Ablock 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?

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)

now from equation (3)

k = 26.25 N/m

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

The spring's force constant is 26.25 N/m

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

F = Force ( N )

k = Spring Constant ( N/m )

Δx = Extension ( m )

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

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 !

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:

Grade: College

Subject: Physics

Chapter: Elasticity