A spring of spring constant `5 xx 10^(2) `Nm is streched initially by `5 cm` from the unstriched position . Then the work required to streach is further by another `5 cm` is

Options

A. 12.50 Nm
B. 18.75 Nm
C. 25.00 Nm
D. 6.25 Nm

Answer

B

Explanation

Work done in stretching the spring initially by 5 cm,
`w_1=1/2kxxx_1^2=1/2xx5xx10^3xx(5xx10^(-2))^2=6.25` J
Now, work done in stretching the spring by 10 cm, i.e., 5 cm+5 cm
`w_2=1/2k(x_1+x_2)^2=1/2xx5xx10^3(5xx10^(-2)+5xx10^(-2))`=25 J
Net work done=`w_2-w_1`=25-6.25=18.75 J=18.75 N m