Work

Learning Goals

After this lesson, you should be able to:

• identify the system, force, and displacement used in a work calculation;
• calculate work by a constant force using a dot product;
• determine whether work is positive, negative, or zero;
• add the work done by multiple forces;
• interpret work as signed area on a force–position graph;
• apply the net work–kinetic energy theorem; and
• design an experiment comparing measured force–position area with kinetic-energy change.

1. Work Is Energy Transfer Across a Boundary

Before calculating work, identify the system. A force exerted by something outside the system can transfer energy across its boundary while the point of application moves.

“Work done by A on B” names both the force agent and the receiving system. Changing the system boundary can change whether the same interaction is described as external work or as a change among internal energy stores.

2. Constant-Force Work

For a constant force \(\vec F\) and displacement \(\Delta\vec r\),

Here \(\theta\) is the angle between the force and displacement vectors, placed tail to tail. Equivalently, use the parallel component:

4. Positive, Negative, and Zero Work

The sign describes the relation between one force and displacement. It is not a statement that the force itself is “positive” or “negative.”

5. Units and Scalar Addition

Work is a scalar. Add the signed work values of individual forces algebraically, not by vector addition:

A newton-meter used for work is called a joule; do not confuse it with the same dimensional combination used to report torque.

6. Worked Example: Angled Pull

Question: A person pulls a cart \(8.0\,\mathrm m\) horizontally with a constant \(50\,\mathrm N\) force directed \(30^\circ\) above horizontal. Find the work done by the pulling force.

Answer: The pulling force does about \(350\,\mathrm J\) of positive work. Its vertical component does no work during the horizontal displacement.

7. Work by Several Forces

Suppose the same cart also experiences \(20\,\mathrm N\) of friction opposite the \(8.0\,\mathrm m\) displacement. Weight and normal force are perpendicular to the horizontal motion.

Zero work by a force does not mean that the force is absent; it can be perpendicular to the displacement.

8. Force–Position Graphs

For one-dimensional motion with a position-dependent force, work is the signed area under the \(F_x\)-versus-\(x\) graph:

In AP Physics 1, areas are commonly evaluated using rectangles, triangles, trapezoids, or estimated grid regions. The integral notation summarizes the area accumulation; a graph can be analyzed algebraically without calculus.

9. Worked Example: Piecewise Force Graph

Question: From \(x=0\) to \(2.0\,\mathrm m\), \(F_x\) rises linearly from 0 to \(6.0\,\mathrm N\). From \(x=2.0\) to \(5.0\,\mathrm m\), it remains at \(6.0\,\mathrm N\). Find the work.

Answer: The force does \(24\,\mathrm J\) of positive work. If a region lay below the axis, its area would be subtracted.

10. Net Work–Kinetic Energy Theorem

The net work done by all external forces on a particle or modeled object equals its change in translational kinetic energy:

Positive net work increases kinetic energy, negative net work decreases it, and zero net work leaves speed unchanged. Direction may still change when the net force is perpendicular to motion.

11. Worked Example: Find Final Speed

Question: A \(4.0\,\mathrm{kg}\) cart initially moves at \(2.0\,\mathrm{m/s}\). The net work on it is \(42\,\mathrm J\). Find its final speed.

Answer: The final speed is \(5.0\,\mathrm{m/s}\). Energy alone does not specify the final velocity direction.

12. When a Force Does Zero Work

• A normal force on an object moving along a stationary horizontal surface is perpendicular to displacement.
• An ideal centripetal force in uniform circular motion is perpendicular to instantaneous velocity and displacement.
• A force exerted while its point of application does not move transfers no mechanical work by that force.

These statements depend on the chosen system and actual displacement. A normal force from a moving platform, for example, can transfer energy.

13. Path and Force Dependence

For one constant uniform force, work depends on the net displacement through \(\vec F\cdot\Delta\vec r\). For friction, the direction continually opposes local sliding, so work commonly depends on total path length. For a variable force, the force value along the entire path matters.

Do not replace path length with straight-line displacement unless the force model permits it.

14. Investigation: Compare Force Area with \(\Delta K\)

Use a cart, force sensor, motion sensor, and level track.

1. Measure cart mass and record force and position during a push.
2. Plot measured force component \(F_x\) against position.
3. Estimate work using signed trapezoidal areas.
4. Measure initial and final speeds over the same position interval.
5. Calculate \(\Delta K=\tfrac12m(v_f^2-v_i^2)\).
6. Compare work and \(\Delta K\) with uncertainties and discuss friction, sensor zero, and timing alignment.

The compared interval and system definition must match in both calculations.

15. Common Traps and Key Takeaways

• Use the angle between force and displacement, not automatically an angle from the horizontal.
• Only the parallel force component does work.
• Work is a signed scalar; force and displacement remain vectors.
• Add signed graph areas, subtracting regions below the axis.
• Do not confuse work by one force with net work.
• Zero work does not imply zero force; the force may be perpendicular to displacement.
• Net work determines the change in speed through \(\Delta K\), not the direction of velocity.