Question

A $60$kg student is standing atop a spring in an elevator as it accelerates upward at $3.0$m/s². The spring constant is$2500$ N/m. By how much is the spring compressed?

Short answer

The spring compressed$∆s=31$cm

Step-by-step solution

Step 1. Given Information

The mass $m=60$kg

Acceleration $a=3.0$m/s

The spring constant$k=2500$N/m

Step 2. Hooke's law

When a spring is compressed, a force is exerted by its material to restore it to its equilibrium position and its force is called restoring force $F_{sp}$. Hooke's law states the relationship between the restoring force and the displacement $∆s$

$F_{sp}=k∆s$

Step 3. How much is spring compressed

The spring force is exerted in an opposite direction to the weight of the student, so the net force that is exerted on the student is

$$\begin{gathered} F_{net}=F_{sp}-mg \\ ma=k∆s-mg \\ ∆s=\frac{m(a+g)}{k} \end{gathered}$$

Now, put the values

$$\begin{gathered} ∆s=\frac{m(a+g)}{k} \\ =\frac{(60kg)(3m/s^2+9.8m/s^2)}{2500N/m} \\ =0.31m \\ =31cm \end{gathered}$$