A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretched nor compr

Question

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 11.0 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.

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Helga 4 years 2021-08-24T16:28:15+00:00 1 Answers 17 views 0

Answers ( )

  1. Amity
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    2021-08-24T16:29:22+00:00
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    Answer:

    A.    A = 0.913 m

    B.    amax = 132.24m/s^2

    C.    Fmax = 324.01N

    Explanation:

    When the block is moving at the equilibrium point , its velocity is maximum.

    A. To find the amplitude of the motion you use the following formula for the maximum velocity:

    v_{max}=A\omega          (1)

    vmax = maximum velocity = 11.0 m/s

    A: amplitude of the motion = ?

    w: angular frequency = ?

    Then, you have to calculate the angular frequency of the motion, by using the following formula:

    \omega=\sqrt{\frac{k}{m}}           (2)

    k: spring constant = 355 N/m

    m: mass of the object = 2.54 kg

    \omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

    Next, you solve the equation (1) for A and replace the values of vmax and w:

    A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

    The amplitude of the motion is 0.913m

    B. The maximum acceleration of the block is given by:

    a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

    The maximum acceleration is 132.24 m/s^2

    C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

    F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

    It is also possible to calculate the maximum force by using:

    Fmax = k*A = (355N/m)(0.913m) = 324.01N

    The maximum force exertedbu the spring on the object is 324.01 N

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