Question

A force was applied to an elastic spring so that it stretches by 2cm. If the force constant is 40N/m, find the force applied.

Answer 1

Answer :

Elongation of spring = 2cm = 0.02m

Spring constant = 40N/m

We have to find magnitude of applied force$.$

Relation b/w applied force and elongation of spring is given by,

፨ F = kx

• F denotes force
• k denotes spring constant
• x debotes elingation of spring

⇒ F = kx

⇒ F = 40 × 0.02

⇒ F = 0.8N

☃ God_Bless :)

Answer 2

Given ,

Force or Spring constant (k) = 40 N/m

Stretch (x) = 2 cm or 2 × (10)^{-2} m

We know that ,

$\boxed{ \sf{Force \: applied \: (F) = kx}}$

Thus ,

F = 40 × 2 × (10)^{-2}

F = 8 × (10)^{-1}

F = 0.8 N

Therefore ,

The force applied on spring is 0.8 N