frac{1}{2} k x_m^2 - \mu m g x_m \Rightarrow \frac{1}{2} k x_m^2 + \mu m g x_m - \frac{1}{2} m v^2 = 0         12(5.25×103)xm2+(5×103)xm−1.25×104=0\frac{1}{2} (5.25 \times 10^3) x_m^2 + (5 \times 10^3) x_m - 1.25 \times 10^4 = 0 * Solving quadratic equation for positive root xmx_m:         xm=−μmg+μ2m2g2+kmv2k=1.35 mx_m = \frac{-\mu m g + \sqrt{\mu^2 m^2 g^2 + k m v^2}}{k} = 1.35\,\text{m}

  • General Non-Conservative System Relation:     Ef−Ei=WncE_f - E_i = W_{nc}     where WncW_{nc} is the net work done by non-conservative forces over the trajectory.

Power

  • Definitions:

    • Power is the time rate at which work is done or energy is transferred.

    • Average Power (PavP_{av}):         Pav=WtP_{av} = \frac{W}{t}

    • Instantaneous Power (PP):         P=dWdt=F⋅drdt=F⋅vP = \frac{dW}{dt} = \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = \mathbf{F} \cdot \mathbf{v}         where v\mathbf{v} is instantaneous velocity.

  • Units and Dimensions:

    • Scalar quantity with dimensions [ML2T−3][M L^2 T^{-3}].

    • SI Unit: Watt (W\text{W}), equivalent to 1 J s−11\,\text{J\,s}^{-1}.

    • Horsepower unit: 1 hp=746 W1\,\text{hp} = 746\,\text{W}.

    • Commercial Energy Unit: Kilowatt-hour (kWh\text{kWh}):         1 kWh=103 W×3600 s=3.6×106 J1\,\text{kWh} = 10^3\,\text{W} \times 3600\,\text{s} = 3.6 \times 10^6\,\text{J}

  • Worked Example 5.10 (Elevator Motor Power):

    • Elevator + passengers mass m=1800 kgm = 1800\,\text{kg}, upward constant speed v=2.0 m s−1v = 2.0\,\text{m\,s}^{-1}. Retarding frictional force Ff=4000 NF_f = 4000\,\text{N}.

    • Total downward force:         F=mg+Ff=(1800×10)+4000=22000 NF = m g + F_f = (1800 \times 10) + 4000 = 22000\,\text{N}

    • Minimum power delivered by motor:         P=Fv=(22000 N)(2.0 m s−1)=44000 W=44 kWP = F v = (22000\,\text{N}) (2.0\,\text{m\,s}^{-1}) = 44000\,\text{W} = 44\,\text{kW}

    • In horsepower:         P=44000746≈59 hpP = \frac{44000}{746} \approx 59\,\text{hp}

Collisions in One and Two Dimensions

  • Momentum Conservation Principle:

    • During collision over time Δt\Delta t, mutual impulsive forces act: F12=−F21\mathbf{F}_{12} = -\mathbf{F}_{21} (Newton's Third Law).

    • Δp1+Δp2=(F12+F21)Δt=0⇒\Delta \mathbf{p}_1 + \Delta \mathbf{p}_2 = (\mathbf{F}_{12} + \mathbf{F}_{21}) \Delta t = 0 \Rightarrow Total linear momentum is conserved in all collisions.

  • Classification of Collisions:

    1. Elastic Collision: Total kinetic energy is conserved before and after collision (Ki=KfK_i = K_f). Deformation is completely restored.

    2. Inelastic Collision: Kinetic energy is not conserved (Kf<KiK_f < K_i); energy transforms into heat, sound, or mechanical deformation.

    3. Completely Inelastic Collision: Colliding bodies stick together post-collision and move with a common final velocity vfv_f.

  • One-Dimensional Completely Inelastic Collision:

    • Mass m1m_1 moving at v1iv_{1i} collides with stationary mass m2m_2 (v2i=0v_{2i} = 0).

