Answer
(a) k = 26,700 N/m
(b) F = 800 N
(c) W = 21.4 J
F = 1070 N
Work Step by Step
(a) When we do work on a spring, the work is stored as elastic potential energy in the spring.
$\frac{1}{2}kx^2 = 12.0~J$
$k = \frac{24.0~J}{(0.0300~m)^2}$
$k = 26,700~N/m$
(b) $F = kx$
$F = (26,700~N/m)(0.0300~m)$
$F = 800~N$
(c) To compress the spring 4.00 cm:
$W = \frac{1}{2}kx^2$
$W = \frac{1}{2}(26,700~N/m)(0.0400~m)^2$
$W = 21.4~J$
$F = kx$
$F = (26,700~N/m)(0.0400~m)$
$F = 1070~N$
