Work Sign and Direction of Motion

Why the Sign of Work Matters

A force can act on an object without doing positive work. The sign depends on how the force is directed compared with the object's displacement. A force that helps the displacement does positive work; a force that resists it does negative work. If there is no displacement, the force does no work on the object, even if the force itself is large.

This distinction matters whenever several forces act at once. For example, while an object is being lifted, the lifting force and gravity point in opposite directions. Their work therefore has opposite signs. The sign belongs to the work done by a particular force, so it is important to specify which force is being considered.

Definition and Sign Rule

Definition: For a constant force \(\mathbf{F}\) acting through a displacement \(\mathbf{d}\), the work done by that force is \(W=\mathbf{F}\cdot\mathbf{d}\). If the angle between the force and displacement is \(\theta\), this is \(W=|\mathbf{F}||\mathbf{d}|\cos\theta\). Work is measured in joules (J), where \(1\text{ J}=1\text{ N}\cdot\text{m}\).

The dot product selects the component of the force in the direction of the displacement. When force and displacement point in the same direction, that component is positive. When they point in opposite directions, it is negative. When the force is perpendicular to the displacement, its component along the displacement is zero.

\begin{aligned} 0\leq \theta<90^\circ &\quad\Longrightarrow\quad W>0,\\ \theta=90^\circ &\quad\Longrightarrow\quad W=0,\\ 90^\circ<\theta\leq 180^\circ &\quad\Longrightarrow\quad W<0. \end{aligned}

There is another way for work to be zero: the displacement can be zero. A force on a stationary object does no work on that object during the interval, because the force has not acted through any displacement.

In one-dimensional problems, choose a positive direction and give the force and displacement signed values in that direction. The work formula becomes \(W=F\Delta x\). Its sign follows from multiplying those signed values. For example, if right is positive, a force directed left is negative, and a displacement to the left is also negative. A leftward force acting through a leftward displacement therefore does positive work.

Why Direction Determines the Sign

For a constant force, only the force component parallel to the displacement contributes to work. The parallel component is \( |\mathbf{F}|\cos\theta\), so multiplying it by the displacement length gives \( |\mathbf{F}||\mathbf{d}|\cos\theta\). Since the lengths are nonnegative, the sign of the work is the sign of \(\cos\theta\). Cosine is positive for an angle below \(90^\circ\), zero at \(90^\circ\), and negative for an angle above \(90^\circ\). This establishes the sign rule.

In a straight-line example, the same reasoning is especially direct: \(W=F\Delta x\). If force and displacement have the same sign, their product is positive; if their signs differ, the product is negative; and if either is zero, the product is zero. These are not separate rules for pushing, lifting, or lowering. They are applications of the same comparison between force direction and displacement direction.

Worked Examples

Worked Example: Pushing a Box and Accounting for Friction

A box moves \(3\text{ m}\) to the right. A person pushes it with a constant force of \(12\text{ N}\) to the right, while friction exerts a \(5\text{ N}\) force to the left. Find the work done by each force and the total work done by these two forces.

Choose right as the positive direction. The displacement is \(\Delta x=+3\text{ m}\). The pushing force is \(F_{\text{push}}=+12\text{ N}\), so

W_{\text{push}}=F_{\text{push}}\Delta x =(+12)(+3)=+36\text{ N}\cdot\text{m}=+36\text{ J}.

The pushing force points along the displacement, so its work is positive. Friction points left, so \(F_{\text{friction}}=-5\text{ N}\). Its work is

W_{\text{friction}}=F_{\text{friction}}\Delta x =(-5)(+3)=-15\text{ N}\cdot\text{m}=-15\text{ J}.

The total work by the two specified forces is the sum of their individual works:

W_{\text{total}}=W_{\text{push}}+W_{\text{friction}} =36+(-15)=21\text{ J}.

Check: The pushing force assists the rightward motion, so \(+36\text{ J}\) has the expected sign. Friction opposes that motion, so \(-15\text{ J}\) has the expected sign. The total is also \(21\text{ J}\), since the net force is \(12-5=7\text{ N}\) to the right and \((7)(3)=21\text{ J}\).

Worked Example: Lifting an Object

An object moves upward \(2\text{ m}\). During this displacement, a person exerts an upward force of \(40\text{ N}\), and gravity exerts a downward force of \(40\text{ N}\). Find the work done by each force.

Choose upward as positive. Then \(\Delta y=+2\text{ m}\), the person's force is \(F_{\text{person}}=+40\text{ N}\), and gravity's force is \(F_{\text{gravity}}=-40\text{ N}\). For the person's force,

W_{\text{person}}=F_{\text{person}}\Delta y =(+40)(+2)=+80\text{ N}\cdot\text{m}=+80\text{ J}.

