AP Physics 1: Work by Gravity on Curved Paths

Understanding how gravity performs work is fundamental to analyzing the motion of objects on roller coasters, slides, and any other curved path. While calculating work often seems tied to straight-line motion, gravity possesses a special property that simplifies analysis tremendously: the work it does depends only on an object's change in vertical height, not on the intricate, winding path it takes. Mastering this concept unlocks the power of energy conservation, allowing you to solve complex curved motion problems with surprising ease.

Defining Work: The Foundation

In physics, work is defined as the process of energy transfer to or from an object via the application of force along a displacement. The crucial point is that work is done only by the component of the force that acts parallel to the object's displacement. The mathematical definition for a constant force moving an object in a straight line is $W=Fd\cos \theta$, where $F$ is the force magnitude, $d$ is the displacement magnitude, and $\theta$ is the angle between the force and displacement vectors. This equation is a scalar dot product: $W=\vec{F}\cdot \vec{d}$. Work is measured in joules (J), where 1 J = 1 N·m.

This formula works perfectly for constant forces along straight lines. However, most interesting paths, like a roller coaster track or a winding slide, are curved. To calculate work along such a path, we must consider it as a series of tiny, straight-line segments. For each infinitesimal displacement $d\vec{r}$, the tiny amount of work done by a force $\vec{F}$ is $dW=\vec{F}\cdot d\vec{r}$. The total work is the sum—or integral—of these tiny contributions along the entire path: $$W={\int }_{\text{path}}\vec{F}\cdot d\vec{r}.$$ This is the general definition of work done by a force along any path.

Work Done by Gravity Along a Curved Path

Let's apply this general definition specifically to the force of gravity, ${\vec{F}}_g=m\vec{g}$, where $\vec{g}$ is the acceleration due to gravity vector (magnitude $g$, directed downward). We'll analyze an object moving from an initial point $A$ to a final point $B$ along an arbitrary curved path.

We set up a coordinate system with the +y-axis pointing upward. Therefore, the gravitational force vector is ${\vec{F}}_g=-mg\hat{j}$. The infinitesimal displacement vector along the curve is $d\vec{r}=dx\hat{i}+dy\hat{j}$. The dot product for the infinitesimal work is: $$d{W}_g={\vec{F}}_g\cdot d\vec{r}=(-mg\hat{j})\cdot (dx\hat{i}+dy\hat{j})=-mgdy.$$ Notice that the $dx$ component vanishes because $\hat{j}\cdot \hat{i}=0$. This is the critical step: the horizontal component of the motion does not contribute to the work done by gravity.

To find the total work, we integrate $d{W}_g$ from the starting y-coordinate ($y_i$) to the final y-coordinate ($y_f$): $$W_g={\int }_A^{B}d{W}_g={\int }_{y_i}^{y_f}-mgdy=-mg{\int }_{y_i}^{y_f}dy=-mg(y_f-y_i).$$

This yields the central result: The work done by gravity is $W_g=-mg\Delta y$, where $\Delta y=y_f-y_i$ is the change in vertical height.

The Conservative Force and Path Independence

The result $W_g=-mg\Delta y$ has profound implications. Notice that the final expression depends only on the initial and final vertical positions ($y_i$ and $y_f$). It does not contain $x$, nor does it involve any details about the specific curved path taken—whether it was a direct drop, a gentle slope, or a loop-the-loop. Forces with this property are called conservative forces. Gravity is a quintessential example.

This means the work done by gravity on an object moving between two points is the same for any path connecting those points. Let's prove it with two paths between points $A$ and $B$:

• Path 1 (Curved Slide): A child slides down a frictionless, winding slide from $A$ to $B$. The work done by gravity is $W_{g,1}=-mg(y_f-y_i)$.
• Path 2 (Vertical Drop + Horizontal Push): Imagine the child instead falls vertically from $A$ to the height of $B$, then moves horizontally to point $B$. For the vertical segment, $\Delta y$ is the same, so work is $-mg\Delta y$. For the horizontal segment, gravity ($-mg\hat{j}$) is perpendicular to the displacement ($dx\hat{i}$), so it does zero work. Total work: $W_{g,2}=-mg\Delta y+0$.

