Question

When a spring is stretched by a distance $x$, it exerts a force, given by $F=(-5x-16x^3)N$. The work done, when the spring is stretched from $0.1m$ to $0.2m$ is

• A. $8.7\times 10^{-2}J$
• B. $12.2\times 10^{-2}J$
• C. $8.7\times 10^{-1}J$
• D. $12.2\times 10^{-1}J$

Answer 1

Answer: (A)

$F=-5x-16x^3=-(5+16x^2)x=-kx$
$\therefore k=5+16x^2$
Work done, $W=\frac{1}{2}{k}_2{x}_2^2-\frac{1}{2}{k}_1{x}_1^2$
$=\frac{1}{2}[5+16(0.2{)}^2](0.2{)}^2=\frac{1}{2}[5+16(0.1{)}^2](0.1{)}^2$
$=2.82\times 4\times 10^{-2}-2.58\times 10^{-2}$
$=8.7\times 10^{-2}J$