Question

A certain spring stretches 3 cm when a load of 15 n is suspended from it. how much will the spring stretch if 30 n is suspended from it (and it doesn't reach its elastic limit)?

Answer 1

Extension of the spring is 6 cm

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Further explanation

Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.

$\boxed {F = k \times \Delta x}$

F = Force ( N )

k = Spring Constant ( N/m )

Δx = Extension ( m )

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The formula for finding Young's Modulus is as follows:

$\boxed {E = \frac{F / A}{\Delta x / x_o}}$

E = Young's Modulus ( N/m² )

F = Force ( N )

A = Cross-Sectional Area ( m² )

Δx = Extension ( m )

x = Initial Length ( m )

Let us now tackle the problem !

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Given:

initial extension of the spring = x = 3 cm

initial load = F = 15 N

final load = F' = 30 N

Asked:

final extension of the spring = x' = ?

Solution:

We will use Hooke's Law to solve this problem:

$F : F' = kx : kx'$

$F : F' = x : x'$

$15 : 30 = 3 : x'$

$1 : 2 = 3 : x'$

$x' = 2 \times 3$

$\boxed {x' = 6 \texttt{ cm}}$

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Learn more

• Young's modulus : brainly.com/question/6864866
• Young's modulus for aluminum : brainly.com/question/7282579
• Young's modulus of wire : brainly.com/question/9755626

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Answer details

Grade: College

Subject: Physics

Chapter: Elasticity

Answer 2

Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
$k= \frac{F}{\Delta x}$
where F is the force applied, and $\Delta x$ is the stretch of the spring with respect to its equilibrium position. Using the data, we find
$k= \frac{15 N}{3.0 cm}=5.0 N/cm$

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
$\Delta x = \frac{F}{k}= \frac{30 N}{5.0 N/cm}=6 cm$