An unusual spring has a restoring force of magnitude F = (2.00 N/m) x + (1.00 N/

Question

An unusual spring has a restoring force of magnitude F = (2.00 N/m)x + (1.00 N/m2)x 2, where x is the stretch of the spring from its equilibrium length. A 3.00-kg object is attached to this spring and released from rest after stretching the spring 1.40 m. If the object slides over a frictionless horizontal surface, how fast is it moving when the spring returns to its equilibrium length? An unusual spring has a restoring force of magnitude F = (2.00 N/m)x + (1.00 N/m2)x 2, where x is the stretch of the spring from its equilibrium length. A 3.00-kg object is attached to this spring and released from rest after stretching the spring 1.40 m. If the object slides over a frictionless horizontal surface, how fast is it moving when the spring returns to its equilibrium length?

Explanation / Answer

PE = integral of F dx = 0.5*2*x^2 + 1/3*1*x^3

Ei = Ef

0.5*2*1.4^2 + 1/3*1*1.4^3 = 0.5*3*v^2

v=1.38 m/s

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