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2.7 Physics for Engineers: Kinematics, Newton's Laws, Work-Energy, and Momentum
PRC Civil Engineer Licensure Examination CELE
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1 Chapter 1: Examination Overview, RA 544 & Professional Practice
1.1 CELE Structure, Format, Scoring, and 2026 Subject Order 1.2 Republic Act No. 544: The Civil Engineering Law and Board Regulations 1.3 Civil Engineering Ethics, Contracts, and Professional Practice
2 Chapter 2: Engineering Mathematics, Numerical Methods & Physics
2.1 Advanced Algebra, Progressions & Word Problems 2.2 Trigonometry and Analytical Geometry 2.3 Differential and Integral Calculus 2.4 Differential Equations and Numerical Methods 2.5 Series, Hyperbolic Functions, Multivariable Calculus, and Laplace Transforms 2.6 Numerical Methods: Linear Systems, Interpolation, Curve Fitting, and ODE Solvers 2.7 Physics for Engineers: Kinematics, Newton's Laws, Work-Energy, and Momentum
3 Chapter 3: Engineering Data Analysis & Engineering Economy
3.1 Engineering Data Analysis and Statistical Applications 3.2 Parameter Estimation, Linear Regression, Correlation, and Queuing Models 3.3 Engineering Economy: Time Value of Money and Interest Formulas 3.4 Project Evaluation, Depreciation, and Benefit-Cost Analysis
4 Chapter 4: Surveying Fundamentals, Construction Layout & Route Surveying
4.1 Errors, Corrections, and Pacing/Taping 4.2 Leveling, Earthworks, and Mass Haul Diagrams 4.3 Traverse Computations, Balancing, and Area Determinations 4.4 Route Surveying: Horizontal and Vertical Curves
5 Chapter 5: Highway & Transportation Engineering
5.1 Highway Geometric Design and Sight Distances 5.2 Traffic Engineering, Flow Fundamentals, and Pavement Design 5.3 Highway Policy, Route Location, Drainage, and Earthwork Slopes 5.4 Transportation Planning, Network Analysis, and Intermodal Systems
6 Chapter 6: Construction Materials, Methods, Management & Safety
6.1 Materials for Construction: Concrete, Formwork, Masonry, Steel, and Wood Products 6.2 Construction Equipment, Formwork, and Construction Methods 6.3 Construction Occupational Safety and Health: RA 11058 and DOLE DO 13 6.4 Construction Project Management, CPM/PERT, and Resource Scheduling 6.5 Construction Principles: Design Requirements, Project Organization, Cost Control, and Value Engineering 6.6 Quantity Surveying, Unit Cost Analysis, and Project Cost Estimates 6.7 Construction Trades and Specialties: Plumbing, Plastics, Doors and Windows, Finishes, and Electrical
7 Chapter 7: Fluid Properties, Hydrostatics & Buoyancy
7.1 Fluid Properties, Units, and Pressure Measurement 7.2 Hydrostatic Pressure on Plane and Curved Surfaces 7.3 Buoyancy, Flotation, and Accelerated Fluid Masses
8 Chapter 8: Hydrodynamics, Pipe Flow, Open Channels & Water Supply
8.1 Fluid Dynamics: Continuity and Energy Equations 8.2 Flow Measurement: Orifices, Weirs, and Venturi Meters 8.3 Fluid Flow in Pipes, Friction Losses, and Pipe Networks 8.4 Open Channel Flow, Manning's Equation, and Hydraulic Jump 8.5 Water Supply: Demand, Storage, Pumping, Distribution, and Treatment
9 Chapter 9: Soil Properties, Seepage & Stresses in Soil Mass
9.1 Soil Composition, Weight-Volume Phase Relationships 9.2 Soil Classification Systems (USCS & AASHTO) and Atterberg Limits 9.3 Soil Permeability, Seepage, Flow Nets, and Effective Stress 9.4 Stresses in Soil Mass, Stress Distribution, and Immediate and Secondary Settlement
