Question

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m. While at this equilibrium position, the mass is then given an initial push downward at v = 5.1 m/s. The block oscillates on the spring without friction.
1)What is the spring constant of the spring?
2)What is the oscillation frequency?
3)After t = 0.42 s what is the speed of the block?
4)What is the magnitude of the maximum acceleration of the block?
5)At t = 0.42 s what is the magnitude of the net force on the block?

Answer 1

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

    k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

      k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

    $\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

   $f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

   $f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

  $f =0.94\ Hz$

c)  $v_b = -v cos \omega t$

    $v_b = -5.1 \times cos (5.92 \times 0.42)$

    $v_b = 4.04\ m/s$

d)  $a_{max} = v \omega$

    $a_{max} = 4.04 \times 5.92$

    $a_{max} =23.94\ m/s^2$

e)  $Y =- A sin (\omega t)$

    $A = \dfrac{v}{\omega}$

    $A = \dfrac{4.04}{5.92}$

        A = 0.682 m

    $Y =- 0.682 \times sin (5.92 \times 0.42)$

    $Y =- 0.42$

Force =$m \omega^2 |Y|$

          =$6.4 \times 5.92^2\times 0.42$

F = 94.20 N