Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

Respuesta :

Answer:

2N/cm

Step-by-step explanation:

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

[tex]F=kx[/tex]

where, [tex]F[/tex] is the force which is stretching or compressing the spring,

[tex]k[/tex] is the spring constant; and

[tex]x[/tex] is the distance the spring is stretched.

Substituting the given values to find the elastic constant  [tex]k[/tex] to get:

[tex]F=kx[/tex]

[tex]4=k(2)[/tex]

[tex]k=\frac{4}{2}[/tex]

[tex]k=2[/tex]

Therefore, the elastic constant is 2 Newton/cm.

ACCESS MORE