10 Chapter 10: Consolidation, Soil Strength, Earth Pressures & Foundations
10.1 Soil Compaction and Soil Compressibility/Consolidation 10.2 Shear Strength of Soils and Mohr-Coulomb Failure Criteria 10.3 Lateral Earth Pressures (Rankine & Coulomb) and Retaining Walls 10.4 Shallow and Deep Foundation Bearing Capacity and Slope Stability
11 Chapter 11: Engineering Mechanics & Statics
11.1 Force Systems, Resultants, and Equilibrium of Rigid Bodies 11.2 Centroids, Moments of Inertia, and Friction 11.3 Cables, Trusses, and Internal Forces in Members
12 Chapter 12: Strength of Materials
12.1 Axial Stress, Strain, and Deformation 12.2 Torsion, Flanged Bolt Couplings, and Thin-Walled Pressure Vessels 12.3 Beam Stresses: Flexural and Transverse Shear 12.4 Beam Deflections, Combined Stresses, and Mohr's Circle
13 Chapter 13: Structural Analysis & NSCP Load Criteria
13.1 Indeterminate Structures: Approximate and Classical Methods 13.2 Moment Distribution Method and Slope Deflection 13.3 Influence Lines for Determinate and Indeterminate Structures 13.4 NSCP Minimum Design Loads: Dead, Live, Wind, and Earthquake
14 Chapter 14: Reinforced Concrete Design (NSCP 2015 / USD)
14.1 Ultimate Strength Design (USD) Principles and Singly Reinforced Beams 14.2 Doubly Reinforced Beams and T-Beams 14.3 Shear and Diagonal Tension, Stirrup Design, and Development Length 14.4 Reinforced Concrete Columns and Isolated Footing Design
15 Chapter 15: Steel, Timber & Prestressed Design and Materials Testing
15.1 Structural Steel Tension Members and Connections 15.2 Structural Steel Compression Members and Beam Design 15.3 Timber Design Principles, Working Stress, and Sawn Lumber Beams 15.4 Prestressed Concrete Fundamentals and Prestress Losses 15.5 Construction Materials Testing and Concrete Mix Proportioning
2.7 Physics for Engineers: Kinematics, Newton's Laws, Work-Energy, and Momentum
Key Takeaways Explain This Topic AI
For constant acceleration, v² = v₀² + 2a·s and s = v₀t + ½at²; velocity is ds/dt and acceleration is dv/dt.
On an incline with friction, a block sliding down accelerates at a = g(sin θ − μ cos θ).
The work done stretching a spring from x₁ to x₂ is ½k(x₂² − x₁²), the area under its force-deformation line.
Impulse equals change in momentum, and in any collision without external impulse the total momentum is conserved.
The coefficient of restitution e = (v₂′ − v₁′)/(v₁ − v₂) equals 1 for a perfectly elastic and 0 for a perfectly plastic impact.
Last updated: October 2026
2.7 Physics for Engineers: Kinematics, Newton's Laws, Work-Energy, and Momentum
"Physics for Engineers" is one of the 15 AMSTHC areas in the 2022 TOS. It has five one-item competencies:
Apply Newton's laws of motion to real-world problems.
Find the work done by variable forces, including Hooke's law.
Use conservation of energy.
Assess problems in linear and angular momentum.
Develop calculus-based solutions in statics and kinematics.
These dynamics skills also underpin hydraulics (momentum of jets), structures (impact and vibration) and transportation (braking and stopping distance).
Kinematics of Particles
Calculus definitions.
v = d s d t , a = d v d t = d 2 s d t 2 = v d v d s v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2 s}{dt^2} = v\frac{dv}{ds} v = d t d s , a = d t d v = d t 2 d 2 s = v d s d v
The last form is useful when acceleration is given as a function of position.
