Question

A 750-kg automobile is moving at 26.2 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill?

Answer 1

To solve this problem it is necessary to apply to the concepts related to energy conservation. For this purpose we will consider potential energy and kinetic energy as the energies linked to the body. The final kinetic energy is null since everything is converted into potential energy, therefore

Potential Energy can be defined as,

$PE = mgh$

Kinetic Energy can be defined as,

$K= \frac{1}{2} mv^2$

Now for Conservation of Energy,

$KE_i+PE_i = PE_f$

$\frac{1}{2}mv_i^2+mgh_1 = mgh_2$

$\frac{1}{2} (750kg) (26.2m/s)^2 + (750)(9.8)(5) = (750)(9.8)h_2$

$h_2 = 40.0224m$

Therefore the highets position the car reaches above the bottom of the hill is 40.02m