Work Formula Calculator

Work is energy transferred by a force acting through a displacement:

$W = Fd\cos\theta$

Symbols and SI units:

• $W$ — work, joules (J), where $1\ \text{J} = 1\ \text{N·m}$
• $F$ — magnitude of the constant force, newtons (N)
• $d$ — magnitude of the displacement, metres (m)
• $\theta$ — angle between the force and the displacement, degrees or radians

When the force is along the displacement, $\cos 0° = 1$ and the formula collapses to $W = Fd$. When the force is perpendicular, $\cos 90° = 0$ and the work is zero — which is why carrying a suitcase horizontally does no work against gravity.

The assumption people forget: $W = Fd\cos\theta$ requires a constant force along a straight path. For a varying force, work is the integral $W = \int \vec{F} \cdot d\vec{s}$.

Work is a signed scalar, not a vector:

• $0° \le \theta < 90°$: positive work, energy is added to the object
• $\theta = 90°$: zero work
• $90° < \theta \le 180°$: negative work, energy is removed (friction, braking)

Net work is the sum over every force acting, or equivalently $W_{\text{net}} = F_{\text{net}}d\cos\theta$. It links directly to kinetic energy:

$W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$

with $m$ in kilograms and $v$ in m/s.

Lifting at constant speed: the applied force equals the weight $mg$, so $W = mgh$ in joules, and gravity does $-mgh$, giving zero net work and no change in speed.

The assumption people forget: an object can be under a large force and still have zero work done on it — a wall pushed hard but not moved receives no work at all.

Units check: a newton times a metre is a joule, so an answer left in N·m is already in the right unit and needs no conversion.

• Using force times distance travelled — the formula needs displacement, and only the component of force along it.
• Omitting $\cos\theta$ — an angled push does less work than its full magnitude suggests.
• Treating work as a vector — it is a scalar with a sign, so works from several forces simply add.
• Forgetting $g$ in lifting problems — to lift $15$ kg you must supply $mg = 147$ N, not $15$ N.
• Confusing work with power — power is work per unit time, $P = W/t$, in watts.
• Reporting the wrong unit — the joule, not the newton. A newton-metre used as torque is a different quantity despite the identical dimensions.
• Double-counting a force — either sum the work done by each force separately, or use the net force once, never both.
• Thinking a curved path invalidates the formula — it does not, as long as the force magnitude and the angle to the path stay constant; otherwise integrate along the path.