Work Done Against Gravity

Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PE_g)put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PE_g gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

Converting Between Potential Energy and Kinetic Energy

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PE_g to KE without explicitly considering the intermediate step of work. (See Example 2.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.

More precisely, we define the change in gravitational potential energy ΔPE_g to be

where, for simplicity, we denote the change in height by h rather than the usual Δh. Note that h is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is

Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the work.

Using Potential Energy to Simplify Calculations

The equation ΔPE_g = mgh applies for any path that has a change in height of h, not just when the mass is lifted straight up. (See Figure 2.) It is much easier to calculate mgh (a simple multiplication) than it is to calculate the work done along a complicated path. The idea of gravitational potential energy has the double advantage that it is very broadly applicable and it makes calculations easier. From now on, we will consider that any change in vertical position h of a mass m is accompanied by a change in gravitational potential energy mgh, and we will avoid the equivalent but more difficult task of calculating work done by or against the gravitational force.

Work Done By The Gravitational Force

The concept of potential energy (and the change in potential energy) can also be used to quickly determine the work done by the gravitational force. For example, in Figure 1a, when the mass moves upwards the gravitational force acts in the opposite direction to motion. This means that even though the gravitational potential energy increased, the work done by the gravitational force was negative. That is, the work done by the gravitational force is the negative of the change in potential energy,

[latex]W_{\text{g}} = -\Delta U_{\text{g}}[/latex]

In Figure 1b, when the mass moves downwards, the gravitational force acts in the same direction as the motion. In this case, the change in potential energy is negative but the work down by gravity is positive (which follows from [latex]W_{\text{g}} = -\Delta U_{\text{g}}[/latex]).

It turns out that this concept holds for any conservative force, not just the gravitational force (you will see another conservative force – the spring force – in the next section). In general, the work done by a conservative force ([latex]W_{\text{cons}}[/latex]) is equal to the negative of the change in potential energy related to that force ([latex]U[/latex]),

[latex]W_{\text{cons}} = -\Delta U[/latex]

Example 2: Finding the Speed of a Roller Coaster from its Height

(a) What is the final speed of the roller coaster shown in Figure 4 if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m/s?

Strategy

The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal force, which is perpendicular to the direction of motion and does no work. The net work on the roller coaster is then done by gravity alone. The loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy. This can be written in equation form as -ΔPEg = ΔKE. Using the equations for PE_g and KE, we can solve for the final speed v, which is the desired quantity.

Solution for (a)

Here the initial kinetic energy is zero, so that [latex]\boldsymbol{\Delta\textbf{KE}=\frac{1}{2}mv^2}.[/latex] The equation for change in potential energy states that ΔPE_g = mgh. Since h is negative in this case, we will rewrite this as ΔPE_g = –mg|h| to show the minus sign clearly. Thus,

[latex]\boldsymbol{-\Delta\textbf{PE}_{\textbf{g}}=\Delta\textbf{KE}}[/latex]

becomes

[latex]\boldsymbol{mg|h|\:=}[/latex][latex]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{mv^2}.[/latex]

Solving for v, we find that mass cancels and that

[latex]\boldsymbol{v=\sqrt{2g|h|}}.[/latex]

Substituting known values,

[latex]\begin{array}{lcl} \boldsymbol{v} & \boldsymbol{=} & \boldsymbol{\sqrt{2(9.80\textbf{ m/s})(20.0\textbf{ m})}} \\ {} & \boldsymbol{=} & \boldsymbol{19.8\textbf{ m/s.}} \end{array}[/latex]

Solution for (b)

Again -ΔPE_g = ΔKE. In this case there is initial kinetic energy, so [latex]\boldsymbol{\Delta\textbf{KE}=\frac{1}{2}mv^2-\frac{1}{2}{mv_0}^2}.[/latex] Thus,

[latex]\boldsymbol{mg|h|\:=}[/latex][latex]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{mv^2-}[/latex][latex]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{{mv_0}^2}.[/latex]

Rearranging gives

[latex]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{mv^2=mg|h|+}[/latex][latex]\boldsymbol{\frac{1}{2}}[/latex][latex]\boldsymbol{{mv_0}^2}.[/latex]

This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass again cancels, and

[latex]\boldsymbol{v=\sqrt{2g|h|+{v_0}^2}}.[/latex]

This equation is very similar to the kinematics equation[latex]\boldsymbol{v=\sqrt{{v_0}^2+2ad}},[/latex]but it is more general—the kinematics equation is valid only for constant acceleration, whereas our equation above is valid for any path regardless of whether the object moves with a constant acceleration. Now, substituting known values gives

[latex]\begin{array}{lcl} \boldsymbol{v} & \boldsymbol{=} & \boldsymbol{\sqrt{2(9.80\textbf{ m/s}^2)(20.0\textbf{ m})+(5.00\textbf{ m/s})^2}} \\ {} & \boldsymbol{=} & \boldsymbol{20.4\textbf{ m/s.}} \end{array}[/latex]

Discussion and Implications

First, note that mass cancels. This is quite consistent with observations made in Chapter 2.7 Falling Objects that all objects fall at the same rate if friction is negligible. Second, only the speed of the roller coaster is considered; there is no information about its direction at any point. This reveals another general truth. When friction is negligible, the speed of a falling body depends only on its initial speed and height, and not on its mass or the path taken. For example, the roller coaster will have the same final speed whether it falls 20.0 m straight down or takes a more complicated path like the one in the figure. Third, and perhaps unexpectedly, the final speed in part (b) is greater than in part (a), but by far less than 5.00 m/s. Finally, note that speed can be found at any height along the way by simply using the appropriate value of h at the point of interest.

We have seen that work done by or against the gravitational force depends only on the starting and ending points, and not on the path between, allowing us to define the simplifying concept of gravitational potential energy. We can do the same thing for a few other forces, and we will see that this leads to a formal definition of the law of conservation of energy.