Energy Diagrams

61 Energy Diagrams

Finally, let use what we have learned to find an expression for the force in terms of the associated potential energy.

Exercise 40.3: Force and Potential Energy

For simplicity, we will solve this problem in 1D only. Consider the work $dW$ done by a force as an object moves a distance $dx$ . Write the work–energy theorem for this displacement, and then “divide” by $dx$ to find an expression for the force in terms of its potential energy.

The above result is another “must know.” Fortunately, as you saw above, it’s not really new information. If you remember that work is minus the change in potential energy, then you can quickly figure out that $F=-dU/dx$ . In three dimensions, this is generalized to

$\vec F = -\nabla U = - \left[ (\partial U/\partial x)\hat x + (\partial U/\partial y)\hat y + (\partial U/\partial z)\hat z$ .

If “upside triangle” stands for the gradient of the potential energy function. You can verify that this expression for the force works as you might expect.

Exercise 40.4:

Verify that if $\vec F = -\nabla U$ , then the work done from a small displacement $\Delta \vec x$ is equal to minus the change in the potential energy.

Talk about gradients, maximizing $\Delta U$ by making sure $\Delta \vec x$ is along $\vec F$ .

61 Energy Diagrams

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