5.2.1 Gravitational Potential Energy

Consider a mass m rising at a constant speed through a vertical height of Δh. This mass is being pulled down by the gravitational force ${{F}_{g}}=mg$ . So to keep it rising at a constant speed, an upward external force ${{F}_{{ext}}}$ of magnitude mg is required.

Since KE is unchanged, the work done by F_ext must be equal to the change in GPE of the mass.

$\begin{aligned}\Delta GPE&=(F\Delta s)\\&={{F}_{{ext}}}\Delta h\\&=mg\Delta h\end{aligned}$

We have thus derived the formula for GPE. From now on, instead of calculating the work done against/by the gravitational force, we can simply calculate the gain/loss of GPE.

Example

A 2.0 kg mass being pulled from rest by a 30 N force through a vertical height of 4.0 m and horizontal distance of 3.0 m. Calculate the final KE of the mass. For convenience, take $g=10\text{ m }{{\text{s}}^{{-2}}}$ .

Solution

There are two possible approaches.

Approach 1

We note that the mass will be experiencing the gravitational pull ${{F}_{g}}=mg=(2.0)(10)=20\text{ N}$ . So

$\displaystyle \begin{aligned}\text{Work done by 30 N}+\text{ Work done by }{{F}_{g}}&=\Delta KE\\(30\times 5.0)\text{ }+\text{ }(-20\times 4.0)\text{ }&=\Delta KE\\\Delta KE&=70\text{ J}\end{aligned}$

Approach 2

We note that the mass has gained GPE and KE. So

$\displaystyle \begin{aligned}\text{Work done by 30 N}&=\Delta KE+\Delta GPE\\(30\times 5.0)\text{ }&=\Delta KE+(20\times 4.0)\\\Delta KE&=70\text{ J}\end{aligned}$

External vs Field Forces

In the 1^st approach, we are applying the work-energy theorem in its original form.

$F\Delta s=\Delta KE$

We are treating the gravitational force as just any ordinary external force. That’s why the LHS involves mg and the RHS does not have the GPE term.

In the 2^nd approach, we have adapted the work-energy theorem to incorporate changes in GPE (besides changes in KE).

$F\Delta s=\Delta KE+\Delta GPE$

Instead of treating gravity as a force (doing work on the mass), we treat gravity as a field with an associated GPE. That’s why the RHS now includes the GPE term. That’s also why the LHS must not involve mg anymore.

Both approaches are equally valid. But the superiority of the potential energy approach will be more apparent in future when we incorporate other more complicated forms of potential energies of other types of fields, where the field forces are not constant.

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Video Explanation

Derivation of the GPE Formula

Concept Test

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