Both paths yield identical work. This path-independence is the defining feature of a conservative force and is why we can define gravitational potential energy as $U_g=mgy$. The work done by gravity is simply the negative of the change in this potential energy: $W_g=-\Delta U_g$.

Applications to Real-World Curved Paths

This principle directly explains the physics of many common systems.

Roller Coasters: Consider a coaster car descending the first hill. No matter the hill's shape—steep or shallow, curved or straight—the work done by gravity from the top (height $h$) to the bottom (height 0) is $W_g=-mg(0-h)=mgh$. This work is converted into kinetic energy, explaining why the speed at the bottom depends only on the height lost, not the hill's design (assuming no friction). This is why roller coaster designers focus on height: it determines the energy budget for the entire ride.

Playground Slides: A child gains the same kinetic energy by sliding down a straight metal slide or a long, spiraling tube slide of the same vertical drop, if friction is negligible. The curvy path is longer, so the gravitational force acts over a longer distance, but because the force is less parallel to the path on the shallower sections, the integrated product $\vec{F}\cdot d\vec{r}$ remains constant. The longer path does not result in more work; it results in a slower acceleration and a longer ride time.

Projectile Motion: The work done by gravity on a projectile from launch to landing is determined solely by its net vertical displacement. A ball thrown upward and landing at the same height has $\Delta y=0$, so gravity does zero net work over the entire flight. Its kinetic energy at launch equals its kinetic energy at landing, which is why the launch and landing speeds are equal (ignoring air resistance).

Common Pitfalls

1. Confusing Total Work with Work Along Part of a Path.

• Pitfall: Thinking gravity does no work at the instant a projectile is at the peak of its trajectory because its velocity is horizontal.
• Correction: While the instantaneous power ($P=\vec{F}\cdot \vec{v}$) is zero at the peak, work is accumulated over the entire path. The work calculation depends on displacement, not velocity. Gravity did negative work on the way up and will do positive work on the way down. The correct analysis uses the overall $\Delta y$.

2. Including Horizontal Displacement in the Work Formula.

• Pitfall: Attempting to use $W=Fd\cos \theta$ with the total path length $d$ for a curved slope.
• Correction: The formula $W=Fd\cos \theta$ only applies for constant force and straight-line displacement. For curves, you must use the integral form or, for gravity specifically, the shortcut $W_g=-mg\Delta y$. The "d" in the shortcut is the vertical drop, not the path length.

3. Forgetting the Negative Sign in $W_g=-mg\Delta y$.

• Pitfall: Writing $W_g=mg\Delta y$ and getting the sign wrong, which leads to errors in the work-energy theorem.
• Correction: Remember the sign convention. If an object falls ($\Delta y$ is negative), the formula $W_g=-mg(\text{negative})$ yields a positive work: gravity adds kinetic energy. If an object rises ($\Delta y$ is positive), gravity does negative work, removing kinetic energy. The sign conveys the energy transfer direction.

4. Applying the Path-Independence Rule When Other Forces Are Present.

• Pitfall: Assuming the net work on an object sliding down a curved track is path-independent.
• Correction: Gravity's work is path-independent. However, if forces like friction or applied pushes are present, their work does depend on the path length. The total net work will differ for different paths. The principle of path-independence applies only to the work done by a conservative force like gravity alone.

Summary

• The work done by the force of gravity on an object of mass $m$ is given by $W_g=-mg\Delta y$, where $\Delta y$ is the change in vertical height. This is derived from the integral of ${\vec{F}}_g\cdot d\vec{r}$ along any path.
• Gravity is a conservative force because the work it does depends only on the starting and ending points (vertical positions), not on the specific path taken between them.
• This path-independence allows us to define gravitational potential energy ($U_g=mgy$), where $W_g=-\Delta U_g$.
• This principle simplifies the analysis of motion on any curved path where gravity is the only force doing work (e.g., frictionless slides, roller coasters), as the change in kinetic energy is directly linked to the change in height.
• A common and powerful application is recognizing that for a frictionless roller coaster, the speed at the bottom of a hill depends only on the height lost, not on the hill's shape.