Answer:
(a). The spring compressed is [tex]\dfrac{ma+mg}{k}[/tex].
(b). The acceleration is 1.5 g.
Explanation:
Given that,
Acceleration = a
mass = m
spring constant = k
(a). We need to calculate the spring compressed
Using balance equation
[tex]kx-mg=ma[/tex]
[tex]x=\dfrac{ma+mg}{k}[/tex]....(I)
The spring compressed is [tex]\dfrac{ma+mg}{k}[/tex].
(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,
The compression is given by
[tex]x=2.5\times x_{0}[/tex]
Here, acceleration is zero
So, [tex]x=2.5\times\dfrac{mg}{k}[/tex]
We need to calculate the acceleration
Put the value of x in equation (I)
[tex]2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}[/tex]
[tex]2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)[/tex]
[tex]a=2.5g-g[/tex]
[tex]a=1.5g[/tex]
Hence, (a). The spring compressed is [tex]\dfrac{ma+mg}{k}[/tex].
(b). The acceleration is 1.5 g.