Question

One end of a light spring with spring constant k1 1,250 N/m Is attached to the celling. A second light spring Is attached to the lower end, with spring constant k2 1,650 N/m. An object of mass 1.50 kg is attached to the lower end of the second spring.

(a) By how much does the pair of springs stretch (In m)? 9E-4 Because the springs are light, you can treat them as essentially massless, so the only downward force acting on them is mg. Use |Flkx to find the amount of stretch,x, for each spring and then find the sum. m
(b) What is the effective spring constant (in N/m) of the spring system?

Answer 1

To solve this problem we will apply the concepts related to Hooke's Law for the mentioned system and Newton's second law for the description of the Weight. In turn, applying the equivalent systems of the springs we will find the constant of the equivalent spring of the whole system. To find the net displacement of the system, it will be necessary to start with point B and then return again to point A.

PART B) Let's start with the spring constant, which when connected in series (analog to a circuit for example) would be given by the function

$\frac{1}{k_{eq}} = \frac{1}{k_1}+\frac{1}{k_2}$

$\frac{1}{k_{eq}} = \frac{k_1+k_2}{k_1k_2}$

The net value would be

$k_{eq}=\frac{k_1k_2}{k_1+k_2}$

Therefore the equivalent constant is

$k_{eq} = \frac{(1250)(1650)}{1250+1650}$

$k_{eq} = 711.206N/m$

PART A) By equilibrium the force caused by the weight must be equivalent to the force exerted by the spring therefore

$F_w = F_s$

$mg = K_{eq} x$

Here,

m = mass

g = Acceleration due to gravity

x = Displacement

Rearranging to find the displacement,

$x = \frac{mg}{K_{eq}}$

$x = \frac{(1.5)(9.8)}{711.206}$

$x = 0.020669m = 2.06cm$