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CalcREX offers 20+ physics calculators including kinematics, force, energy, momentum, velocity, acceleration, and more.

How do Newton's laws of motion work?

Newton's three laws: (1) An object stays at rest or in uniform motion unless acted on by a net force, (2) F = ma relates force, mass, and acceleration, (3) every action has an equal and opposite reaction.

What is the difference between elastic and inelastic collisions?

In elastic collisions, both momentum and kinetic energy are conserved. In inelastic collisions, momentum is conserved but kinetic energy is not.

How do you calculate projectile motion?

Separate horizontal (constant velocity) and vertical (gravity-accelerated) components. Key equations: x = v0*cos(θ)*t, y = v0*sin(θ)*t - 0.5*g*t², Range = v0²*sin(2θ)/g.

Physics Calculators - CalcREX

Newton's Laws of Motion & Applications

Understanding the fundamental laws that govern all motion - from falling apples to orbiting planets.

1 The Three Laws of Motion

First Law - Law of Inertia

An object at rest stays at rest, and an object in motion stays in motion with the same speed and direction, unless acted upon by an unbalanced external force.

Inertia is the tendency of an object to resist changes in its state of motion. Mass is the quantitative measure of inertia - the greater the mass, the greater the resistance to acceleration.

Second Law - F = ma

The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

F_net = m × a

F_net = Net force (N)m = Mass (kg)a = Acceleration (m/s²)

This is the most widely used law in physics. The direction of acceleration is always in the direction of the net force.

Third Law - Action & Reaction

For every action, there is an equal and opposite reaction. When object A exerts a force on object B, object B simultaneously exerts an equal force in the opposite direction on object A.

The action-reaction pair always acts on different objects. This is why they don't cancel out.

2 Free Body Diagrams (FBDs)

A free body diagram isolates a single object and shows all forces acting on it as vectors. This is the essential first step in solving any force problem.

Figure 1: Free body diagram of a block on a horizontal surface. Weight (W = mg) acts downward, normal force (N) acts upward, friction (f = μN) opposes motion, and applied force pushes the block right.

Steps to Draw a Free Body Diagram

1Identify the object and draw it as a simple shape (box or dot).

2Identify every force acting on the object (not forces it exerts).

3Draw each force as an arrow pointing in the correct direction.

4Label all forces and choose a coordinate system.

3 Equilibrium - When Forces Balance

An object is in static equilibrium when at rest with zero net force. Dynamic equilibrium means constant velocity with zero net force.

ΣF = 0

ΣF_x = 0 (horizontal forces balance)ΣF_y = 0 (vertical forces balance)

Worked Example 1

Block on a Horizontal Surface with Friction

Problem: A 15 kg box is pushed along a horizontal floor with a force of 80 N. The coefficient of kinetic friction is μ_k = 0.30. Find (a) the normal force, (b) friction force, and (c) acceleration.

Step 1: Known values

m = 15 kg, F_app = 80 N, μ_k = 0.30, g = 9.81 m/s²

Step 2: Vertical equilibrium (ΣF_y = 0)

N − mg = 0

N = mg = 15 × 9.81

N = 147.15 N

Step 3: Friction force

f_k = μ_k × N = 0.30 × 147.15

f_k = 44.15 N

Step 4: Horizontal - ΣF_x = ma

F_app − f_k = ma

80 − 44.15 = 15a

a = 2.39 m/s²

Answers: (a) N = 147.15 N, (b) f_k = 44.15 N, (c) a = 2.39 m/s² to the right.

Worked Example 2

Block on an Inclined Plane

Problem: A 10 kg block slides down a frictionless 30° incline. Find the acceleration and normal force.

Figure 2: On an incline, gravity decomposes into mg sin θ (along surface, causes sliding) and mg cos θ (perpendicular, balanced by normal force).