The person's force and the displacement point upward, so this work is positive. For gravity,

W_{\text{gravity}}=F_{\text{gravity}}\Delta y =(-40)(+2)=-80\text{ N}\cdot\text{m}=-80\text{ J}.

Gravity points down while the object moves up, so gravity does negative work. The sum of these two works is \(+80+(-80)=0\text{ J}\).

Check: The forces balance: \(+40+(-40)=0\text{ N}\). Their work values also cancel for the same displacement. The zero total does not mean that neither force did work; each did nonzero work, with opposite signs.

Worked Example: Lowering a Load with a Supporting Rope

A load is lowered \(5\text{ m}\) at constant speed. During the motion, gravity exerts a \(60\text{ N}\) downward force and the rope exerts a \(60\text{ N}\) upward tension. Find the work done by each force.

Choose downward as positive. The displacement is \(\Delta y=+5\text{ m}\). Gravity points downward, so \(F_{\text{gravity}}=+60\text{ N}\). Its work is

W_{\text{gravity}}=F_{\text{gravity}}\Delta y =(+60)(+5)=+300\text{ N}\cdot\text{m}=+300\text{ J}.

The rope points upward, opposite the displacement, so \(F_{\text{rope}}=-60\text{ N}\). The rope's work is

W_{\text{rope}}=F_{\text{rope}}\Delta y =(-60)(+5)=-300\text{ N}\cdot\text{m}=-300\text{ J}.

The total work by these forces is \(+300+(-300)=0\text{ J}\).

Check: Gravity acts in the direction the load moves, giving positive work. The rope resists the downward motion, giving negative work. Their forces sum to zero, consistent with constant speed in this stated situation; their work values also sum to zero.

Worked Example: Pushing Against a Wall That Does Not Move

A person pushes horizontally on a rigid wall with a force of \(150\text{ N}\), but the wall does not move. Find the work done by the person's force on the wall during this interval.

The force is nonzero, but the displacement is \(\Delta x=0\text{ m}\). Therefore,

W=F\Delta x=(150)(0)=0\text{ N}\cdot\text{m}=0\text{ J}.

Check: The work formula gives zero because the displacement is zero. It would be incorrect to infer positive work merely from the fact that the person is pushing hard: work on the wall depends on displacement as well as force.

Keep the Force and the Work Being Asked About Clear

When multiple forces act on an object, ask whose work is being calculated. “The work done by gravity” and “the work done by the person lifting” are different quantities. They can have different signs during the same motion. The total work is found by adding the work done by each force included in the question.

Also separate the direction of a force from the direction of motion. Lowering does not automatically mean that every force does negative work. In the load example, gravity points in the same direction as the downward displacement and therefore does positive work. The rope points against the displacement and does negative work. The sign is determined force by force, not by labeling the whole situation “lifting” or “lowering.”

Takeaway: For each force, compare its direction with the displacement. Same direction gives positive work; opposite direction gives negative work; perpendicular force or zero displacement gives zero work.

A Common Sign Mistake

A common error is to assign a negative sign to a force just because the object is moving downward. The sign of a force is set by the chosen positive direction, not by the motion. For instance, in the lowering example, choosing downward as positive gives \(\Delta y=+5\text{ m}\) and \(F_{\text{gravity}}=+60\text{ N}\). The incorrect calculation \(W_{\text{gravity}}=(+60)(-5)=-300\text{ J}\) mixes sign conventions: it treats downward displacement as negative while keeping downward force positive.

There are two consistent choices. With downward positive, use \(F_{\text{gravity}}=+60\text{ N}\) and \(\Delta y=+5\text{ m}\), giving \(+300\text{ J}\). With upward positive, use \(F_{\text{gravity}}=-60\text{ N}\) and \(\Delta y=-5\text{ m}\), also giving \(+300\text{ J}\). Either convention works, but force and displacement must use the same coordinate choice.

Check Your Understanding

For each question, identify the direction of the force and displacement before deciding the sign of the work.

A \(10\text{ N}\) force acts to the right while an object moves \(4\text{ m}\) to the right. Is the work positive, negative, or zero? Calculate it.
An object moves left while a force acts right. What is the sign of the work done by that force?
A force acts vertically downward while an object moves horizontally. What is the work done by that force?
A person lowers an object at constant speed. Explain why gravity can do positive work while the supporting force does negative work.
A force acts on an object that remains at rest. What work does that force do on the object during that interval?