Integration of a function can also be done by determining the area under a curve for the specified function. xx FxFx AA To find the total area under the curve we sum all the small area segments. For infinitesimally small area segments:  A – Small segment of total area defined by rectangle. F x – height of segment  A.  x – length of segment  A. The integral gives us the same result as obtained for the summation with more precision. The smaller  x is the more precisely we can match the curve. FxFx x If we have a plot of force vs. position we can obtain an estimate of the amount of work done to move from an initial position to a final position, by finding the total area under the segment of the curve between the two positions.

Work done by a Spring A spring has a rest length that is defined when the spring is unloaded (no force applied). The spring resists the applied force by exerting a force in the opposite direction. This force is called a restoring force, since it attempts to return the spring to its rest length (equilibrium). We can mathematically model this using Hooke’s Law.  x – Spring displacement [m] k – spring constant [N/m] F – Restoring Force [N] - How far the spring is stretched or compressed from its initial length. All lengths are measured relative to the equilibrium position of the spring (x = 0). Equilibrium Compressed Stretched F F When a force is applied to stretch or compress the spring, what does the spring do? Opposes applied force

-x x FsFs x FsFs -F s  = 180 o, so cos  = -1 Same! The work done by a spring was derived from both the definition of work and the area under the curve. Both methods resulted in the same expression. The amount of work done to stretch or compress a spring depends on the displacement of the spring from equilibrium. For a pre-stretched (or pre-compressed) spring it is necessary to look at the energy for the initial and final location and then subtract to determine how much the energy changed by. Let us look at the work required to stretch or compress a spring.

Let us now look at the work done by a net force.  = 0 o, so cos  = 1 Maximum kinetic energy that an object can have. Change in kinetic energy due to a change from an initial velocity v 0 to a final velocity v. Work – kinetic energy theorem. In order to change the kinetic energy and hence the speed of an object you must do work on the object. [KE] = J = Nm = kg m 2 /s 2 Energy and work have the same units.

A cart on an air track is moving at 0.5 m/s when the air is suddenly turned off. The cart comes to rest after traveling 1 m. The experiment is repeated, but now the cart is moving at 1 m/s when the air is turned off. How far does the cart travel before coming to rest? 1. 1 m 2. 2 m 3. 3 m 4. 4 m 5. 5 m 6. impossible to determine If v o is doubled, than v o 2 is four times as great, and therefore d 2 must be 4 times greater than d 1.