Step 1: Decompose gravity

Along incline: mg sin 30° = 10 × 9.81 × 0.5 = 49.05 N

Perpendicular: mg cos 30° = 10 × 9.81 × 0.866 = 84.96 N

Step 2: Normal force (ΣF⊥ = 0)

N = mg cos 30° = 84.96 N

Step 3: Acceleration along incline

mg sin θ = ma → a = g sin 30°

a = 4.905 m/s²

Answers: a = 4.905 m/s² down the incline, N = 84.96 N. The acceleration depends only on angle and g - not mass!

Worked Example 3

Atwood Machine - Two Connected Masses

Problem: Two masses m₁ = 5 kg and m₂ = 3 kg are connected by a string over a frictionless pulley. Find the acceleration and string tension.

Figure 3: Atwood Machine - heavier mass accelerates down, lighter mass accelerates up with same magnitude.

Step 1: Newton's Second Law for each mass

m₁: m₁g − T = m₁a ... (1)

m₂: T − m₂g = m₂a ... (2)

Step 2: Add equations to find acceleration

a = (m₁ − m₂)g / (m₁ + m₂) = (5−3)(9.81) / (5+3)

a = 2.45 m/s²

Step 3: Find tension

T = m₂(g + a) = 3(9.81 + 2.45)

T = 36.79 N

Answers: a = 2.45 m/s², T = 36.79 N. The tension lies between m₁g (49.05 N) and m₂g (29.43 N).

Kinematics - Projectile Motion, Free Fall & Acceleration

Describing motion without worrying about forces - the mathematics of movement.

1 The Kinematic Equations (SUVAT)

These four equations relate displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t) for constant acceleration.

v = u + at

Missing: s

s = ut + ½at²

Missing: v

v² = u² + 2as

Missing: t

s = ½(u + v)t

Missing: a

Tip: To pick the right equation, identify which variable is not given and not asked for - then choose the equation that doesn't contain it.

2 Free Fall

Free fall is motion under gravity alone with a = g = 9.81 m/s² (downward). Air resistance is neglected. All objects fall at the same rate regardless of mass.

Worked Example 1

Object Dropped from a Height

Problem: A ball is dropped from rest from a 45 m building. Find (a) the time to reach the ground, and (b) impact velocity.

Step 1: Known values (downward positive)

u = 0, a = 9.81 m/s², s = 45 m

Step 2: Time - s = ut + ½at²

45 = 0 + ½(9.81)t²

t² = 45/4.905 = 9.174

t = 3.03 s

Step 3: Velocity - v = u + at

v = 0 + 9.81 × 3.03

v = 29.72 m/s ≈ 107 km/h

Answers: (a) t = 3.03 s, (b) v = 29.72 m/s downward.

3 Projectile Motion

A projectile moves under gravity only after launch. The key insight: horizontal and vertical motions are independent.

Horizontal (x)

No acceleration (a_x = 0)

v_x = v₀ cos θ = constant

x = v₀ cos(θ) × t

Vertical (y)

Acceleration = −g (upward positive)

v_y = v₀ sin θ − gt

y = v₀ sin(θ)t − ½gt²

Figure 4: Projectile trajectory (parabola). At the peak, v_y = 0. Maximum range at θ = 45°.

Key Projectile Formulas

Max Height: H = v₀² sin²θ / (2g) Time of Flight: T = 2v₀ sin θ / g Range: R = v₀² sin(2θ) / g

Worked Example 2

Projectile Launched at an Angle

Problem: A football is kicked at 25 m/s at 40° above horizontal. Find (a) max height, (b) time of flight, (c) range.

Step 1: Resolve velocity

v₀ₓ = 25 cos 40° = 19.15 m/s

v₀ᵧ = 25 sin 40° = 16.07 m/s

Step 2: Maximum height

H = v₀²sin²θ / (2g) = 625 × 0.413 / 19.62

H = 13.16 m

Step 3: Time of flight

T = 2v₀ sin θ / g = 2(16.07)/9.81

T = 3.28 s

Step 4: Range

R = v₀²sin(2θ)/g = 625 × sin 80° / 9.81

R = 62.73 m

Answers: (a) H = 13.16 m, (b) T = 3.28 s, (c) R = 62.73 m.

