The concept of gravitational potential energy would be of little interest or use if it applied only to vertical free fall. Let’s begin to expand the idea. FIGURE 10.7 shows a particle of mass $m$ moving at an angle while acted on by gravity. How much work does gravity do?

FIGURE 10.7 Gravity does work on a particle moving at an angle.

Gravity is a constant force. In Chapter 9 you learned that, in general, the work done by a constant force is $W=\vec{F}\cdot \text{Δ}\vec{r}$. If we write both $\vec{F}{}_{\text{G}}$ and $\text{Δ}\vec{r}$ in terms of components, and use the Chapter 9 result for calculating the dot product with components, we find that the work done by gravity is

Because $\vec{F}{}_{\text{G}}$ has no $x$-component, the work depends only on the vertical displacement $\text{Δ}y$.

Consequently, if we define the system to be object $\text{+}$ earth, the change in the system’s gravitational potential energy depends only on the object’s vertical displacement. This is true not only for motion along a straight line, as in Figure 10.7, but also for motion along a curved trajectory because a curve can be represented as the limit of a very large number of very short straight-line segments.

For example, FIGURE 10.8 shows an object sliding down a curved, frictionless surface. The change in gravitational potential energy of the object $+$ earth system depends only on $\text{Δ}y$, the distance the object descends, not on the shape of the curve. But now there’s an additional force—the normal force of the surface. Does this force affect the system’s energy? No! The normal force is always perpendicular to the box’s instantaneous displacement, and you learned in Chapter 9 that forces perpendicular to the displacement do no work. Forces always perpendicular to the motion do not affect the system’s energy. They can be ignored during an energy analysis.

FIGURE 10.8 For motion on any frictionless surface, only the vertical displacement $\text{Δ}y$ affects the energy.

EXAMPLE 10.3 The speed of a sled

Christine runs forward with her sled at $2.0\text{m/s}$. She hops onto the sled at the top of a 5.0-m-high, very slippery slope. What is her speed at the bottom?

MODEL Let the system consist of the earth and the sled, which we model as a particle. Because the slope is “very slippery,” we’ll assume that friction is negligible. The slope exerts a normal force on the sled, but it is always perpendicular to the motion and does not affect the energy.

VISUALIZE FIGURE 10.9a shows a before-and-after pictorial representation. We are not told the angle of the slope, or even if it is a straight slope, but the change in potential energy depends only on the vertical distance Christine descends and not on the shape of the hill. Figure 10.9b is an energy bar chart in which we see an initial kinetic and potential energy being transformed into entirely kinetic energy as Christine goes down the slope.

FIGURE 10.9 Pictorial representation and energy bar chart of the sled $\text{+}$ earth system.

Full Alternative Text

Full Alternative Text

SOLVE The energy analysis is just like in Example 10.2; the fact that the object is moving on a curved surface hasn’t changed anything. The energy principle, written as a conservation statement, is

$$\begin{array}{ccc}{K}_{\text{i}}+U_{\text{Gi}}& =& \frac{1}{2}m{v}_0{}^2+mg{y}_0\hfill \\ & =& K_{\text{f}}+U_{\text{Gf}}=\frac{1}{2}m{v}_1{}^2+0\end{array}$$

Her speed at the bottom is

$$\begin{array}{ccc}\hfill v_1& \hfill =\hfill & \sqrt{v_0{}^2+2g{y}_0}\hfill \\ \hfill & \hfill =\hfill & \sqrt{(2.0{\text{m/s)}}^2+2(9.80{\text{m/s}}^2)(5.0\text{m})}\hfill \\ \hfill & \hfill =\hfill & 10\text{m}\text{/}\text{s}\hfill \end{array}$$

REVIEW $10\text{m/s}\approx 20\text{mph}$ is fast but believable for a $5\text{m}\approx 15\text{ft}$ descent.