Question

A spring of spring constant $5\times 10^{3}N{m}^{-1}$ is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is

• A. 12.50 N-m
• B. 18.75 N-m
• C. 25.00 N-m
• D. 6.25 N-m

Answer 1

Answer: (B)

Work done $W_1=\frac{1}{2}k\times x_1^2$
$=\frac{1}{2}\times 5\times 10^3\times (5\times 10^{-2}{)}^2$
$=6.25J$
$W_2=\frac{1}{2}k(x_1+x_2{)}^2$
$=\frac{1}{2}\times 5\times 10^3(5\times 10^{-2}+5\times 10^{-2}{)}^2$
$=25J$
Net work done $=W_2-W_1$
$=25-6.25$
$18.75J=18.75N-m$