A block is suspended from the ceiling by a long, thin strip of tungsten metal. The strip behaves as a spring. To produce a 0.25m horizontal deflection of the block, a force of 6.5 N is required. Calculate the spring constant for the tungsten strip.

Answer :

F = kx

6.5 N = K 0.25

6.5 / 0.25 = 26 N/m

Hope this helps
johanrusli

The spring constant for the tungsten strip is 26 N/m

[tex]\texttt{ }[/tex]

Further explanation

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

[tex]\boxed {F = k \times \Delta x}[/tex]

F = Force ( N )

k = Spring Constant ( N/m )

Δx = Extension ( m )

[tex]\texttt{ }[/tex]

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

[tex]\boxed {E = \frac{F / A}{\Delta x / x_o}}[/tex]

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 !

[tex]\texttt{ }[/tex]

Given:

extension of strip = x = 0.25 m

magnitude of force = F = 6.5 N

Asked:

the spring constant for the tungsten strip = k = ?

Solution:

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

[tex]F = k x[/tex]

[tex]6.5 = k \times 0.25[/tex]

[tex]k = 6.5 \div 0.25[/tex]

[tex]\boxed {k = 26 \texttt{ N/m}}[/tex]

[tex]\texttt{ }[/tex]

Learn more

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

[tex]\texttt{ }[/tex]

Answer details

Grade: College

Subject: Physics

Chapter: Elasticity

${teks-lihat-gambar} johanrusli

Other Questions