# Doubts about work-energy theorem and negative gravitational potential energy

A ball is dropped from a certain height. The work done on the ball by the gravitational force is

$$W = F \cdot h = mgh$$

and the gravitational potential energy is

$$U = -mg h$$

The difference in kinetic energy is

$$\Delta K = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 = \frac{1}{2} m v_2^2$$

where $$v_1 = 0$$ since ball starts falling from rest. The total energy change is

$$W_1 + U_1 + K_1 = W_2 + U_2 + K_2$$

where $$W_1 = 0$$, $$U_1 = -mgh$$, $$K_1 = 0$$, $$W2 = mgh$$, $$U_2 = 0$$, and $$K_2 = \frac{1}{2} m v_2^2$$. The sum is then

$$0 + (-mgh) + 0 = mgh + \frac{1}{2} m v_2^2$$

which equals

$$-2mgh = \frac{1}{2} m v_2^2$$

Questions:

• What am I exactly doing wrong here?
• We either use work or potential energy and not both? But aren't they both energies?
• Why make a distinction if you could just use the work done by gravity?
• Why don't we calculate the work done by gravity as $$W = F h = mgh$$?
• Why do we use the difference in potential energy: $$U_2 - U_1 = 0 - (-mgh) = +mgh$$?
• Does gravity perform work on the particle?

TL;DR There are few things wrong in how you understand the work-energy theorem:

• the work $$W$$ does not enter the energy diagram from both sides;
• if you include gravitational potential energy $$U$$ separately in the energy diagram, then you should not include it again as work $$W$$;

## The work-energy theorem

The work-energy theorem says that the total work done on an object equals change in its kinetic energy

$$\Delta K = K_2 - K_1 = W \qquad \text{or} \qquad \boxed{K_1 + W = K_2}$$

where $$W$$ is total work done on an object. This means work done by all forces acting on the object! Note that by definition, work is a scalar value that can be either positive or negative.

One thing people usually forget is that the work done by gravitational and elastic (spring) forces equals negative change in the potential energy:

$$W_g = -\Delta U_g \qquad \text{and} \qquad W_e = -\Delta U_e$$

where $$U_g = mgy$$ is gravitational potential energy and $$U_e = \frac{1}{2} k x^2$$ is elastic potential energy. Since difference $$\Delta$$ always means final minus initial value, the work by gravitational and elastic forces is defined as

$$W_g = U_{g,1} - U_{g,2} \qquad \text{and} \qquad W_e = U_{e,1} - U_{e,2}$$

If we use this in the work-energy theorem we get

$$\boxed{K_1 + U_{g,1} + U_{e,1} + W_\text{other} = K_2 + U_{g,2} + U_{e,2}}$$

where $$W_\text{other}$$ is work done by forces other than gravitational and elastic forces.

## Example: A freely-falling ball

As you have correctly identified, the change in kinetic energy can be negative, but the kinetic energy itself cannot. In example of dropping the ball from a height $$h$$, the energy diagram would be

$$\underbrace{\frac{1}{2} m 0^2}_{K_1} + \underbrace{mgh}_{U_{g,1}} = \underbrace{\frac{1}{2} m v_1^2}_{K_2} + \underbrace{mg0}_{U_{g,2}}$$

or in a general case

$$\frac{1}{2} m v_1^2 + mgy_1 = \frac{1}{2} m v_2^2 + mgy_2$$

which might be more familiar in this form

$$\boxed{v_2^2 = v_1^2 - 2g(y_2 - y_1)}$$

You must be careful what you use for $$y_1$$ and $$y_2$$ - subscript 1 denotes the starting point and 2 denotes the ending point.

## Work by gravitational force equals negative $$\Delta U_g$$

As to why work done by gravitational force is negative change of potential energy, we start from the definition of work:

$$W = \vec{F} \cdot \vec{x} \qquad \text{or} \qquad W = \int \vec{F} \cdot d\vec{x}$$

where both force and displacement are vectors, which means they have both magnitude and direction. The work is defined as a scalar product of force and displacement, which means that only portion of force parallel to displacement does work, whereas perpendicular portion does no work. The scalar product finds that portion of work parallel to displacement. If force is constant and displacement is a straight line you can use the (simpler) left-hand side equation, but in all other cases you should use the right-hand side equation.

Let positive $$\hat{\jmath}$$ axis point upwards (away from Earth's center), then the gravitational force is

$$\vec{F}_g = -mg\hat{\jmath}$$

where $$\hat{\jmath}$$ is just a unit vector (magnitude equals to one) which defines direction of the force. The work done by gravitational force $$\vec{F}_g$$ over some distance $$\Delta \vec{y}$$ is

$$W_g = \vec{F}_g \cdot \Delta \vec{y} = -mg\hat{\jmath} \cdot (y_2 - y_1) \hat{\jmath} = -(mgy_2 - mgy_1) = -\Delta U_g$$

where $$\hat{\jmath} \cdot \hat{\jmath} = 1$$ is a scalar product between two unit vectors.

• But why don't you calculate the work done by gravity $W = F h = mgh$ and why do you use the difference in potential energy then? $U_2 - U_1 = 0 - - mgh = + mgh$ I mean gravity is performing work on the particle? Jan 26, 2022 at 8:42
• @bananenheld Check if my most recent edit answers your question. You are correct, in case of freely falling object, the gravitational force does positive work: $W_g = -\Delta U_g = -(U_{g,2}-U_{g,1})$ and $W_g = -(mg0-mgh) = mgh$ which increases kinetic energy of the body. You must be careful what you put for $y_1$ and $y_2$ in the above equation - 1 means starting point, 2 means ending point. Jan 26, 2022 at 9:00
• Thanks for your very detailed answer. I understand it better now. But I still get wrong minus signs. I'm working on a basic problem, with a ball on a spring, then shoots of it in the air. Then I get for a point of the ball in the air: $K_2 - K1 = W = - \Delta U$ $1/2 mv^2 - 0 = W = - (U_f - U_b) = - (U_{air} - U_{ground}) = - (U_{grav} - U_{spring}) = - ((-mgy - - 0) - 1/2 k \Delta y^2) = mgy + 1/2 k \Delta y^2$ But I should get: $1/2 mv^2 + mgy = 1/2 k \Delta y^2$, where $k \Delta y^2$ is amount that the spring got compressed. The minus signs confuse me. Jan 26, 2022 at 10:11
• Work done by a spring is defined in the same way, i.e. as negative change of the elastic potential energy $W_e = -\Delta U_e$. It is difficult to answer your (new) question in comments. Feel free to ask a new question with this new system. Please focus your question on concept rather than actual numbers.. Jan 26, 2022 at 10:56

You have made a fundamental mistake in writing

The potential energy is: $$U = - mg h$$

What you should have written is the change in potential energy is: $$\Delta U = U_{\rm final} -U_{\rm initial} = - mg h$$ and if you assume $$U_{\rm initial} = 0$$.

You have also double counted in that in your equation $$W_1 + U_1 + T_1 = W_2 + U_2 + T_2$$, the $$W_1$$ and $$W_2$$ refers to work other than that done by the gravitational force as work done by the gravitational forces is already accounted for in the change in potential energy.

$$U_{\rm initial} + K_{\rm initial} = U_{\rm final} + K_{\rm final} \Rightarrow K_{\rm final} - K_{\rm initial} = U_{\rm initial}-U_{\rm final}$$

$$\Rightarrow \frac 12 m v_{\rm final}^2 - 0 = \frac 12 m v_{\rm final}^2 = -(-mgh) = mgh$$