Question 7.9: The relationship (7.52) between the potential energy functio...
The relationship (7.52) between the potential energy function and a conservative force is illustrated for two situations. (i) Given V(x), find the conservative force F(x) ; and, conversely, (ii) given a conservative force F(x) , find the potential energy function V(x). When V(x) is known, the work done is readily determined by (7.45 ).
\mathbf{F}(\mathbf{x})=-\nabla V(\mathbf{x}) (7.52)
\mathscr{W}=-\Delta V, (7.45 )
(i) If V(x) is given, then the force F(x) derived from (7.52) is a conservative force. For example , consider
V(\mathbf{x})=-\frac{c y^2}{2} + d, (7.54a)
where c and d are constants. Then (7.47) gives \nabla V=-c y \mathbf{j}, and (7.52) shows that the force \mathbf{F}=-\boldsymbol{\nabla} V=c y \mathbf{j} is conservative.
\nabla V(\mathbf{x}) \equiv \frac{\partial V}{\partial x} \mathbf{i} + \frac{\partial V}{\partial y} \mathbf{j} + \frac{\partial V}{\partial z} \mathbf{k} (7.47)
(ii) Conversely, suppose we know that a force F(x) is conservative. Then (7.52) holds and the potential energy function V(x) may be found by integration of (7.53). For example, we know that F(x) = cyj is a conservative force, hence (7.53) become
\frac{\partial V}{\partial x}=-F_x, \quad \frac{\partial V}{\partial y}=-F_y, \quad \frac{\partial V}{\partial z}=-F_z . (7.53)
\frac{\partial V}{\partial x}=0, \quad \frac{\partial V}{\partial y}=-c y, \quad \frac{\partial V}{\partial z}=0 . (7.54b)
The first and last of these equations show that V is independent of x and z. Hence, V(x) = V(y) is at most a function of y ; and the second equation in (7.54b) becomes d V / d y=-c y. The solution of this equation is given by (7.54a) in which d is an arbitrary integration constant that may be chosen to meet any convenient purpose . For example, we may wish to define V(0) = 0; then d = 0 for this choice.
We recall that F = cyj is the same force encountered earlier in (7.26a) and for which the work done in (7.26b) is independent of the path. An easier calculation for the work done by a conservative force is now provided by the rule (7.45). Let the potential energy be given by (7.54a), for example. Then, by (7.45), the work done on any path between the end points \mathbf{x}_1 = (0,0) and \mathbf{x}_2 = (1 , a) is
\mathbf{F}=c y \mathbf{j} (7.26a)
\mathscr{W}=\int_6 \mathbf{F} \cdot d \mathbf{x}=\int_{(0,0)}^{(1, a)} c y d y=\int_0^a c y d y=\frac{c a^2}{2} (7.26b)
\mathscr{W}=-\Delta V=-\left[V\left(\mathbf{x}_2\right) – V\left(\mathbf{x}_1\right)\right]=-\left[-\frac{c a^2}{2} + d-d\right]=\frac{c a^2}{2}, (7.54c)
precisely the result derived differently in (7.26b). Clearly, the physical dimensions of potential energy are those of work: [V]=[\mathscr{W}]=[F L].