Worked Example 3

Car Braking - Stopping Distance

Problem: A car at 90 km/h (25 m/s) brakes at 8 m/s². Find (a) stopping time and (b) stopping distance.

Step 1: Known values

u = 25 m/s, v = 0, a = −8 m/s²

Step 2: Time - v = u + at

0 = 25 − 8t → t = 25/8

t = 3.125 s

Step 3: Distance - v² = u² + 2as

0 = 625 − 16s

s = 39.06 m

Answers: (a) t = 3.125 s, (b) s = 39.06 m. Doubling speed quadruples stopping distance (s ∝ u²).

Energy & Momentum Conservation - Collisions

Two of the most powerful conservation laws in physics - energy and momentum are never lost, only transformed.

1 Conservation of Energy

Energy cannot be created or destroyed - only transformed. Total mechanical energy (kinetic + potential) stays constant when only conservative forces act.

KE₁ + PE₁ = KE₂ + PE₂

½mv₁² + mgh₁ = ½mv₂² + mgh₂

With friction: KE₁ + PE₁ = KE₂ + PE₂ + W_friction (some energy converts to heat).

Worked Example 1

Roller Coaster - Energy Conservation

Problem: A 500 kg roller coaster starts from rest at the top of a 40 m hill. Find its speed at the bottom and at a second 25 m hill. (No friction.)

Figure 5: Energy conservation on a roller coaster - total mechanical energy stays constant throughout.

Step 1: Speed at bottom (h = 0)

mgh₁ = ½mv² → v = √(2gh₁) = √(2 × 9.81 × 40)

v = 28.01 m/s ≈ 100.8 km/h

Step 2: Speed at second hill (h₂ = 25 m)

v₂ = √(2g(h₁ − h₂)) = √(2 × 9.81 × 15)

v₂ = 17.15 m/s ≈ 61.7 km/h

Answers: Bottom speed = 28.01 m/s. Second hill speed = 17.15 m/s. Mass cancels - heavier cars reach the same speed!

2 Conservation of Momentum

Total momentum of an isolated system remains constant. Momentum is always conserved in collisions.

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Total momentum before = Total momentum after

3 Types of Collisions

Property

Elastic

Inelastic

Perfectly Inelastic

Momentum

Conserved ✓

Kinetic Energy

Conserved ✓

NOT conserved ✗

After Collision

Objects bounce apart

Some deformation

Objects stick together

Example

Billiard balls

Car crash (rebound)

Bullet in block

Worked Example 2

Elastic Collision - Equal Mass Billiard Balls

Problem: A 0.17 kg billiard ball at 5 m/s strikes an identical stationary ball head-on (elastic). Find both velocities after collision.

Step 1: Conservation of momentum

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

0.17(5) + 0 = 0.17v₁' + 0.17v₂'

5 = v₁' + v₂' ... (1)

Step 2: Conservation of kinetic energy

½m₁v₁² = ½m₁v₁'² + ½m₂v₂'²

25 = v₁'² + v₂'² ... (2)

Step 3: Equal-mass shortcut - velocities exchange

v₁' = 0 m/s (first ball stops)

v₂' = 5 m/s (second ball moves at original speed)

Verification: momentum = 0.85 kg⋅m/s before and after ✓. KE = 2.125 J before and after ✓.

Answer: Equal-mass elastic collision: moving ball stops, stationary ball takes its velocity. This is the "Newton's Cradle" effect.

Worked Example 3

Perfectly Inelastic - Bullet in Block

Problem: A 0.01 kg bullet at 400 m/s embeds in a 2 kg block at rest. Find (a) combined velocity, (b) kinetic energy lost.

Step 1: Conservation of momentum (stick together)

m₁v₁ = (m₁ + m₂)v'

0.01(400) = (2.01)v'

v' = 1.99 m/s

Step 2: Energy lost

KE_before = ½(0.01)(400²) = 800 J

KE_after = ½(2.01)(1.99²) = 3.98 J

Energy lost = 796.02 J (99.5%!)

