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

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop

when heated, the valance electrons increased their speed and transferred the heat from electron to electron.   i just took the test.