F.B Posted November 20, 2005 Posted November 20, 2005 I am stuck on part b of this question. 2. A 0.20 kg mass is hung from a vertical spring of force constant 55 N/m. When the spring is released from its unstretched equilibrium position, the mass is allowed to fall. Use the law of conservation of energy to determine a)The speed of the mass after it falls 1.5 cm b)The distance the mass will fall before reversing direction. How do we determine the maximum stretch. I know how to do a) but i dont kno how to do b). The answer for a is 0.48 m/s. Can someone please help me.
Klaynos Posted November 21, 2005 Posted November 21, 2005 When something is fully stretched just like a pendulum at the top of it's arc it's velocity is 0, knowing that you should be able to do pretty much the reverse of what you did for part a
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