A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?
ok i used hooks law
$$ 4 in = \frac 13 ft, 6 in = \frac 12 ft $$ Edit^
$$f(x) = kx, k = \frac{10lb}{\frac 13ft}$$
so $$\int_{\frac 13}^{\frac 12} (30x)dx$$
$$\frac{30x^2}{2} = 15x^2$$
From $\frac 13$ to $\frac 12$ which = $\frac{25}{12}$ ft-lb...... but the book says $\frac{15}{4}$ ft-lb!!!!
You are asked to calculate the work for stretching it from its natural length to 6 inches, so $W = \int_0^{1/2} 30x \, dx$.