Answers: (a) v' = 1.99 m/s. (b) 796 J lost to heat, sound, deformation. Perfectly inelastic collisions lose the most kinetic energy.

Waves - Interference, Diffraction & Doppler Effect

How energy propagates through space without the transfer of matter.

1 Wave Fundamentals

v = fλ

v = wave speed (m/s), f = frequency (Hz), λ = wavelength (m)

Figure 6: Transverse wave. Wavelength (λ) = distance between two consecutive crests. Amplitude (A) = maximum displacement from equilibrium.

Transverse waves oscillate perpendicular to propagation (light, water surface). Longitudinal waves oscillate parallel to propagation (sound). Sound in air travels at approximately 343 m/s at 20°C.

2 Interference

When two waves overlap, the principle of superposition applies: the resultant displacement equals the sum of individual displacements.

Constructive Interference

Waves in phase - crests align with crests. Path difference = nλ (n = 0, 1, 2...). Amplitudes add → louder/brighter.

Destructive Interference

Waves out of phase - crests align with troughs. Path difference = (n + ½)λ. Amplitudes cancel → silence/darkness.

Young's Double-Slit Experiment

Bright fringes: d sin θ = nλ (n = 0, ±1, ±2...) Dark fringes: d sin θ = (n + ½)λ Fringe spacing: Δy = λL / d

3 Diffraction

Diffraction is the bending and spreading of waves around obstacles or through openings. Most noticeable when the opening size ≈ wavelength. A single slit of width a gives minima at:

a sin θ = nλ (n = ±1, ±2, ...)

a = slit width, θ = angle to minimum, n = order (≠ 0)

Wide slit → narrow central maximum (less spreading). Narrow slit → wide central maximum (more spreading). The central maximum is always twice the width of the other maxima.

4 The Doppler Effect

The apparent change in frequency when source and observer are in relative motion. You hear higher pitch as an ambulance approaches, lower as it recedes.

f' = f × (v ± v_o) / (v ∓ v_s)

f' = observed frequency, f = source frequencyv = speed of sound, v_o = observer speed, v_s = source speed

Sign convention: Top signs (+v_o, −v_s) when moving toward each other (frequency ↑). Bottom signs when moving apart (frequency ↓).

Worked Example 1

Doppler Effect - Approaching Ambulance

Problem: Ambulance siren: 700 Hz. Speed: 30 m/s toward stationary observer. Speed of sound: 343 m/s. Find heard frequency (a) approaching, (b) receding.

Step 1: Approaching

f' = f × v / (v − v_s) = 700 × 343 / (343 − 30)

f' = 700 × 343 / 313

f' = 767.1 Hz (higher pitch)

Step 2: Receding

f' = f × v / (v + v_s) = 700 × 343 / (343 + 30)

f' = 700 × 343 / 373

f' = 643.7 Hz (lower pitch)

Answers: Approaching: 767.1 Hz. Receding: 643.7 Hz. The shift is 123.4 Hz - very noticeable.

Worked Example 2

Young's Double-Slit - Fringe Spacing

Problem: Light of wavelength 550 nm passes through slits separated by 0.25 mm. Screen is 2 m away. Find (a) fringe spacing, and (b) angle to second bright fringe.

Step 1: Fringe spacing

Δy = λL / d = (550 × 10⁻⁹ × 2) / (0.25 × 10⁻³)

Δy = 4.4 × 10⁻³ m = 4.4 mm

Step 2: Angle to 2nd-order bright fringe (n = 2)

d sin θ = nλ

sin θ = nλ/d = 2(550 × 10⁻⁹) / (0.25 × 10⁻³)

sin θ = 0.0044

θ = 0.252° ≈ 0.25°

Answers: (a) Fringe spacing = 4.4 mm. (b) θ₂ = 0.25°. The small angle confirms the fringes are closely spaced - this is why you need a screen far from the slits to observe them clearly.

