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 5cm, `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))^2=25` J
Net work done = `w_2-w_1=25-6.25`
=18.75 J = 18.75 N m