A 4.4×102 kg block on a horizontal frictionless surface is attached to a spring

Question

A 4.4×102 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 340 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). When the displacement is x = 1.2×102 m, the kinetic energy of the block is closest to: A 4.4×102 block on a horizontal frictionless surface is attached to a spring whose force constant is 340 . The block is pulled from its equilibrium position at = 0 to a displacement = +0.080 and is released from rest. The block then executes simple harmonic motion along the -axis (horizontal). When the displacement is = 1.2×102 , the kinetic energy of the block is closest to:

A: 1.7 J

B: 0.55 J

C: 1.1 J

D: 0.73 J

E: 2.2 J

Explanation / Answer

Total energy = Kinetic energy + Potential energy

Total energy = 0.5*k*A^2 = 0.5*340*0.08^2 = 1.088 J

Potential energy U = 0.5*k*x^2 = 0.5*340*0.012^2 =0.02448J

Kinetic energy is KE = TE-U = 1.088-0.02448 = 1.06352 J = 1.1 J (Closest value )

SO the answer is C) 1.1 J

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