From $x=x_1$ to $x=0$ force is along displacement

ie work done $=\large\frac{1}{2}$$kx_1^2$

from $x=0$ to $x=x_2$ force is opposite of displcement so work done is $-\large\frac{1}{2}$$kx_2^2$

$\therefore$ Total work done $=\large\frac{500}{2}$$ \bigg[(0.15)^2-(0.08)^2\bigg]$

$\qquad= 4.03\;J$

Hence a is the correct answer.

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