A 3-kg mass attached to a spring of force constant 27 N/m is free to oscillate o

Question

A 3-kg mass attached to a spring of force constant 27 N/m is free to oscillate on a horizontal , frictionless surface. The mass is displaced 5cm to the right of its equilibrium position and set into motion with a leftward push of speed 9cm/s.

Find the kinetic energy of the mass, as well as the magnitude of its acceleration, when the potential energy of the spring is 0.01J. (Hint: it is not necessary to determine the time)

I got (1/2)(3kg)(0.09m/s)^2 + 0.01 J = 0.02215J, which was completely wrong.

Please provide a step by step !!!!!!! and explanation

Explanation / Answer

We can solve this by energy conservation method, we know that total mechanical energy is conserved everytime unless an external force is applied during the motion, so, initial mechanical energy = .5*k*x^2 + .5*m*v^2 (ie. P.E + K.E) = .5*27*.05^2 + .5*3*.09^2 = 0.0459J So, at the time P.E is 0.01J, so, P.E = 0.5*kx^2 = 0.01 so, x = sqrt(0.01*2/k) = 2.72 cm so, force acting = kx = 27*.0272 = 0.735 N = ma so, a = 0.735/m = 0.735/3 = 0.245 m/s2

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