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A 1.55 kg frictionless block is attached to an ideal spring with force constant 340 N/m . Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 14.0 m/s.
A.) Find the amplitude of the motion.
B.) Find the maximum acceleration of the block.
C.) Find the maximum force the spring exerts on the block.

Sagot :

Answer:

Explanation:

Let amplitude be A . At displacement A , velocity will be zero , so

1/2 k A² = 1/2 m v²

340 A² = 1.55 x 14²

A = .9452 m

= 94.52 cm

B ) angular frequency of oscillation ω = √ ( k / m )

= √ ( 340 / 1.55 )

ω = 14.81 radian /s

maximum acceleration = ω²A

= 14.81² x 0.9452

= 207.32 m/s²

C) Maximum force = mass x maximum acceleration

= 1.55 x 207.32

= 321.36  N .

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