Question

A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.What is the spring constant k?b.How long is the spring when a 3.0kg mass is suspended from it?

Answer 1

(a) 392 N/m

Hook's law states that:

$F=k\Delta x$ (1)

where

F is the force exerted on the spring

k is the spring constant

$\Delta x$ is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

- The stretching of the spring is

$\Delta x=15 cm-10 cm=5 cm=0.05 m$

Solving eq.(1) for k, we find the spring constant:

$k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m$

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

$F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N$

And so, the stretch of the spring is

$\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm$

And since the initial lenght of the spring is

$x_0 = 10 cm$

The final length will be

$x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm$

Answer 2

(a) The spring constant of the spring is 392 N/m

(b) Length of the spring is 17.5 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 length of spring = Lo = 10 cm

mass of object = m = 2.0 kg

extension of the spring = x = 15 - 10 = 5 cm = 0.05 m

mass of second object = m' = 3.0 kg

Asked:

a. spring constant of the spring = k = ?

b. length of spring = L = ?

Solution:

Part a.

$F = kx$

$mg = kx$

$k = mg \div x$

$k = 2.0 ( 9.8 ) \div 0.05$

$\boxed {k = 392 \texttt{ N/m}}$

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Part b.

$F' = kx'$

$m' g = k x'$

$x' = ( m' g ) \div k$

$x' = ( 3.0 (9.8) ) \div 392$

$x' = 0.075 \texttt{ m} = 7.5 \texttt{ cm}$

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$L = Lo + x'$

$L = 10 + 7.5$

$\boxed {L = 17.5 \texttt{ cm}}$

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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