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Guide 8-2. Solving Conservation of Energy Problems

 Solving Problems with Conservation of Energy Here are the steps involved in solving a conservation of energy problem. State what's given and what you are to find. Identify your system and specifically list what objects are included in the system. List the forces external to the system. Indicate whether or not these forces do positive, zero, or negative work on the system. Conservation of energy problems always deal with a beginning point and an ending point. These are termed the initial and final states. Decide what you will use for these states and list those choices. Then, in your solution, label related quantities with the subscripts i and f for initial and final. Identify the forms of energy in the system and how the energy terms change from the initial to the final state. Is ΔK positive, negative, or zero? How about ΔUg and ΔUe? Draw a diagram in which you show the initial and final states and specify the positive axis directions and the origin. Choose the origin this way:  a) If there's an elastic potential energy change in the problem, pick the origin at the relaxed position of the spring. This will ensure that the elastic potential energy of the relaxed spring is 0. b) If the only potential energy change is gravitational, the choice of origin is arbitrary but is often selected to be the lowest point.  Select +y to be up, as this will help you avoid sign difficulties later. Write the general conservation of energy equation, Wext = ΔEsys. Apply what you know to the solution of the problem.  Substitute terms for initial and final energy changes on the right-hand side of the equation. Then expand the energy changes in terms of initial and final terms: Kf, Ki, Ugf, Ugi, Uef, and Uei. Solve for the unknown symbolically. Apply the usual checks of signs, units, and sense. Substitute values and units and reduce.

Example 1

Problem: Consider a block sliding down a frictionless, inclined plane. The problem is to find the speed vf of the box of mass m after the box has slid a particular distance, d, along the plane. We'll assume the box is moving initially with speed, vi. The angle of the plane above the horizontal is θ. For this example, we'll number the steps to correspond to the numbers above.

 1. Given:  d, m, vi, θ, g     Goal: Find vf in terms of the givens. 2. System: block, Earth 3. External force: normal (does no work) 4. initial state: highest position of block     final state: lowest position of block 5. ΔUg is negative since the block decreases in elevation;     ΔK is positive since the block speeds up The diagram below shows (among other things): the direction of +y (up) the lowest position of the block designated as yf= 0 6.

7,8. Now we're ready for equations.

 We first write the general conservation of energy law. Wext is 0, because the normal force does no work on the block. The energy changes in the system are kinetic and gravitational. Expand the delta notation in terms of initial and final quantities. The final gravitational energy is 0, because we specified the lowest position of the object to be 0. Substitute expressions for the remaining energy terms. Solve for the unknown. Use the geometry of the plane to substitute for yi in terms of given symbols. Select the positive root since we're only interested in the speed.

9. Checks:  The units reduce to m/s. The sign of the quantity under the radical is positive, so the square root yields a real number. Kf - Ki is positive since vf > vi. Ugf - Ugi = -mgyi is negative.

10. There are no values to substitute.

Example 2

Problem: A ball of mass 1.25 kg is placed in front of a spring-loaded platform mounted on a table 0.500 m above the floor as shown to the right. The spring constant of the spring is 84.5 N/m, and the spring is compressed 0.1435 m. The uncompressed position of the right edge of the spring is at the right edge of the table. At t = 0, the ball is quickly released. Determine the speed of the ball after the spring is released and the ball leaves the edge of the table.

 1. Given: m = 1.25 kg    k = 84.5 N/m xi = -0.1435 m     xf = 0 m  (Note that xi is negative due to the way the coordinate system is set up.) vi = 0 m/s     Goal:  Find vf. 2. System: spring, ball (The Earth and gravity needn't be included because the gravitational potential energy of the ball doesn't change.) 3. External forces: normal and weight (neither does work) 4. initial state: spring fully compressed     final state: spring fully decompressed 5. ΔUe is negative since the spring relaxes;     ΔK is positive since the ball speeds up The diagram below shows: the direction of +x is the the right the position of the relaxed spring is xf= 0 6.

7-10. Now for equations...

 Write the general conservation of energy law. Wext is 0, because the normal force and gravity do no work on the block. The energy changes in the system are kinetic and elastic. Expand the delta notation in terms of initial and final quantities. The initial kinetic energy and final elastic potential energy are 0. Substitute expressions for the remaining energy terms. Solve for the unknown. Substitute values with units. Select the positive root since we're only interested in the speed.

Checks:  The units m[N/(m∙kg)]½ reduce to m/s. The sign of the quantity under the radical is positive, so the square root yields a real number.

Here are more example problems (presented at a WX session).