Question

A 1500kg car is approaching the hill shown in the figure at 11.0m/s when it suddenly runs out of gas. What is the car's speed after coasting down the other side? (picture in comments)

Answer 1

hint:
the gravitational field is a conservative field, so the total mechanical energy, namely potential energy plus kinetic energy, has to remain constant.
the initial kinetic energy of the car is equal to:
\[KE=\frac{1}{2}m{v^2} = \frac{1}{2}1500 \times {11^2} = ...?\]
when the car is at the top of the hill, its mechanical energy is:
\[KE - mgh = KE - 1500 \times 9.81 \times 5 = ...?\]
when the car is at the right side of the hill, then its mechanical energy will be:
\[KE - mg{h_1} + mg{h_2} = KE - 1500 \times 9.81 \times 5 + 1500 \times 9.81 \times 10 = ...?\]
and that amount of energy is all kinetic energy. So we can write:
\[KE - mg{h_1} + mg{h_2} = \frac{1}{2}mv_2^2\]
Here v_2 is the requested speed

Answer 2

In essence the car simply 'falls' 5m over the route

thus it loses mgh of PE

this is added to the KE

Answer 3

I got 8.66

Answer 4

The final speed HAS to be faster thant he initial, since it has gon down hill

Your answer is less than the original - maybe oyu need to ADD the acquired KE to the initial value?

Answer 5

v^2 = u^2 + 2as

u= 11m/s
a= g
s = 5m