# Two identical, side-by-side springs with spring constant 280 N/m support a 3.00 kg hanging box....

## Question:

Two identical, side-by-side springs with spring constant 280 N/m support a 3.00 kg hanging box. By how much is each spring stretched?

## Hooke's Law:

Within the limits of a spring's elasticity, it will obey Hooke's law, which states that the relationship between the force applied to the spring and the distance it stretches is linear. That is, applying twice as much force to a spring causes it to stretch twice as far.

Givens:

• The spring constant of each spring: {eq}k = 280\ \rm{N/m} {/eq}
• The mass of the box: {eq}m = 3.00\ \rm{kg} {/eq}

The weight of the box can be found by multiplying its mass by the acceleration due to gravity: {eq}(g= 9.8\ \text{m/s}^2) {/eq}. This box's weight is:

$$F_g = mg = (3\ \rm{kg})(9.8\ \rm{m/s^2}) = 29.4\ \rm{N}$$

Being identical to each other, each spring will support half the weight of the box: {eq}F_{1/2} = \dfrac{29.4\ \rm{N}}{2} = 14.7\ \rm{N} {/eq}

Each spring will obey Hooke's law, which gives the extension distance of the spring, {eq}\Delta x {/eq},

based on the force, {eq}F {/eq}, applied to the spring and the spring constant, {eq}k {/eq}. The equation is: {eq}\Delta x = \dfrac{F}{k} {/eq}. Therefore the stretch in either spring is:

$$\Delta x = \dfrac{14.7\ \rm{N}}{280\ \rm{N/m}} = \bf{0.0525\ \text{m}}$$

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.