10/10 Spring Potential Energy  Text: Chapter 6 Energy  HW 10/10 “Spring Energy” due Monday 10/14  Exam 2 Thursday, 10/ Wit Wit 114 (only if needed) Please send if other time needed

The Kinetic Bucket... KE + PE g + PE s = E

PE Gravity Bucket... KE + PE g + PE s = E hh

A new bucket... KE + PE g + PE s = E xx

Force and Springs Force is proportional to the displacement from “relaxed” length. Relaxed Length 10 N 20 N 30 N xx xx xx xx xx xx

HOOKE’S LAW F = - k  x FORCE SPRING CONSTANT DISPLACEMENT

Force constant “k”  Tells us how “stiff” the spring is.  Stiff spring = large k  Units: N/m (“Newton per Meter”)  Meaning: “How many Newton’s of force for each meter of stretch or compression”

That Pesky Minus Sign note: minus sign just tells us that the force opposes the displacement Use your head, not the math F = - k  x

xx An external agent (the hand) pushes on a mass and compresses a spring with spring constant k a distance  x. KE + PE g + PE s = E i KE + PE g + PE s = E f

xx An external agent (the hand) pushes on a mass and compresses a spring with spring constant k a distance  x. KE + PE g + PE s = E i KE + PE g + PE s = E f +WORK

xx Work = F  x Hooke’s Law F = k  x Work = (k  x)  x = k  x 2 ??? KE + PE g + PE s = E i KE + PE g + PE s = E f F = k  x F = 0 F ave = 1/2 k  x Work = F ave  x Work = (1/2 k  x)  x = 1/2 k  x 2 PE s

Energy stored in a spring (PE s )  energy is stored whether spring is compressed or stretched   x measured from relaxed spring length  amount stored given by: PE s = 1/2 k  x 2

 k =0.2 AB 10kg A 10kg block compresses a spring 0.5m and is then released... KE + PE g + PE s = E

 k =0.2 AB  x=0.5m k=16000N/m How much energy is stored in the spring? 10kg KE + PE g + PE s = E

 k =0.2 AB  x=0.5m k=16000N/m We release the block. 10kg KE + PE g + PE s = E

 k =0.2 AB  x=0.5m k=16000N/m How much kinetic energy does the block have at point A? 10kg KE + PE g + PE s = E 10kg

 k =0.2 AB  x=0.5m k=16000N/m What is the speed of the block at point A? 10kg KE + PE g + PE s = E 10kg

 k =0.2 AB  x=0.5m k=16000N/m How far will the block slide past point B before it comes to a stop? 10kg KE + PE g + PE s = E 10kg

h=? 2 m  K = 0.3 v = 0  x = 0.5 m k = 160 N/m A BC Practice Problem: A block of mass M = 10 kg slides from rest down a frictionless ramp and drops a distance h. After crossing a rough patch the block compresses a spring with k = 160 N/m an amount  x = 0.5 m and comes to a momentary stop. Find h.