Electricity - Ohm's Law, Circuits & Electromagnetic Induction

Understanding current flow, circuit analysis, and how changing magnetic fields create electricity.

1 Ohm's Law & Basic Relationships

Ohm's Law relates voltage, current, and resistance in a conductor.

V = I × R

V = Voltage (V), I = Current (A), R = Resistance (Ω)

Power formulas: P = VI = I²R = V²/R

2 Series and Parallel Circuits

Series Circuit

Components end-to-end. Same current through all.

R_total = R₁ + R₂ + R₃

V_total = V₁ + V₂ + V₃

I_total = I₁ = I₂ = I₃

Parallel Circuit

Components across same two points. Same voltage.

1/R_T = 1/R₁ + 1/R₂ + 1/R₃

I_total = I₁ + I₂ + I₃

V_total = V₁ = V₂ = V₃

Series Circuit

+──[ R₁ ]──[ R₂ ]──[ R₃ ]──+ | | [V] Same current I [V] | | +────────────────────────────+

Parallel Circuit

+──[ R₁ ]──+ | | +────+──[ R₂ ]──+────+ | | | | [V] +──[ R₃ ]──+ [V] | | +─────────────────────+

Parallel total resistance is always less than the smallest individual resistor.

Worked Example 1

Series-Parallel Combination Circuit

Problem: A 12 V battery connects to R₁ = 4 Ω in series with R₂ = 6 Ω ∥ R₃ = 3 Ω. Find total resistance, current, voltage drops, and branch currents.

Step 1: Parallel combination

1/R_p = 1/6 + 1/3 = 3/6

R_parallel = 2 Ω

Step 2: Total resistance

R_total = R₁ + R_p = 4 + 2 = 6 Ω

Step 3: Total current

I = V/R = 12/6 = 2 A

Step 4: Voltage drops

V_R₁ = 2 × 4 = 8 V

V_parallel = 2 × 2 = 4 V

Check: 8 + 4 = 12 V ✓ (KVL)

Step 5: Branch currents

I_R₂ = 4/6 = 0.667 A

I_R₃ = 4/3 = 1.333 A

Check: 0.667 + 1.333 = 2 A ✓ (KCL)

Answers: R_total = 6 Ω, I = 2 A, V_R₁ = 8 V, V_parallel = 4 V, I_R₂ = 0.667 A, I_R₃ = 1.333 A.

3 Kirchhoff's Laws

KCL - Junction Rule

Total current entering a junction equals total current leaving. (Conservation of charge.)

ΣI_in = ΣI_out

KVL - Loop Rule

Sum of voltage gains and drops around any closed loop = 0. (Conservation of energy.)

ΣV = 0 (around any closed loop)

4 Electromagnetic Induction

A changing magnetic flux through a circuit induces an EMF. This powers generators, transformers, and wireless charging.

Faraday's Law: ε = −N × ΔΦ/Δt

ε = induced EMF (V), N = number of turnsΔΦ/Δt = rate of change of magnetic flux (Wb/s)

Lenz's Law: The induced current flows in a direction that opposes the change in flux that caused it. This is the reason for the negative sign in Faraday's Law and is a consequence of energy conservation.

Worked Example 2

Electromagnetic Induction - Generator Coil

Problem: A 200-turn coil has an area of 0.05 m². The magnetic field through it changes from 0.3 T to 0 T in 0.02 s. Find the induced EMF.

Step 1: Calculate change in flux

ΔΦ = ΔB × A = (0 − 0.3) × 0.05 = −0.015 Wb

Step 2: Apply Faraday's Law

ε = −N × ΔΦ/Δt = −200 × (−0.015) / 0.02

ε = 150 V

Answer: The induced EMF is 150 V. The faster the field changes or the more turns in the coil, the greater the induced voltage.

Worked Example 3

Electrical Power & Energy Cost

Problem: A 2000 W electric heater runs for 3 hours. If electricity costs $0.12/kWh, find (a) the energy consumed and (b) the cost.

