Answer
(a) The spring stretches 4.9 cm
(b) $k = 143~N/m$
(c) The work done by the applied forces to stretch the spring is $0.0876~J$
Work Step by Step
(a) Since $F= kx$, then $k = \frac{F}{x}$
Let $x$ be the distance the spring stretches with a force of 7.0 N:
$\frac{5.0~N}{3.5~cm} = \frac{7.0~N}{x}$
$x = \frac{7.0~N}{5.0~N}~(3.5~cm)$
$x = 4.9~cm$
The spring stretches 4.9 cm
(b) We can find the spring constant:
$k = \frac{F}{x} = \frac{5.0~N}{0.035~m} = 143~N/m$
(c) We can find the elastic potential energy stored in the spring:
$U_s = \frac{1}{2}kx^2$
$U_s = \frac{1}{2}(143~N/m)(0.035~m)^2$
$U_s = 0.0876~J$
The work done by the applied forces to stretch the spring is $0.0876~J$.
