When a mass of 2kg is hung off spring, the spring stretches by 4cm. What is spring's constant? a)...
Question:
When a mass of {eq}\displaystyle 2kg {/eq} is hung off spring, the spring stretches by {eq}\displaystyle 4cm. {/eq} What is spring's constant?
{eq}\displaystyle a) 5 N/cm \\b) 5 N/m \\c) 5 cm/N \\d) 5 m/N {/eq}
Hooke's Law:
Spring is an elastic object. Elastic objects are those objects, which when deformed, restore their shape once the forces causing the deformation are removed. This is possible because of a restoring force that acts within the elastic object in a direction that tends to restore the original shape of the object (hence the name).
According to Hooke's law, the magnitude of the restoring force of a stretched or compressed spring is directly proportional to the extension/compression of the spring.
Answer and Explanation: 1
We are given:
- The mass hung from the spring, {eq}m=2\;\rm kg {/eq}
- The extension in the spring, {eq}\Delta x=4\;\rm cm {/eq}
At the stretched length, the restoring force, {eq}F {/eq}, of the spring must be able to balance the weight, {eq}W {/eq}, of the mass.
{eq}F_r=W {/eq}.
The weight of an object is given by the following equation:
{eq}W=mg {/eq}
- where {eq}m {/eq} is the mass and {eq}g=9.8\;\rm m/s^2 {/eq} is the acceleration due to gravity.
According to Hooke's Law, the magnitude of the restoring force developed by a spring when it is stretched/compressed by a length, {eq}\Delta x {/eq}, is given by the equation:
{eq}F=K\Delta x {/eq}
Here,
- {eq}K {/eq} is the spring constant.
For the given system, we have:
{eq}\begin{align*} F_r&=W\\ \Rightarrow K\Delta x&=mg\\ \Rightarrow K&=\dfrac{mg}{\Delta x}\\ &=\dfrac{2\;\rm kg \times 9.8\;\rm m/s^2 }{4\;\rm cm}\\ &=\boxed{\mathbf{4.9\;\rm N/cm}} \end{align*} {/eq}
Learn more about this topic:
from
Chapter 4 / Lesson 19After 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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