# A 8.0 cm long spring is attached to the ceiling. When a 1.5 kg mass is hung from it, the spring...

## Question:

A {eq}8.0 \ cm {/eq} long spring is attached to the ceiling. When a {eq}1.5 \ kg {/eq} mass is hung from it, the spring stretches to a length of {eq}14 \ cm {/eq}.

a) What is the spring constant {eq}k {/eq}?

b) How long is the spring when a {eq}3.0 \ kg {/eq} mass is suspended from it?

## Hooke's Law:

In physics, Hooke's law describes the effect of an applied force to the change of the length of the spring. According to this law, the change in the length is directly proportional but is in the opposite direction of the applied force. The proportionality constant of this relation is the spring constant which is a characteristic of the spring material.

(a) To solve this problem, we use Hooke's law equation which is given by

{eq}\displaystyle F = -k \Delta x {/eq}

where

• {eq}F {/eq} is the force
• {eq}k {/eq} is the spring constant
• {eq}\Delta x {/eq} is the change in the length of the spring

In this case, the force is given by the weight of the load mass.

{eq}\displaystyle F = mg = (1.5\ \rm kg)(-9.8\ \rm m/s^2) = -14.7\ \rm N {/eq}

If the spring stretches from 8.0 cm to 14 cm, then {eq}\Delta x = 14\ \rm cm - 8.0\ \rm cm = 6.0\ \rm cm = 0.06\ \rm m {/eq}. The spring constant is therefore,

{eq}\displaystyle k = -\frac{F}{ \Delta x} = -\frac{-14.7\ \rm N}{ 0.06\ \rm m} = 245\ \rm N/m {/eq}

(b) If the mass is changed to {eq}3.0\ \rm kg {/eq}, the change in the length would be

{eq}\displaystyle \Delta x = -\frac{F}{k} = -\frac{(3.0\ \rm kg)(-9.8\ \rm m/s^2)}{ 245\ \rm N/m} = 0.12\ \rm m = 12\ \rm cm {/eq}

Therefore, the final length of the spring is

{eq}\displaystyle L = 8.0\ \rm cm + \Delta x = 8.0\ \rm cm + 12\ \rm cm = 20\ \rm cm {/eq}

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

from

Chapter 4 / Lesson 19
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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.