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 General Physics General Physics Help Forum Dec 18th 2017, 11:58 AM #1 Junior Member   Join Date: Dec 2017 Posts: 7 Stretched Spring & Block Problem A mass m=0.25kg that is attached to a spring with spring constant k=100 N/m is stretched 10cm to the right of its equilibrium position. After a time t=1s, the mass is how far from the equilibrium position? I am using the work of the spring and the kinetic energy of the block to solve this and I can't get the right answer. Should I be using F=kx also? Thank you!   Dec 18th 2017, 12:54 PM #2 Senior Member   Join Date: Aug 2010 Posts: 434 Well, not "f= kx" because you have already said k= 100, positive. I would have used "f= -kx". You are told that k= 100 N/m so f= ma= md^2x/dt^2= -100x so that d^x/dt^ -100x/m where m is the mass of the object, 0.25, do that d^2x/dt^2= -400x. Knowing that the derivative of sin(at) is acos(at) and the derivative of cos(at) is -asin(at) so that the second derivative of sin(at) is -a^2 sin(at) and the second derivative of cos(at) is -a^2cos(at) so both sin(20t) and cos(20t) satisfy that differential equation. The general solution is x(t)= C sin(20t)+ D cos(20t). Use the facts that x(0)= 0.10 (10 cm= 0.10 m) and dx/dt(0)= 0 to determine C and D.   Dec 18th 2017, 01:01 PM #3 Junior Member   Join Date: Dec 2017 Posts: 7 So this is more than a work energy problem? I was just using the work done by the spring and setting it equal to the KE of the block.   Dec 18th 2017, 02:14 PM #4 Senior Member   Join Date: Jun 2016 Location: England Posts: 1,005 They are looking for the oscillation in position back and forth about the equilibrium point. I guess you could use potential energy converting to kinetic energy and back, since kinetic energy = 1/2V^2 this could give you the velocity... But the force=mass times acceleration route (as used by HallsOfIvy) seems much easier. we know that f=k.x Where: f=force, x=displacement, k=spring constant. from f=m.a and f=k.x, you have: a.m = k.x Where a=acceleration, m=mass, x=displacement, k=spring constant. The complication is that the force (and thus the acceleration) changes with the amount of displacement. Note that HallsOfIvy has used -k, this is to make the acceleration negative when the displacement is positive, so the acceleration is acting in the opposite direction to the displacement, to reduce the stretch of the spring. __________________ ~\o/~ Last edited by Woody; Dec 18th 2017 at 02:19 PM.  Tags block, problem, spring, stretched Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post help Energy and Work 2 Feb 19th 2017 11:10 AM SMA777 Energy and Work 0 Oct 15th 2011 05:49 PM fireemblem13 Kinematics and Dynamics 1 Dec 30th 2009 07:23 PM Black Kinematics and Dynamics 1 Dec 15th 2009 10:05 PM mathieumg Periodic and Circular Motion 3 May 15th 2009 11:06 PM