Answer
(a) The effective spring constant is $6.0\times 10^{10}~N/m$
(b) The block is compressed $8.0~nm$
(c) The work done by the applied forces is $1.92\times 10^{-6}~J$
Work Step by Step
(a) We can find the effective spring constant:
$F = kx$
$k = \frac{F}{x}$
$k = \frac{120~N}{2.0\times 10^{-9}~m}$
$k = 6.0\times 10^{10}~N/m$
The effective spring constant is $6.0\times 10^{10}~N/m$
(b) We can find the distance that the block is compressed:
$x = \frac{F}{k} = \frac{480~N}{6.0\times 10^{10}~N/m} = 8.0\times 10^{-9}~m = 8.0~nm$
The block is compressed $8.0~nm$
(c) We can find the elastic potential energy stored in the block when it is compressed:
$U_s = \frac{1}{2}kx^2$
$U_s = \frac{1}{2}(6.0\times 10^{10}~N/m)(8.0\times 10^{-9}~m)^2$
$U_s = 1.92\times 10^{-6}~J$
The work done by the applied forces is $1.92\times 10^{-6}~J$
