Answer
(a) F = 800 N
(b) The additional work required is 240 J.
The maximum force is 1600 N.
Work Step by Step
Let $k$ be the sum of the force constants of the two springs. We can use the work to find $k$.
$\frac{1}{2}kx^2 = W$
$k = \frac{2W}{x^2}$
$k = \frac{(2)(80.0~J)}{(0.200~m)^2}$
$k = 4000~N/m$
(a) $F = kx = (4000~N/m)(0.200~m)$
$F = 800~N$
(b) $W_{tot} = \frac{1}{2}kx^2$
$W_{tot} = \frac{1}{2}(4000~N/m)(0.400~m)^2$
$W_{tot} = 320~J$
The additional work required is 320 J - 80.0 J, which is 240 J.
$F = kx = (4000~N/m)(0.400~m)$
$F = 1600~N$
The maximum force is 1600 N.
