Question

A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring constant 1.08 kN/m . None of the objects in the elevator is moving relative to the elevator, but the spring is stretched by 6.0 cm .?

Answer 1

Answer:

The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².

Explanation:

By Hooke's law, $F = ke$

F is the force on a spring, k is the spring constant and e is the extension or compression.

From the question,

$F = (1.08\text{ kN/m}) \times (6.0 \times 10^{-2}\text{ m}) = 64.8 \text{ N}$

This is the force on the mass suspended on the spring. Its acceleration, a, is given by

$F = ma$

$a = \dfrac{F}{m}$

$a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2$

This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of

12.96 - 9.8 m/s² = 3.16 m/s²