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When a 3.80 kg object is hung vertically on a certain light spring that obeys Hooke's law, the...

Question:

When a {eq}\displaystyle 3.80\text{ kg object } {/eq} is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches {eq}\displaystyle 2.60 cm. {/eq}

If the {eq}\displaystyle 3.80\text{ kg object } {/eq} is removed, how far will the spring stretch if a {eq}\displaystyle 1.50kg {/eq} block is hung on it?

{eq}\displaystyle \boxed{\space}cm. {/eq}

Spring Force:

  • Hooke's Law is applicable to springs up to the proportional limit. In this limit, the change in the length of the spring is directly proportional to the magnitude of restoring force developed in the spring.
  • The relation between the spring force {eq}(F){/eq}, spring constant {eq}(k){/eq}, and the change in the length of the spring {eq}(x){/eq} are given as:

{eq}\hspace{1cm} F= k x {/eq}

Answer and Explanation:


We have the following given data

{eq}\begin{align} \\ M_1 &=3.80\text{ kg} \\[0.3cm] x_1 &=2.60 ~\rm{cm } \\[0.3cm] M_2 &=1.50kg \\[0.3cm] x_1 &=\, ?\\[0.3cm] \end{align}\\ {/eq}

Solution

The restoring force developed in the spring due to an equal magnitude of the externally applied force is directly proportional to the change in the length of the spring. The proportionality constant is known as the spring constant. The relation between the spring force {eq}(F){/eq}, spring constant {eq}(k){/eq}, and the change in the length of the spring {eq}(x){/eq} are given as:

{eq}F= k x {/eq}

Applying this formula, we have

{eq}\begin{align} \\ F&= kx \\[0.3cm] \frac{ F_2 }{F_1 }&= \frac{kx_2 }{kx_1} && \left[ \text{Taking the ratio of forces in two separate conditions } \right] \\[0.3cm] \frac{ M_2g }{M_1g }&= \frac{kx_2 }{kx_1} && \left[ \text{ Here}~ F=Mg \right] \\[0.3cm] \frac{ M_2 }{M_1 }&= \frac{x_2 }{x_1} \\[0.3cm] x_2&= \frac{ M_2 x_1}{M_1 } \\[0.3cm] x_2&= \frac{ (1.50\text{ kg}) (2.60 ~\rm{cm } )}{3.80\text{ kg} } \\[0.3cm] x_2&= \frac{ (1.50) (2.60)}{3.80 } ~\rm{cm } \\[0.3cm] x_2&= 1.02631 ~\rm{cm } \\[0.3cm] x_2&= 1.03 ~\rm{cm } && \left[ \text{Rounding off to three significant figures } \right] \\[0.3cm] \end{align}\\ {/eq}


Therefore, spring stretches by {eq}\displaystyle \boxed{\color{blue} { x_2= 1.03 ~\rm{cm } }} {/eq}


Learn more about this topic:

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Practice Applying Spring Constant Formulas

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Chapter 17 / Lesson 11
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In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.


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