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 1.25 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 1.25 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

energy stored in the spring

U = x2 +x3/3

for x = 1.25m

U = 2.2135 J

potential energy is converted into kinetic energy

U = 0.5 mV^2

so substituting values we get

V = 1.21 m/sec

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