Question

Springs: An ornament of mass 40.0 g is attached to a vertical ideal spring with a force constant (spring constant) of 20.0 N/m. The ornament is then lowered very slowly until the spring stops stretching. How much does the spring stretch?

Answer 1

Answer:

x=0.02 m

Explanation:

Given that

Mass , m = 40 g        ( 1 kg = 1000 g)

m = 0.04 kg

K= 20 N/m

We know that spring force ,F is given as

F= k x

x=Stretching in the spring

At equilibrium position

m g= K x

$x=\dfrac{mg}{K}$     (take g= 10 m/s²)

$x=\dfrac{0.04\times 10}{20}\ m$

x=0.02 m

Therefore the extension in the spring will be 0.02 m.