A spring has a spring constant of 105 N/m. If you compress the spring of 0.1 m past it’s natural length, what force does the spring apply

Heya!!

For calculate force, lets applicate Hooke's law:

                                                 Δ   Being   Δ

                                                 F = Force = ?

                                         k = Constant = 105 N/m

                                          x = Elongation = 0,1 m

⇒ Let's replace according the formula:

⇒ Resolving

Result:

The force is 10,5 Newtons (N).

Good Luck!!

f=ma newton

f=100x2.5=250n