Mechanical work

First name the force whose work is required: rope tension, gravity or friction, for example. During the same motion one force may perform positive work while another performs negative work. For a constant force and straight displacement, $W=Fs\cos\alpha$. Here $F$ and $s$ are positive magnitudes; the angle between their directions determines the sign. Only the sum of all contributions gives net work.[1]

For extended bodies, follow the force’s application point. Centre-of-mass displacement alone does not capture every contribution in deformation or rotation. The simple calculations below use a particle or a body without rotation.[1]

Resolve force into a component along the displacement and a perpendicular component. Only the first contributes to this work. The perpendicular rope component may change the normal force and hence friction, but it still contributes no direct work during purely horizontal displacement. The angle α is between the selected force and the relevant displacement of its application point. A rope angle measured from another reference direction must first be converted.[1]

Zero work does not require zero force. A perpendicular force can change direction without changing speed. Likewise, a large force does not automatically mean large work: without displacement of its application point, that force performs no mechanical work.[1]

The general definition sums contributions along the actual path. Each small piece uses the tangential force component at that location. Do not multiply an initial or final value of a varying force by the whole distance. In a one-dimensional force–position graph, signed area under the force component gives work; areas below the position axis subtract.[1]

For a constant force vector, take it outside the integral: $\int_A^B\mathrm d\vec r=\vec r_B-\vec r_A$. This yields the dot product with endpoint displacement. For straight motion without reversal, its magnitude equals distance s. Along a curved path, total distance generally exceeds the magnitude of endpoint displacement.

At $\alpha=0^\circ$, $W=+Fs$; at 180°, $W=-Fs$; at 90°, $W=0$. An inward force during circular motion is instantaneously perpendicular to the tangent and does no work. An additional tangential force can still do work and change speed.[1]

Original sign check: a body slides 2 m horizontally. A 30 N pulling force contributes +60 J; a constant opposing friction force of 10 N contributes −20 J. Weight and a perpendicular normal force contribute zero along this horizontal path. The sum is +40 J.

Holding a bag still gives $W=0$ for the holding force at its stationary application point. Muscles nevertheless consume chemical energy. In ideal steady horizontal carrying, vertical support is also perpendicular to displacement; lifting, setting down and real oscillations are separate intervals.

Consider a 2.0 kg crate sliding horizontally rightward. A constant horizontal pull of 30 N acts against 10 N sliding friction. Weight and normal force balance vertically. Over 2.0 m, neither vertical force performs work. The horizontal force balance and the work balance describe the same process from different viewpoints.[2]

Force $F$ is measured in newtons and displacement $s$ in metres. Their projected product gives work $W$ in $\mathrm{N\,m}=\mathrm J$. A force component remains a force measured in newtons, not joules. When $\cos\alpha=0$, division cannot determine a unique force or displacement.[1]

Use degree mode for degree angles and radian mode for radian angles. Work and torque can both have newton-metre units but are different quantities: work uses a dot product; torque uses a cross product. Energy is therefore expressed as joules.

A rearrangement does not automatically yield a physically admissible magnitude. Negative work with $\alpha=0^\circ$ contradicts positive F and s. With perpendicular force, $W=0$ does not determine distance: any distance gives the same work.

For a linear spring, force grows with extension. Added elastic potential energy between initial and final extension follows from integrating the external stretching force; it is not final force times displacement. The second example explicitly starts with a pre-extended spring.[3]

In the particle model, net work changes kinetic energy. Add every work contribution on the same body with its sign. Work by one pulling force equals net work only when the remaining contributions sum to zero.[2]

During steady lifting, external work is $+mg\Delta h$ and gravitational work is $-mg\Delta h$. Net work is zero because speed is unchanged, while potential energy of the body–Earth system rises. External lifting work and gravitational work therefore have opposite signs.[3]

For friction between real bodies, conversion to internal energy matters. A rolling or deformable object may require more than a particle balance: include rotation, deformation and contact-point motion.

During steady lifting, the gravitational potential energy of the body–Earth system increases. Lifting force and gravity perform opposite work on the raised body; the example balances both and checks that kinetic energy is unchanged.[3]

Before substituting, check whether force and direction stay constant and s describes the relevant displacement. Integrate a spring’s varying force. For constant-magnitude friction along a stationary track, use total travelled distance with negative sign, even if the body returns to its starting point.

Area under Fx(x) is useful for one-dimensional problems. Regions below the x-axis subtract. A force-versus-time curve gives impulse, not work; displacement or velocity is still needed.

Identify which force works on which object. For constant force use its dot product with displacement; for varying force use the path integral. Add positive and negative work contributions before finding an energy change. Force–position area is work; force–time area is impulse.