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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

Respuesta :

ChizuruChan

Heya!!

For calculate force, lets applicate Hooke's law:

                                                [tex]\boxed{F=k * x}[/tex]

                                                 Δ   Being   Δ

                                                 F = Force = ?

                                         k = Constant = 105 N/m

                                          x = Elongation = 0,1 m

⇒ Let's replace according the formula:

[tex]\boxed{F=105\ N/m * 0,1\ m}[/tex]

⇒ Resolving

[tex]\boxed{F = 10,5 \ N }[/tex]

Result:

The force is 10,5 Newtons (N).

Good Luck!!

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