Answer
a). By Hooke's law $F=-kx$
When x=0, F=0
So x=0 is the point where the spring exactly balances the weight of the mass, resulting in a force of magnitude zero.
b). The force , |F|=kx.
Also, |F|=ma
So, $a= \frac{kx}{m}$
when x=0.05m,
|F|=kx=$150\times0.05=7.5N$
$a=\frac{7.5}{0.5}=15m/s^{2}$
when x=0.15m,
|F|=kx=$150\times 0.15=22.5N$
$a=\frac{22.5}{0.5}=45m/s^{2}$
Work Step by Step
a). By Hooke's law $F=-kx$
When x=0, F=0
So x=0 is the point where the spring exactly balances the weight of the mass, resulting in a force of magnitude zero.
b). The force , |F|=kx.
Also, |F|=ma
So, $a= \frac{kx}{m}$
when x=0.05m,
|F|=kx=$150\times0.05=7.5N$
$a=\frac{7.5}{0.5}=15m/s^{2}$
when x=0.15m,
|F|=kx=$150\times 0.15=22.5N$
$a=\frac{22.5}{0.5}=45m/s^{2}$
