## Essential University Physics: Volume 1 (4th Edition)

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: $W \approx 6.6 \ kJ$