What kind of variation is there in the mechanical energy as the cart rolls down the ramp? Does this agree with your prediction? Explain.

 Em₀ = U = m g h ,  Em_{f} = K = ½ m v²

Explanation:

When a car is on a ramp it has a certain amount of mechanical energy. At the highest point of the ramp the mechanical energy is fully potential given by

          Em₀ = U = m g h

As part of this energy descends down the ramp, part of this energy is transformed into kinetic energy and has one part of each, even though the sum remains the initial energy

            Em = K + U = ½ m v² + mg y

        y <h

when it reaches the bottom of the ramp it has no height therefore there is no potential energy, all of it has been transformed into kinetic energy

          Em_{f} = K = ½ m v²

This energy transformation is in the case that the friction force is zero.

If there is a friction force, it performs work against the low car, it is reflected in an increase in the internal energy (temperature) of the car. In this case the energy in the lower part is less than the initial one by a factor

          = - fr L

therefore the numeraire values of the velocity are lower, due to the energy lost by friction.

do you have a formula possibly