Question

A spring is hung vertically with a 425g mass attached to it. The mass is at rest. If the mass causes the spring to stretch 0.67 m, what is the spring constant?

Answer 1

Answer:

6.22 N/m

Explanation:

From Hooke's law we deduce that F=kx where F is the applied force and k is the spring constant while x is the extension or compression of the spring. Making k the subject of the above formula then

$k=\frac {F}{x}$

We also know that the force F is equal to mg where m is the mass of an object and g is acceleration due to gravity hence substituting F with mg we get that

$k=\frac {mg}{x}$

Substituting m with 425 g which is equivalent to 0.425 kg and g with 9.81 then 0.67 for x we get that

$k=\frac {mg}{x}=\frac {0.425\times 9.81}{0.67}=6.222761194 N/m\approx 6.22\ N/m$

Therefore, the spring constant is approximately 6.22 N/m