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 $0.1 \,m$ to $0.2\, m $ 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, $W=\frac{1}{2}k_{2}x^{2}_{2}-\frac{1}{2}k_{1}x^{2}_{1}$
$=\frac{1}{2}[5+16(0.2)^{2}](0.2)^{2}=\frac{1}{2}[5+16(0.1)^{2}] (0.1)^{2}$
$=2.82 \times 4 \times 10^{-2}-2.58\times 10^{-2}$
$=8.7 \times 10^{-2}\,J$