Question

Hi, please provide the correct answer ASAP. circle the answers clearly. This is the 2nd time I've had to ask this so plz. come through.

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-/12.5 POINTS EXAMPLE 13.1 Simple Harmonic Motion on a Frictionless Surface GOAL Calculate forces and accelerations for a horizontal spring system. PROBLEM A 0.350-kg object attached to a spring of force constant 1.30 x 10- N/m is free to move on a frictionless horizontal surface. If the object is released from rest at x = 0.100 m, find the force on it and its acceleration at x = 0.100 m, x = 0.0500 m, x = 0 m, x = -0.0500 m, and x = -0.100 m STRATEGY Substitute given quantities into Hooke's law to find the forces, then calculate the accelerations with Newton's second law. The amplitude A is the same as the point of release from rest, x = 0.100 m. SOLUTION Write Hooke's force law. Fs = -kx Substitute the value for k and take x = A = 0.100 m, finding the force at that point. Fmax = -KA = -(1.30 x 102 N/m)(0.100 m) -13.0 N Solve Newton's second law for a and substitute to find the acceleration at X = A. ma = Fmax a Fmax = -13.0 N = -37.1 m/s2 m 0 .350 kg Repeat the same process for the other four points, assembling a table. Position (m) 0.100 0.050 Force (N) -13.0 -6.50 Acceleration (m/s) -37.1 -18.6 0 +18.6 +37.1 -0.050 -0.100 +6.50 +13.0 LEARN MORE

Answer 1

Question Third option is correct which says .. Hooke's law will not be applicable if spring is stretched beyond elastic limit. Question m = 0.440kg , k = 1.00 X 10 N/m n = 0.100 m, using Fa-kx, F=-1.00 Xió XO.100 N F:-10.0N Then using Newton's and law, f=ma acceleration a = F = -10.0 3 - 22.7 mys3 0.440_ Now at x = 0.0som Fe-1.00 X 10 X 0.05 N [F=-5:00N and a = F = ano (= -1.umpse)

... at n=0, F = 0 and a=o at u= 0.050 m, F +5.00 N la= +11.4 m/821 at no 0.100m F - HO.ONI la = +22.7 m/s? Exercise a= +9.20m/s² s0 F = ma = 0.440x9,20 = F = 4.05 N & using farka ºg -f =-4.05 - roox 10² = -0.0405m)