Constant acceleration:
v = v 0 + a t , s = v 0 t + 1 2 a t 2 , v 2 = v 0 2 + 2 a s v = v_0 + at, \qquad s = v_0 t + \tfrac{1}{2}at^2, \qquad v^2 = v_0^2 + 2as v = v 0 + a t , s = v 0 t + 2 1 a t 2 , v 2 = v 0 2 + 2 a s
Example (calculus-based). A particle moves with s = t 3 − 6 t 2 + 9 t s = t^3 - 6t^2 + 9t s = t 3 − 6 t 2 + 9 t meters.
Velocity: v = 3 t 2 − 12 t + 9 = 3 ( t − 1 ) ( t − 3 ) v = 3t^2 - 12t + 9 = 3(t-1)(t-3) v = 3 t 2 − 12 t + 9 = 3 ( t − 1 ) ( t − 3 ) , so it stops at t = 1 t = 1 t = 1 and t = 3 s t = 3\text{ s} t = 3 s .
Acceleration: a = 6 t − 12 a = 6t - 12 a = 6 t − 12 , which is zero at t = 2 s t = 2\text{ s} t = 2 s .
Positions: s ( 0 ) = 0 s(0) = 0 s ( 0 ) = 0 , s ( 1 ) = 4 s(1) = 4 s ( 1 ) = 4 , s ( 3 ) = 0 s(3) = 0 s ( 3 ) = 0 , s ( 4 ) = 4 s(4) = 4 s ( 4 ) = 4 . The total distance traveled in 4 s is 4 + 4 + 4 = 12 m 4 + 4 + 4 = 12\text{ m} 4 + 4 + 4 = 12 m , even though the displacement is only 4 m 4\text{ m} 4 m .
Projectile motion (no air resistance) splits into constant horizontal velocity and constant vertical acceleration g g g :
x = v 0 cos θ t , y = v 0 sin θ t − 1 2 g t 2 x = v_0\cos\theta\, t, \qquad y = v_0\sin\theta\, t - \tfrac{1}{2}gt^2 x = v 0 cos θ t , y = v 0 sin θ t − 2 1 g t 2
On level ground, the range is R = v 0 2 sin 2 θ / g R = v_0^2 \sin 2\theta / g R = v 0 2 sin 2 θ / g , a maximum at θ = 45 ∘ \theta = 45^\circ θ = 4 5 ∘ . The maximum height is H = ( v 0 sin θ ) 2 / ( 2 g ) H = (v_0\sin\theta)^2/(2g) H = ( v 0 sin θ ) 2 / ( 2 g ) .
Curvilinear motion. Normal (centripetal) acceleration is a n = v 2 / ρ a_n = v^2/\rho a n = v 2 / ρ , the basis of highway superelevation.
Newton's Laws and Friction
A body remains at rest or in uniform motion unless acted on by a net force.
∑ F = m a \sum F = ma ∑ F = ma , applied along each axis.
Action and reaction are equal and opposite.
D'Alembert's principle treats − m a -ma − ma as an inertia force, turning a dynamics problem into a statics problem.
Block on an incline with angle θ \theta θ and kinetic friction coefficient μ \mu μ :
Sliding down: a = g ( sin θ − μ cos θ ) a = g(\sin\theta - \mu\cos\theta) a = g ( sin θ − μ cos θ ) .
Sliding up after an initial push: deceleration = g ( sin θ + μ cos θ ) = g(\sin\theta + \mu\cos\theta) = g ( sin θ + μ cos θ ) .
Example. A crate slides down a 30 ∘ 30^\circ 3 0 ∘ chute with μ = 0.25 \mu = 0.25 μ = 0.25 . Then a = 9.81 ( 0.500 − 0.25 × 0.866 ) = 2.78 m/s 2 a = 9.81(0.500 - 0.25 \times 0.866) = 2.78\text{ m/s}^2 a = 9.81 ( 0.500 − 0.25 × 0.866 ) = 2.78 m/s 2 . Starting from rest, after 5 m 5\text{ m} 5 m its speed is v = 2 ( 2.78 ) ( 5 ) = 5.27 m/s v = \sqrt{2(2.78)(5)} = 5.27\text{ m/s} v = 2 ( 2.78 ) ( 5 ) = 5.27 m/s .
