A2.00 kg, frictionless block s attached to an ideal spring with force constant 550 n/m. at t = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 10.0 m/s. for related problem-solving tips and strategies, you may want to view a video tutor solution of angular frequency, frequency, and period in shm. find: -a) the amplitude. b) the maximum acceleration of the block. c) the maximum force the spring exerts on the block.

a)    A = 0.603 m , b) a = 165.8 m / s² , c)  F = 331.7 N

Explanation:

For this exercise we use the law of conservation of energy

Starting point before touching the spring

    Em₀ = K = ½ m v²

End Point with fully compressed spring

    = = ½ k x²

    Emo =

    ½ m v² = ½ k x²

    x = √(m / k)    v

    x = √ (2.00 / 550)   10.0

    x = 0.603 m

This is the maximum compression corresponding to the range of motion

     A = 0.603 m

b) Let's write Newton's second law at the point of maximum compression

    F = m a

    k x = ma

    a = k / m x

    a = 550 / 2.00 0.603

    a = 165.8 m / s²

With direction to the right (positive)

c) The value of the elastic force, let's calculate

    F = k x

    F = 550 0.603

   F = 331.65 N

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