    • Momentum conservation:         m1v1i=(m1+m2)vf⇒vf=(m1m1+m2)v1im_1 v_{1i} = (m_1 + m_2) v_f \Rightarrow v_f = \left( \frac{m_1}{m_1 + m_2} \right) v_{1i}

    • Kinetic energy loss ΔK\Delta K:         ΔK=12m1v1i2−12(m1+m2)vf2=12(m1m2m1+m2)v1i2>0\Delta K = \frac{1}{2} m_1 v_{1i}^2 - \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) v_{1i}^2 > 0

  • One-Dimensional Elastic Collision:

    • Conservation equations:         m1v1i=m1v1f+m2v2fm_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}         12m1v1i2=12m1v1f2+12m2v2f2\frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2

    • Relative velocity relation:         v1i+v1f=v2f⇒v2f−v1f=v1iv_{1i} + v_{1f} = v_{2f} \Rightarrow v_{2f} - v_{1f} = v_{1i}

    • Final velocity expressions:         v1f=(m1−m2m1+m2)v1iv_{1f} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) v_{1i}         v2f=(2m1m1+m2)v1iv_{2f} = \left( \frac{2 m_1}{m_1 + m_2} \right) v_{1i}

    • Special Cases:

      • Equal Masses (m1=m2m_1 = m_2): v1f=0v_{1f} = 0, v2f=v1iv_{2f} = v_{1i} (particles exchange velocities completely).

      • Heavy Target (m2≫m1m_2 \gg m_1): v1f≈−v1iv_{1f} \approx -v_{1i}, v2f≈0v_{2f} \approx 0 (light mass rebounds with same speed, heavy target remains at rest).

  • Worked Example 5.11 (Slowing Down Neutrons in Moderators):

    • Fast neutron (m1m_1) collides elastically with stationary moderator nucleus (m2m_2).

    • Fractional kinetic energy lost by neutron (f1f_1):         f1=K1i−K1fK1i=1−(m1−m2m1+m2)2=4m1m2(m1+m2)2f_1 = \frac{K_{1i} - K_{1f}}{K_{1i}} = 1 - \left( \frac{m_1 - m_2}{m_1 + m_2} \right)^2 = \frac{4 m_1 m_2}{(m_1 + m_2)^2}

    • For Deuterium (m2=2m1m_2 = 2 m_1):         f1=4(1)(2)(1+2)2=89≈88.9%f_1 = \frac{4 (1) (2)}{(1 + 2)^2} = \frac{8}{9} \approx 88.9\%

    • For Carbon (m2=12m1m_2 = 12 m_1):         f1=4(1)(12)(1+12)2=48169≈28.4%f_1 = \frac{4 (1) (12)}{(1 + 12)^2} = \frac{48}{169} \approx 28.4\% transferred (71.6%71.6\% retained).

  • Two-Dimensional Collisions:

    • Mass m1m_1 moving along x-axis with v1iv_{1i} hits stationary mass m2m_2.

    • Post-collision speeds v1fv_{1f} and v2fv_{2f} at angles θ1\theta_1 and θ2\theta_2 relative to initial x-axis.

    • Component Momentum Equations:

      • x-axis: m1v1i=m1v1fcos⁡(θ1)+m2v2fcos⁡(θ2)m_1 v_{1i} = m_1 v_{1f} \cos(\theta_1) + m_2 v_{2f} \cos(\theta_2)

      • y-axis: 0=m1v1fsin⁡(θ1)−m2v2fsin⁡(θ2)0 = m_1 v_{1f} \sin(\theta_1) - m_2 v_{2f} \sin(\theta_2)

    • Elastic Energy Equation:         12m1v1i2=12m1v1f2+12m2v2f2\frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2

  • Worked Example 5.12 (Billiard Balls Collision):

    • Equal masses m1=m2m_1 = m_2. Target ball deflected at θ2=37∘\theta_2 = 37^\circ. Elastic collision.

    • Vector momentum conservation: v1i=v1f+v2f\mathbf{v}_{1i} = \mathbf{v}_{1f} + \mathbf{v}_{2f}.

    • Squaring yields: v1i2=v1f2+v2f2+2v1fv2fcos⁡(θ1+37∘)v_{1i}^2 = v_{1f}^2 + v_{2f}^2 + 2 v_{1f} v_{2f} \cos(\theta_1 + 37^\circ).

    • Energy conservation for equal masses: v1i2=v1f2+v2f2v_{1i}^2 = v_{1f}^2 + v_{2f}^2.

    • Equating gives 2v1fv2fcos⁡(θ1+37∘)=0⇒cos⁡(θ1+37∘)=0⇒θ1+37∘=90∘⇒θ1=53∘2 v_{1f} v_{2f} \cos(\theta_1 + 37^\circ) = 0 \Rightarrow \cos(\theta_1 + 37^\circ) = 0 \Rightarrow \theta_1 + 37^\circ = 90^\circ \Rightarrow \theta_1 = 53^\circ

    • General Theorem: When two equal masses undergo a glancing elastic collision with one initially at rest, they move perpendicular to each other after collision (θ1+θ2=90∘\theta_1 + \theta_2 = 90^\circ).