Question

A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at x = +0.240 m, its acceleration is ax = -14.0 m/s2 and its velocity is vx = +4.00 m/s. What is the spring's force constant k?

Answer 1

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

$\omega = \sqrt{\dfrac{k}{m}}$

now from equation (3)

$a_x = \dfrac{k}{m}x$

$k = \dfrac{m a_x}{x}$

$k = \dfrac{0.45 \times (-14)}{0.24}$

k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

Answer 2

The spring's force constant is 26.25 N/m

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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:

mass of block = m = 0.450 kg

extension = x = +0.240 m

acceleration = a_x = -14.0 m/s²

velocity = v_x = +4.00 m/s

Asked:

spring's force constant = k = ?

Solution:

$k = m \omega^2$

$k = m \frac{-a_x}{x}$

$k = 0.450 \times \frac{ - ( -14.0 ) }{0.240}$

$\boxed{k = 26.25 \texttt{ N/m}}$

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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

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Answer details

Grade: College

Subject: Physics

Chapter: Elasticity