A 0.39Kg block on a horizontal frictionless surface is attached ot a spring whos
ID: 2008308 • Letter: A
Question
A 0.39Kg block on a horizontal frictionless surface is attached ot a spring whose force constant is a 570 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 = 0.057 m, the kinetic energy of the block is closest to:A) 0.90 J B) 1.1 J C) 1.0 J D) 0.84 J E) 0.95 J
The answer key says the answer is A, but I am lost on how to get there. Please show the algerbra and concepts involved. Thank You!
Explanation / Answer
PE at any point = 1/2 k x^2
Total E = 1/2 k A^2
KE = E - PE = 1/2 k (A^2 - x^2)
k = 570
A = 0.080
x = 0.057
Substitute to get:
KE = 0.898 J = 0.9 J (A)
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