A 0.20-kg object is attached to a spring with spring constant k = 10 N/m and mov
ID: 1746976 • Letter: A
Question
A 0.20-kg object is attached to a spring with spring constantk = 10 N/m and moves with simple harmonic motion over ahorizontal frictionless surface. At the instant that it isdisplaced from equilibrium by -0.050 m, what is itsacceleration? a. 1 000 m/s2 b. -40 m/s2 c. 0.1 m/s2 d. 2.5 m/s2i put a and i was wrong help A 0.20-kg object is attached to a spring with spring constantk = 10 N/m and moves with simple harmonic motion over ahorizontal frictionless surface. At the instant that it isdisplaced from equilibrium by -0.050 m, what is itsacceleration? a. 1 000 m/s2 b. -40 m/s2 c. 0.1 m/s2 d. 2.5 m/s2 a. 1 000 m/s2 b. -40 m/s2 c. 0.1 m/s2 d. 2.5 m/s2 A 0.20-kg object is attached to a spring with spring constantk = 10 N/m and moves with simple harmonic motion over ahorizontal frictionless surface. At the instant that it isdisplaced from equilibrium by -0.050 m, what is itsacceleration?
Explanation / Answer
we know that : Force on a spring F = 1/2 ( k x2) From Newton's law : F = ma m a = 1/2 ( k x2 ) Acceleration ( a ) = 1/2m ( k x2 ) = [ 1/ ( 2 * 0.2 ) ] ( 10 * ( - 0.05)2 ) = 0.0625 m /s2 ˜ 0.1 m /s2 Hope this helps u! ˜ 0.1 m /s2 Hope this helps u!Related Questions
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