Question

a 57 kg student is standing atop a spring in an elevator that is accelerating upward at 3.3 m/s2. the spring constant is 2000 n/m.By how much is the spring compressed?

Answer 1

Approximately
`0.37\; {\rm m}` (assuming that
`g = (-9.81)\; {\rm m\cdot s^(-2)}}`.)

Step-by-step explanation:

Let
`m` denote the mass of the student. There are two forces on this student:

• Weight
  `m\, g`, which points downwards, and
• Normal force
  `F_{\text{normal}}` from the spring, which points upwards.

The net force on this student would be:

`F_{\text{net}} = m\, g + F_{\text{normal}}`.

It is given that acceleration of the student is
`a = 3.3\; {\rm m\cdot s^(-2)}`. The net force on this student will also be equal to:

`F_{\text{net}} = m\, a`.

Note that acceleration is positive since the student is accelerating upwards.

Thus:

`m\, a = F_{\text{net}} = m\, g + F_{\text{normal}}`.

`m\, a = m\, g + F_{\text{normal}}`.

Rearrange this equation to find
`F_{\text{normal}}`:

`F_{\text{normal}} = m\, a - m\, g`.

In other words, the spring exerts a restoring force of
`F_{\text{normal}} = m\, a - m\, g` on this student. To find the displacement
`x` of the spring from equilibrium, divide the restoring force by the spring constant
`k`:

`\begin{aligned}x &= \frac{F_{\text{normal}}}{k} \\ &= (m\,a - m\, g)/(k) \\ &= (m\, (a - g))/(k)\end{aligned}`.

Note that the gravitational field strength
`g = (-9.81)\; {\rm m\cdot s^(-2)}` is negative because the gravitational field near the Earth points downward- towards the center of the planet.

Substitute in mass
`m = 57\; {\rm kg}`, acceleration
`a = 3.3\; {\rm m\cdot s^(-2)}`, and spring constant
`k = 2000\; {\rm N\cdot m^(-1)}`:

`\begin{aligned}x &= (m\, (a - g))/(k) \\ &= ((57) \, (3.3 - (-9.81)))/(2000)\; {\rm m} \\ &\approx 0.37\; {\rm m}\end{aligned}`.

In other words, the compression of this spring would be approximately
`0.37\; {\rm m}`.