A 20-newton weight is attached to a spring.
causing it to stretch, as shown in the diagram.
What is the spring constant of this spring?

Respuesta :

Answer:

40 N/m

Explanation:

The diagram attached is used to answer the question

We know from Hooke's law that extension is directly proportional to the applied force hence

F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then

[tex]k=\frac {F}{k}[/tex]

From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m

Substituting 20 N for F and 0.5 m for x then

[tex]k=\frac {20}{0.5}=40 N/m[/tex]

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