A system consists of a vertical spring with force constant k = 1,250 N/m, length L = 1.50 m, and...


A system consists of a vertical spring with force constant k = 1,250 N/m, length L = 1.50 m, and object of mass m = 5.00 kg attached to the end. The objects are placed at the level of the point of attachment with the spring un-stretched, at position {eq}y_{i} {/eq} = L, and then it is released so that it swings like a pendulum.

Write Newton{eq}' {/eq}s second law symbolically for the system as the object passes through its lowest point. (Note that at the lowest point, r = L - yf)

Force Constant:

It is the necessary effect to produce unit compression or extension in the elastic object by applying some force. This effect varies with the applied force, modulus of elasticity, and material of the elastic object.

Answer and Explanation: 1

Given Data

  • The force constant of the spring is: {eq}F = 1250\;{\rm{N}}/{\rm{m}} {/eq}.
  • The length of the spring is: {eq}L = 1.50\;{\rm{m}} {/eq}.
  • The mass of the attached object is: {eq}m = 5\;{\rm{kg}} {/eq}.
  • The position of lowest point is: {eq}r = L - {y_f} {/eq}.

The spring force during the swing of object is:

{eq}{F_s} = - k{y_f} {/eq}

The gravity force of the object is given as:

{eq}{F_i} = mg {/eq}

The centripetal force for the swing of object at lowest position is:

{eq}{F_c} = \dfrac{{m{v^2}}}{r} {/eq}

Here, {eq}v {/eq} is the tangential speed of object at lowest position.

Apply the Newton's second law for system of object.

{eq}{F_s} = {F_i} + {F_c} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} - k{y_f} &= mg + \dfrac{{m{v^2}}}{{L - {y_f}}}\\ mg + \dfrac{{m{v^2}}}{{L - {y_f}}} + k{y_f} &= 0 \end{align*} {/eq}

Thus, the symbolic expression for system of object is {eq}mg + \dfrac{{m{v^2}}}{{L - {y_f}}} + k{y_f} = 0 {/eq}.

Learn more about this topic:

Hooke's Law & the Spring Constant: Definition & Equation


Chapter 4 / Lesson 19

After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.

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