Step 1: Energy in kWh

E = P × t = 2 kW × 3 h

E = 6 kWh

Step 2: Cost

Cost = E × rate = 6 × $0.12

Cost = $0.72

Answers: (a) 6 kWh = 21.6 MJ. (b) Cost = $0.72. In Joules: E = 2000 × 10800 = 21,600,000 J.

Thermodynamics - Heat Transfer & Ideal Gas Law

The science of heat, energy transfer, and the behavior of gases - from steam engines to refrigerators.

1 Heat Transfer Methods

Conduction

Heat transfer through direct contact between molecules. Metals are good conductors; wood and air are insulators.

Q/t = kA(T₂ − T₁)/L

Convection & Radiation

Convection: Heat transfer by fluid movement (hot air rises). Radiation: Heat transfer via electromagnetic waves (no medium needed - how the Sun heats Earth).

2 Specific Heat & Calorimetry

The amount of heat needed to change the temperature of a substance depends on its mass, specific heat capacity, and temperature change.

Q = mcΔT

Q = heat energy (J)m = mass (kg), c = specific heat capacity (J/kg⋅K)ΔT = temperature change (K or °C)

Water has one of the highest specific heats: c = 4186 J/(kg⋅K). This is why oceans moderate coastal climates and why water is used in cooling systems.

Water: c = 4186 J/(kg⋅K)

Liquid

Ice: c = 2090 J/(kg⋅K)

Solid

Aluminum: c = 897 J/(kg⋅K)

Metal

Iron: c = 449 J/(kg⋅K)

Metal

Copper: c = 385 J/(kg⋅K)

Metal

Worked Example 1

Heating Water - How Much Energy?

Problem: How much energy is needed to heat 2 kg of water from 20°C to 100°C? If a 1500 W kettle does this, how long does it take?

Step 1: Calculate heat energy

Q = mcΔT = 2 × 4186 × (100 − 20)

Q = 2 × 4186 × 80

Q = 669,760 J ≈ 670 kJ

Step 2: Time for kettle (P = E/t → t = E/P)

t = Q / P = 669,760 / 1500

t = 446.5 s ≈ 7.4 minutes

Answers: Q = 670 kJ. Time ≈ 7.4 minutes. In practice it takes longer due to heat losses to the surroundings.

Worked Example 2

Calorimetry - Mixing Hot and Cold Water

Problem: 0.5 kg of water at 80°C is mixed with 1.5 kg of water at 20°C in an insulated container. Find the final equilibrium temperature.

Step 1: Heat lost = Heat gained (energy conservation)

m_hotc(T_hot − T_f) = m_coldc(T_f − T_cold)

Since both are water, c cancels:

0.5(80 − T_f) = 1.5(T_f − 20)

Step 2: Solve for T_f

40 − 0.5T_f = 1.5T_f − 30

70 = 2T_f

T_f = 35°C

Answer: Final temperature = 35°C. The result is closer to the cold water temperature because there is more cold water (larger mass dominates).

3 Ideal Gas Law

The ideal gas law relates pressure, volume, temperature, and amount of gas. It works well for most gases at moderate temperatures and pressures.

PV = nRT

P = pressure (Pa), V = volume (m³)n = moles, R = 8.314 J/(mol⋅K)T = temperature (Kelvin! Always!)

Always use Kelvin! T(K) = T(°C) + 273.15. The gas laws break down if you use Celsius or Fahrenheit.

For a fixed amount of gas changing between two states:

P₁V₁ / T₁ = P₂V₂ / T₂

This is the Combined Gas Law - useful when n is constant

Boyle's Law: P₁V₁ = P₂V₂

Constant T

Charles's Law: V₁/T₁ = V₂/T₂

Constant P

Gay-Lussac's: P₁/T₁ = P₂/T₂

Constant V

Worked Example 3

Ideal Gas Law - Finding Pressure

Problem: 2 moles of an ideal gas occupy 0.05 m³ at 300 K. Find the pressure. Then the gas is heated to 600 K at constant volume - find the new pressure.

