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Select a mode and enter your values to calculate force, work, or power. Results will appear here.
Solve W = F·d·cos θ and P = W/t for introductory mechanics. Handles inclined planes with friction and work from variable-force paths. For motor efficiency, AC power factor, or watts↔hp conversions, use the Motor Efficiency Calculator.
Force (N) = mass × acceleration — any push or pull that changes motion.
Work (J) = force × distance × cos(angle) — energy transferred when moving an object.
Power (W) = work ÷ time — how fast work is done.
For example, pushing a 10 kg box with 2 m/s² acceleration requires 20 N of force. Moving that box 5 meters horizontally does 100 J of work (20 N × 5 m × cos(0°)). If done in 2 seconds, your power output is 50 W.
Example: Lifting a Box
Lifting a 20 kg box to a 1.5 m high shelf against gravity (9.8 m/s²):
Select a mode and enter your values to calculate force, work, or power. Results will appear here.
Push a 100 kg box across a level floor at constant 200 N for 10 m and you've done 2,000 J of work, even though the box's kinetic energy hasn't changed. Lift the same box vertically by 1 m at constant speed and you do 981 J on it (against gravity), and again the kinetic energy is the same at the end as the start. Where does the work go? In the first case, into friction heat. In the second, into gravitational potential energy. W = Fd cos θ doesn't care about the destination, only about how force aligns with displacement. This force work power calculator handles every angle case, including variable-force integrals where you can't just multiply. For motor sizing with efficiency, AC power factor, or unit conversions between kW and horsepower, see the Motor Efficiency Calculator.
| Unknown | Formula | Required Inputs |
|---|---|---|
| F (force) | F = ma | m, a |
| W (work) | W = Fd cos θ | F, d, θ |
| P (power) | P = W/t = Fv | W and t, or F and v |
| d (distance) | d = W/(F cos θ) | W, F, θ |
| t (time) | t = W/P | W, P |
Work and power problems start with the same step that force problems do: draw the free-body diagram. List every force acting on the body. Mark which ones are aligned with the motion, which oppose it, and which are perpendicular. Pick a coordinate axis along the displacement. The work each force does is then F · d cos θ where θ is the angle between that specific force and the displacement direction.
What's usually constant: the applied force in basic problems, gravity (mg, downward), and the angle of an incline. What changes: position, velocity, kinetic energy. Friction can be either constant (μ N where N is steady) or position-dependent (if N varies along a curved track). For this calculator, the constant-force assumption holds. If your real problem has F = F(x), the closed-form W = Fd cos θ doesn't apply and you need W = ∫F · dx.
Power is the rate of energy transfer. Instantaneous power is P = F · v (the dot product), so for constant force aligned with motion, P = Fv. For a body moving at constant speed up a hill, the power delivered by the motor exactly matches the rate of energy loss to gravity plus friction. That's the steady-state condition you size motors for. If the system is accelerating, you need extra power to feed the rising kinetic energy, which is what makes startup loads exceed cruise loads.
There are really only three equations: F = ma, W = F · d (with cos θ for off-axis forces), and P = W/t = F · v. Everything else is a substitution.
Cosine factor quick reference:
Power has two faces. P = W/t works when you have total work and total time. P = Fv works when you have instantaneous force and velocity at a single moment. They're equivalent because W = Fd and d = vt at constant velocity, so P = Fd/t = Fv. The two-formula choice is about which inputs you have, not about which is "more correct."
For variable force, swap multiplication for an integral. A spring obeys F = kx (Hooke's law), so the work to compress from 0 to x is W = ∫₀ˣ kx' dx' = ½kx². Gravity at orbital distances obeys F = GMm/r², so escape work is ∫F dr from R_earth to infinity. Constant-force shortcuts don't apply, but the integral always does. This calculator uses trapezoidal integration on user-supplied (force, position) data points when you have measured curves rather than analytic functions.
One more decision: when the problem asks for work done by a specific force versus net work. The work-energy theorem is about net work. ΔKE = W_net. If you compute the work done by you on a sliding box and ignore friction, you won't get the kinetic energy change. You'll get something larger. Sum the work from every force to get net.
Work is a signed scalar. Positive work means the force pushed the object along the displacement direction. Negative work means the force pushed against the displacement. Friction always does negative work on a sliding body because it points backward. Gravity does negative work on an object moving up and positive work on an object moving down.
The cosine in W = Fd cos θ carries the sign automatically if you measure θ as the angle between the force vector and the displacement vector, treating both as positive magnitudes. A 30° pull above the horizontal of a sled moving forward gives cos 30° ≈ 0.87, so W is positive and equals 0.87 · F · d. A 30° pull behind the sled (force pointing back, displacement forward) gives θ = 150° and cos 150° = −0.87, so W is negative.