Connected bodies. For a mass m 1 m_1 m 1 on a smooth table pulled by a hanging mass m 2 m_2 m 2 over a frictionless pulley, a = m 2 g / ( m 1 + m 2 ) a = m_2 g/(m_1 + m_2) a = m 2 g / ( m 1 + m 2 ) and the cord tension is T = m 1 a T = m_1 a T = m 1 a .
- Work Done by Variable Forces and Hooke's Law
- W = ∫ s 1 s 2 F d s W = \int_{s_1}^{s_2} F\,ds W = ∫ s 1 s 2 F d s
For a linear spring obeying Hooke's law , F = k x F = kx F = k x , the work to stretch it from x 1 x_1 x 1 to x 2 x_2 x 2 is the area under the force-deformation line:
W = 1 2 k ( x 2 2 − x 1 2 ) W = \tfrac{1}{2}k\left(x_2^2 - x_1^2\right) W = 2 1 k ( x 2 2 − x 1 2 )
Example. A spring with k = 4,000 N/m k = 4{,}000\text{ N/m} k = 4 , 000 N/m is stretched from 0.05 0.05 0.05 to 0.15 m 0.15\text{ m} 0.15 m . W = 0.5 ( 4,000 ) ( 0.0225 − 0.0025 ) = 40 J W = 0.5(4{,}000)(0.0225 - 0.0025) = 40\text{ J} W = 0.5 ( 4 , 000 ) ( 0.0225 − 0.0025 ) = 40 J . Equivalently, average force times displacement gives 1 2 [ 4,000 ( 0.05 ) + 4,000 ( 0.15 ) ] ( 0.10 ) = 400 ( 0.10 ) = 40 J \tfrac{1}{2}[4{,}000(0.05) + 4{,}000(0.15)](0.10) = 400(0.10) = 40\text{ J} 2 1 [ 4 , 000 ( 0.05 ) + 4 , 000 ( 0.15 )] ( 0.10 ) = 400 ( 0.10 ) = 40 J , which is exact for a linear spring.
Gravity. Lifting a weight W W W a height h h h requires W h W h Wh , whatever the path.
Conservation of Energy and Power
When only conservative forces (gravity and springs) do work:
T 1 + V 1 = T 2 + V 2 , T = 1 2 m v 2 , V g = m g h , V e = 1 2 k x 2 T_1 + V_1 = T_2 + V_2, \qquad T = \tfrac{1}{2}mv^2, \qquad V_g = mgh, \qquad V_e = \tfrac{1}{2}kx^2 T 1 + V 1 = T 2 + V 2 , T = 2 1 m v 2 , V g = m g h , V e = 2 1 k x 2
With friction or other losses, use the work-energy principle : T 1 + V 1 + U 1 → 2 , nonconservative = T 2 + V 2 T_1 + V_1 + U_{1\to2,\text{nonconservative}} = T_2 + V_2 T 1 + V 1 + U 1 → 2 , nonconservative = T 2 + V 2 , where friction work is negative.
Example. A 2 kg 2\text{ kg} 2 kg block falls from rest 0.80 m 0.80\text{ m} 0.80 m onto a spring with k = 2,000 N/m k = 2{,}000\text{ N/m} k = 2 , 000 N/m . Find the maximum compression x x x :
m g ( 0.80 + x ) = 1 2 k x 2 ⟹ 19.62 ( 0.80 + x ) = 1,000 x 2 mg(0.80 + x) = \tfrac{1}{2}kx^2 \implies 19.62(0.80 + x) = 1{,}000x^2 m g ( 0.80 + x ) = 2 1 k x 2 ⟹ 19.62 ( 0.80 + x ) = 1 , 000 x 2
Solving the quadratic gives x = 0.135 m x = 0.135\text{ m} x = 0.135 m .
Power is the rate of doing work: P = d W / d t = F v P = dW/dt = Fv P = d W / d t = F v , with 1 hp = 746 W 1\text{ hp} = 746\text{ W} 1 hp = 746 W .
