Answer
a) $W_1=W_2=-0.11J$
b) stays the same
Work Step by Step
(a) Work done along path 1 can be calculated as
$W_1=(\frac{1}{2})K[(x_1^2-x_2^2)+(x_2^2-x_3^2)]$
We plug in the known values to obtain:
$W_1=(\frac{1}{2})(550)[(0)^2-(0.040)^2+(0.040)^2-(0.020)^2]$
$W_1=(-0.44)+(0.33)=-0.11J$
Now work along path 2 is given as
$W_2=(\frac{1}{2})K[(x_1^2-x_4^2)+(x_4^2-x_3^2)]$
We plug in the known values to obtain:
$W_2=(\frac{1}{2})(550)[(0)^2-(0.020)^2+(0.020)^2-(0.020)^2]$
$W_2=-0.1J+0J=-0.11J$
(b) We can say that work done is independent of the mass, hence if the mass is increased the work done will stay the same.