On an inclined plane, two sign traps wait. First: gravity has a component mg sin θ along the surface (downhill) and mg cos θ perpendicular to it. Pick a positive axis up the slope. If the body moves up, gravity does negative work (force down-slope, motion up-slope). If down, gravity does positive work. Second: friction always opposes motion, so its sign relative to your axis flips depending on which way the body is moving.
Power signs follow work signs. Negative power means the body is delivering energy back to the agent, which happens for a regenerative-braking electric vehicle on a downhill or a person catching a falling weight. The motor torque equation P = τω uses the same convention: a generator sees negative shaft power because torque opposes rotation.
Real machines rarely operate in a single steady state. A conveyor accelerates the load, runs at constant speed, then decelerates at the discharge. Power demand differs across all three.
Motor catalogs publish both continuous and short-time ratings precisely because peak power during startup can exceed cruise power by 2x or more. Sizing the motor for cruise alone leads to thermal damage during acceleration. The total work done across a cycle is the integral of power over time, which equals the area under the P vs. t curve. Energy-per-cycle drives the duty rating.
For a problem about lifting a 50 kg load 3 m vertically at constant speed, the force approach goes: F_net = 0 means the lift force equals mg = 490.5 N. Work done by the lifter = F · d = 490.5 × 3 = 1471.5 J. The energy approach goes: ΔPE = mgh = 50 × 9.81 × 3 = 1471.5 J. Same answer, different starting point.
Force methods win when the question asks about a specific force, an applied force at a specific instant, or the reaction at a support. These quantities aren't accessible from energy bookkeeping alone. Sizing a motor needs F or τ, which forces a force-based analysis at the steady-state condition.
Energy methods win when the path is complicated but you only need start and end states. A roller-coaster cart at the top of a 30 m drop arrives at the bottom with v = √(2g · 30) ≈ 24.3 m/s independent of whether the track is straight, curved, or wiggles. Path doesn't matter for conservative forces. Trying the same problem with F = ma needs Newton's laws integrated along the curve, which is painful.
For variable forces, energy almost always wins. Springs, gravity over interplanetary distances, magnetic compression, all are easier as ½kx² or U(r) integrals than as F = ma differential equations. Halliday and Resnick teach force first because it's more intuitive, but most working physicists default to energy for any non-trivial problem.
A cyclist plus bike weighs 80 kg. The grade is 6 percent, which corresponds to arctan(0.06) ≈ 3.43°. The rider climbs at a steady 5 m/s (18 km/h, a moderate uphill pace for a fit amateur). Rolling resistance coefficient is 0.005 (clincher tires on smooth pavement). Aerodynamic drag at 5 m/s is roughly 12 N for an upright rider (CdA ≈ 0.65 m², air density 1.225 kg/m³, F_drag = ½ ρ CdA v² ≈ 9.95 N, rounded up for clothing flap and helmet).
Step 1: gravity component along the slope
F_gravity = mg sin(3.43°) = 80 × 9.81 × 0.0599 = 47.0 N
The cyclist needs to overcome this component or they slide back.
Step 2: rolling resistance
F_roll = μ_r · mg cos(3.43°) = 0.005 × 80 × 9.81 × 0.9982 = 3.92 N
Rolling resistance scales with the normal force, which is mg cos θ on the slope. The cosine correction is tiny at 6 percent (less than 0.2 percent off mg).
Step 3: total resistive force
F_total = 47.0 + 3.92 + 12.0 = 62.9 N
Step 4: power at the rear hub
P = F · v = 62.9 × 5 = 314.5 W
That's the mechanical power leaving the wheel, not the rider's metabolic power.
Drivetrain efficiency for a clean chain is around 97 percent, so the pedal power is about 314.5 / 0.97 ≈ 324 W. Human metabolic efficiency converting food to mechanical pedal power is around 22 to 25 percent, so the rider is burning about 1300 to 1480 W of metabolic energy. Translated to dietary calories: roughly 280 kcal per hour at this output. A 30-minute climb at this pace burns ~140 kcal beyond resting metabolism.
The same rider on flat ground at 5 m/s needs only F_roll + F_drag = 3.92 + 12.0 ≈ 15.9 N, so P ≈ 80 W. The 6 percent grade more than tripled the power demand. That's why hills hurt. Gravity work scales linearly with grade, while drag and roll stay roughly the same.