Step 1: Find initial pressure using PV = nRT

P = nRT / V = (2 × 8.314 × 300) / 0.05

P₁ = 99,768 Pa ≈ 99.8 kPa ≈ 1 atm

Step 2: Heated at constant volume (Gay-Lussac's Law)

P₁/T₁ = P₂/T₂

P₂ = P₁ × T₂/T₁ = 99,768 × 600/300

P₂ = 199,536 Pa ≈ 199.5 kPa ≈ 2 atm

Answers: P₁ ≈ 99.8 kPa (≈ 1 atm). P₂ ≈ 199.5 kPa (≈ 2 atm). Doubling temperature (in Kelvin) at constant volume doubles the pressure - which is why aerosol cans warn against heating.

4 Laws of Thermodynamics

First Law - Energy Conservation

The change in internal energy of a system equals the heat added minus the work done by the system.

ΔU = Q − W

ΔU = change in internal energy, Q = heat added, W = work done by system

Second Law - Entropy Always Increases

Heat flows spontaneously from hot to cold, never the reverse. The total entropy of an isolated system always increases. No heat engine can be 100% efficient.

Efficiency: η = W/Q_H = 1 − Q_C/Q_H

Carnot (maximum): η_max = 1 − T_C/T_H

Worked Example 4

Carnot Efficiency - Heat Engine

Problem: A heat engine operates between a hot reservoir at 500°C and a cold reservoir at 25°C. Find the maximum possible (Carnot) efficiency. If the engine absorbs 1000 J of heat, how much work can it produce?

Step 1: Convert to Kelvin

T_H = 500 + 273 = 773 K

T_C = 25 + 273 = 298 K

Step 2: Carnot efficiency

η_max = 1 − T_C/T_H = 1 − 298/773

η_max = 0.614 = 61.4%

Step 3: Maximum work output

W = η × Q_H = 0.614 × 1000

W = 614 J

Answers: Maximum efficiency = 61.4%. Max work = 614 J. The remaining 386 J must be rejected to the cold reservoir. Real engines always have lower efficiency than Carnot.

Comprehensive Physics Calculators - Solve Any Physics Problem

Select a Calculator

How it works

Common Physics Constants

Physics Topics - In-Depth Guides with Worked Examples

Newton's Laws of Motion & Applications

1 The Three Laws of Motion

2 Free Body Diagrams (FBDs)

3 Equilibrium - When Forces Balance

Block on a Horizontal Surface with Friction

Block on an Inclined Plane

Atwood Machine - Two Connected Masses

Kinematics - Projectile Motion, Free Fall & Acceleration

1 The Kinematic Equations (SUVAT)

2 Free Fall

Object Dropped from a Height

3 Projectile Motion

Projectile Launched at an Angle

Car Braking - Stopping Distance

Energy & Momentum Conservation - Collisions

1 Conservation of Energy

Roller Coaster - Energy Conservation

2 Conservation of Momentum

3 Types of Collisions

Elastic Collision - Equal Mass Billiard Balls

Perfectly Inelastic - Bullet in Block

Waves - Interference, Diffraction & Doppler Effect

1 Wave Fundamentals

2 Interference

3 Diffraction

4 The Doppler Effect

Doppler Effect - Approaching Ambulance

Young's Double-Slit - Fringe Spacing

Electricity - Ohm's Law, Circuits & Electromagnetic Induction

1 Ohm's Law & Basic Relationships

2 Series and Parallel Circuits

Series Circuit

Parallel Circuit

Series-Parallel Combination Circuit

3 Kirchhoff's Laws

4 Electromagnetic Induction

Electromagnetic Induction - Generator Coil

Electrical Power & Energy Cost

Thermodynamics - Heat Transfer & Ideal Gas Law

1 Heat Transfer Methods

2 Specific Heat & Calorimetry

Heating Water - How Much Energy?

Calorimetry - Mixing Hot and Cold Water

3 Ideal Gas Law

Ideal Gas Law - Finding Pressure

4 Laws of Thermodynamics

Carnot Efficiency - Heat Engine