Question

A 500 kg object is hanging from a spring attached to the ceiling. If the spring constant in the spring is 900 N/kg, how far does the spring stretch?

Answer 1

We are given that a 500 kg object is hanging from a spring. To determine the amount the spring is stretched we will use Hook's law, which states the following:

$F=kx$

Where:

$\begin{gathered} F=\text{ force} \\ k=\text{ spring constant} \\ x=\text{ distance stretched} \end{gathered}$

Since the object is hanging the only force acting on the spring is the weight of the object. The weight of the object is:

$F_g=mg$

Where:

$\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}$

Plugging in the values we get:

$F_g=(500\operatorname{kg})(9.8\frac{m}{s^2})$

Solving the operations:

$F_g=4900N$

Now we solve for "x" from Hook's law by dividing both sides by "k":

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

Now we plug in the known values:

$\frac{4900N}{900\frac{N}{m}}=x$

Solving the operations:

$5.4m=x$

Therefore, the spring is stretched by 5.4 meters.