A 1400 kg car is approaching the hill shown in the figure at 13

This is a trick question. Technically, there issome energy lost to friction, but as you've written it there's nofriction given, so we'll assume that the car doesn't lose anyenergy to friction.

In that case, we just use the conservation of energy equation:E = U + T, where U=potential energy and T= kinetic energy.The energy on one side of the hill is equal to the energy onthe other. Remember that the gravitational potential is mgy,then we have:

.5*1400*vf^2= 1400*9.8*5 + 0.5*1400*13^2

v=16.3401346384