Example. A hoist raises 500 kg 500\text{ kg} 500 kg at a constant 0.5 m/s 0.5\text{ m/s} 0.5 m/s . P = m g v = 500 ( 9.81 ) ( 0.5 ) = 2.45 kW P = mgv = 500(9.81)(0.5) = 2.45\text{ kW} P = m gv = 500 ( 9.81 ) ( 0.5 ) = 2.45 kW delivered to the load. At 80% efficiency, the motor needs 3.07 kW 3.07\text{ kW} 3.07 kW .
- Linear Impulse and Momentum
- ∫ F d t = m v 2 − m v 1 \int F\,dt = m v_2 - m v_1 ∫ F d t = m v 2 − m v 1
Impulse equals the change in momentum. When no external impulse acts on a system, its total momentum is conserved.
Collisions. Momentum is conserved along the line of impact:
m 1 v 1 + m 2 v 2 = m 1 v 1 ′ + m 2 v 2 ′ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' m 1 v 1 + m 2 v 2 = m 1 v 1 ′ + m 2 v 2 ′
The coefficient of restitution is:
e = v 2 ′ − v 1 ′ v 1 − v 2 e = \frac{v_2' - v_1'}{v_1 - v_2} e = v 1 − v 2 v 2 ′ − v 1 ′
It equals 1 for a perfectly elastic impact (kinetic energy conserved) and 0 for a perfectly plastic impact (the bodies move together).
Example. A 1,500 kg 1{,}500\text{ kg} 1 , 500 kg car at 20 m/s 20\text{ m/s} 20 m/s strikes a stationary 1,000 kg 1{,}000\text{ kg} 1 , 000 kg car, and they lock together ( e = 0 e = 0 e = 0 ):
Common velocity: v ′ = 1,500 ( 20 ) / 2,500 = 12 m/s v' = 1{,}500(20)/2{,}500 = 12\text{ m/s} v ′ = 1 , 500 ( 20 ) /2 , 500 = 12 m/s .
Kinetic energy falls from 300 kJ 300\text{ kJ} 300 kJ to 1 2 ( 2,500 ) ( 12 ) 2 = 180 kJ \tfrac{1}{2}(2{,}500)(12)^2 = 180\text{ kJ} 2 1 ( 2 , 500 ) ( 12 ) 2 = 180 kJ .
So 120 kJ 120\text{ kJ} 120 kJ is dissipated in deformation.
Angular Momentum and Rotation
For a rigid body rotating about a fixed axis:
∑ M = I α , T = 1 2 I ω 2 , H = I ω \sum M = I\alpha, \qquad T = \tfrac{1}{2}I\omega^2, \qquad H = I\omega ∑ M = I α , T = 2 1 I ω 2 , H = I ω
The mass moment of inertia I I I is, for example, 1 2 m r 2 \tfrac{1}{2}mr^2 2 1 m r 2 for a solid disk and m r 2 mr^2 m r 2 for a thin ring. The angular impulse ∫ M d t \int M\,dt ∫ M d t equals the change in angular momentum. When no external moment acts, I 1 ω 1 = I 2 ω 2 I_1\omega_1 = I_2\omega_2 I 1 ω 1 = I 2 ω 2 .
Rolling without slipping. For a body of radius r r r rolling down an incline, v = ω r v = \omega r v = ω r , and the total kinetic energy is 1 2 m v 2 + 1 2 I ω 2 \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2 2 1 m v 2 + 2 1 I ω 2 . A solid cylinder ( I = 1 2 m r 2 I = \tfrac{1}{2}mr^2 I = 2 1 m r 2 ) rolling from rest down a height h h h reaches:
v = 4 3 g h v = \sqrt{\tfrac{4}{3}gh} v = 3 4 g h
That is slower than a frictionless sliding block, which reaches 2 g h \sqrt{2gh} 2 g h , because part of the energy goes into rotation.