## College Physics (4th Edition)

(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 $0.0876~J$
(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 $0.0876~J$.