We are asked to determine the distance that spring will stretch when a given mass is attached to it.
To do that we will use Hook's law:
[tex]F=kx[/tex]where:
[tex]\begin{gathered} F=\text{ force } \\ k=\text{ spring constant} \\ x=\text{ distance that the spring is stretched} \end{gathered}[/tex]We will determine the constant of the spring "k" first using the fact that the spring is stretched 0.8 meters when a mass of 3kg hangs from it.
Since the only force acting on the spring is the weight of the object we have:
[tex]F=mg[/tex]Where:
[tex]\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now, we substitute and we get:
[tex]mg=kx[/tex]Now, we divide both sides by "x":
[tex]\frac{mg}{x}=k[/tex]Now, we plug in the values:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{0.8m}=k[/tex]Solving the operations:
[tex]36.75\text{ N/m}=k[/tex]Now, we substitute the value of "k":
[tex]F=(36.75\text{ N/m\rparen}x[/tex]Now, we solve for "x":
[tex]\frac{F}{36.75\text{ N/m}}=x[/tex]Now, we substitute the value of the weight of the second object:
[tex]\frac{(14kg)(9.8\frac{m}{s^2})}{36.75\text{ N/m}}=x[/tex]Solving the operations:
[tex]3.733m=x[/tex]Therefore, the spring will stretch by 3.733 meters.
