Answer
6.6 kJ
Work Step by Step
We know that the work is the area under the force versus distance graph. Thus, we find:
$W = \int_{x_1}^{x_2}F(x)dx $
$W = \int_{0}^{10}(\mu_k mg)dx $
$W = \int_{0}^{10}((\mu_0 + ax^2) mg)dx $
$W = mg \int_{0}^{10}(\mu_0 + ax^2)dx $
We take the integral and plug in the known values to obtain:
$ W \approx 6.6 \ kJ$
