An unstretched spring with spring constant 18 N/cm is suspended from the ceiling. A 3.2 kg mass...

Question:

An unstretched spring with spring constant 18 N/cm is suspended from the ceiling. A 3.2 kg mass is attached to the spring and let fall. To the nearest tenth of a centimeter, how far does it stretch the spring?

Spring Constant

The spring's stiffness or stretchability is determined by the parameter called the spring constant. Springs with large values of spring constant tend to be stiffer than those with smaller values.

Answer and Explanation: 1

The spring constant, {eq}k {/eq}, is given by Hooke's law

{eq}F = kx {/eq}

where {eq}F {/eq} is the force that stretches the spring {eq}x {/eq} distance from its equilibrium length.


A mass of {eq}m = 3.2 \text{ kg} {/eq} hangs from the spring, so a weight of {eq}F = mg = (3.2 \text{ kg})(9.8 \text{ m/s}^2) = 31.36\text{ N} {/eq} is pulling the spring down.


From Hooke's law, we can now determine the stretching of the spring when the mass is hung.

{eq}\displaystyle \begin{align*} F&= kx,\ F= 31.36\text{ N}, \ k = 18\text{ N/ cm}\\ 31.36\text{ N} &= (18\text{ N/ cm})x\\ x &= \frac{31.36\text{ N} }{18\text{ N/ cm}}\\ x &\approx \boxed{1.7\text{ cm}} \end{align*} {/eq}


Therefore, the spring is stretched approximately {eq}\boxed{ x \approx 1.7\text{ cm}} {/eq} from the equilibrium.


Learn more about this topic:

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

from

Chapter 4 / Lesson 19
201K

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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