This calculator is designed for physics coursework, homework verification, and conceptual exploration. It uses idealized constant-friction and constant-force models. Real engineering applications (motor sizing, structural design, vehicle dynamics) require additional factors like efficiency losses, dynamic loads, and safety margins that this tool does not model. For professional applications, consult engineering standards and qualified engineers.
Real questions from students stuck on angle factors, unit conversions, and motor sizing.
Work in physics is force times displacement in the direction of the force: W = F · d · cos(θ), where θ is the angle between the force vector and the direction of motion. The unit is the joule (J), which equals one newton-meter. The cosine matters. Push a 100 kg crate horizontally with 200 N over 5 m, and you've done 1000 J of work. Holding the same crate stationary at chest height, no matter how tired your arms get, counts as zero work. The lifting force is vertical; the floor is horizontal. Carry the crate across that floor and you've still done zero physics work, because cos(90°) = 0. Work is a scalar but carries a sign. Force in the direction of motion does positive work. Force opposite to motion (friction, braking) does negative work. The work-energy theorem ties this back to kinetic energy: net work equals the change in KE, W_net = ΔKE = ½mv² − ½mu². For a variable force, work becomes the integral W = ∫F · dx. A spring stretched to displacement x has stored elastic energy ½kx², which is the work done against the spring force.
They're mathematically identical but suited for different situations. Use P = W/t when you know total work and total time (like lifting a load a certain height over some seconds). Use P = Fv when you know instantaneous force and velocity (like a car engine producing thrust at highway speed). Since W = Fd and d = vt, both formulas reduce to the same thing.
The arithmetic checks out. The angle doesn't. Work equals Fd cos θ, and cos(90°) = 0. If your force is perpendicular to the motion (the normal force on a sliding block, or holding a heavy suitcase while walking) no work is done because the force isn't moving the object in the direction of the force. You can push on a wall all day and do zero physics work because the wall doesn't move.
Probably friction or the incline angle. For an inclined conveyor, you need force to overcome both gravity (mg sin θ) AND friction (μ mg cos θ). Forgetting friction can underestimate required power by 20-50%. Also check your efficiency—real motors are 70-90% efficient, so the electrical input power is higher than mechanical output power.
Rearrange W = Fd cos θ to get cos θ = W/(Fd), then take the inverse cosine: θ = arccos(W/(Fd)). If your calculator gives radians, multiply by 180/π to convert to degrees. Watch the domain: arccos only works for values between -1 and +1. If W/(Fd) is outside that range, something's wrong with your inputs.
There are actually three different 'horsepowers': mechanical/imperial hp = 745.7 W, metric horsepower (PS, CV) = 735.5 W, and electrical hp = 746 W exactly. Car engines in the US use mechanical hp; European car specs often use PS (metric). The differences are small (~1.4%) but matter for precise conversions. When in doubt, use 746 W/hp as a round number.
Common ballpark values: rubber on concrete ~0.7, steel on steel (dry) ~0.5, wood on wood ~0.4, steel on steel (oiled) ~0.1, ice ~0.03. For physics homework without a given μ, the problem might want you to assume frictionless (μ = 0) or to solve symbolically. For engineering estimates, use conservative (higher) values to ensure adequate motor sizing.
You need calculus (or the derived formula). Spring force varies as F = kx, so using W = Fd with the final force overestimates work. The correct formula is W = ½kx² for compressing from 0 to x. If you don't have calculus, use this derived result. The calculator handles variable-force integration automatically when you select that mode.
Calculate power: P = Fv = (mg)(v) = (500)(9.8)(2) = 9,800 W ≈ 13 hp. But this is theoretical minimum—you need to add margins: efficiency losses (motor ~85% efficient → need ~11,500 W input), startup torque (motors need extra power to accelerate loads), and safety factor (typically 1.25–1.5×). Specify at least a 15–20 hp motor for this application.
Work is positive when force and motion are in the same direction (you add energy). Work is negative when force opposes motion (you remove energy). Friction always does negative work because it opposes the direction of sliding. When you catch a ball, your hands do negative work—removing kinetic energy from the ball and stopping it.
Use integration: W = ∫F(x)dx. If you have discrete data points (measured force at several positions), use trapezoidal approximation: sum [(F_i + F_{i+1})/2 × Δx] for each segment. The calculator's variable-force mode does this automatically—enter your (position, force) data pairs and it computes total work.
Learn more about force, work, and power from these authoritative physics resources:
Comprehensive physics education on work, energy, and power
Educational resource on force, motion, and Newton's laws
Georgia State University physics reference on work and energy
Official SI unit definitions from the National Institute of Standards and Technology
Disclaimer: This calculator provides educational estimates based on classical mechanics principles. For professional engineering applications, consult qualified engineers and refer to relevant industry standards.