Guide 8-6a. Conservation of Energy Example 1

This is an example of a combined energy-circular motion problem.  The method of solution shown can also be applied to pendulums and swings where you're asked to find the tension in a string or rope.  The problem is number 8-52 on page 236. We'll give complete solutions but provide fewer comments in blue than for previous examples.

Given: vA = 8.0 m/s yA = 1.75 m yB = 0 m = 61 kg r = 12 m Goal:  Find the normal force on the skier at the bottom of the depression. system = skier, Earth external forces: normal (does no work) initial state = skier at point A final state = skier at point B  ΔUg is negative since the skier decreases in elevation;  ΔK is positive since the skier speeds up
	First, use conservation of energy to solve for the speed of the skier at the bottom of the depression.
	Next, do a net force problem to find the normal force.  The acceleration is up (toward the center of the circle) at the bottom of the depression, so the direction of positive is selected pointing up. The net force equation is solved for N.  Then the result for vB² from the previous conservation of energy problem is substituted. The result has units of newtons and is positive as expected for a force magnitude. The value of 1100 N (about 250 lbs) seems high. This is greater than the weight of the skier. But then, the depression is fairly deep.

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