Answer
$E=685.66kJ$
Work Step by Step
Here we have to kind of work, the one to change the potential energy and the one to change the kinetic energy:
$W=\Delta E$
First we are going to calculate the work to change the potential energy:
$W_{p}=mg\Delta z=mg(z_{2}-z_{1})=1300kg*9.81\frac{m}{s^2}*(40m-0m)=510120J$
And now the work to change the kinetic energy:
$W_{k}=\frac{m}{2}\Delta V^2=\frac{m}{2}(V_{2}^2-V_{1}^2)=\frac{1300kg}{2}*[((60\frac{km}{h}*(\frac{1000m}{1km})*(\frac{1h}{3600s}))-((10\frac{km}{h})*(\frac{1000m}{1km})*(\frac{1h}{3600s}))]=175540.12J$
Then the total energy needed is:
$E= W_{t}=W_{p}+W_{k}=510.12kJ+175.54kJ=685.66kJ$
