Answer
$6.3\times10^{-2}\space m$
Work Step by Step
Let's apply Newton's second for the object.
$F=ma$ ' Let's plug known values into this equation.
$F-mg=ma$
$F=m(g+a)$
According to equation 10.2 we can write,
$x=-\frac{F}{k}-(2)$
(1)=>(2),
$x=-\frac{m(g+a)}{k}=-\frac{(5\space kg)(9.8\space m/s^{2}+0.6\space m/s^{2})}{830\space N/m}=-6.3\times 10^{-2}m$
The amount that the spring stretches = $6.3\times10^{-2}m$
