A car is launched by a spring and goes to the top of a loop. Fill in the energy bar chart by drawing a vertical bar(s) in the correct position (both Initial & Final conditions).

A) the initial point all energy is elastic potential and the final point all energy is kinetic

B) a bar graph the two bars have the same height and the sum of their height is the initial energy

C) two bars, one for the  kinetic energy and the other for the gravitational potential energy.

Explanation:

A) For this exercise we must use the energy conservation relations

starting point. When the spring is compressed

        Em₀ = K_e = ½ k x²

end point, at the bottom of the loop

        Em_f = K = ½ m v²

energy is conserved

        Em₀ = Em_f

         ½ k x² = ½ m v²

         v =   x

In a bar graph the initial point all energy is elastic potential and the final point all energy is kinetic

B) intermediate point in a quarter of the radius

In this case we use the lower part of the loop as the starting point and the quarter part of the bow as the end point.

        Em₀ = K

         Em_f = K + U = ½ m v² + m g R

in a bar graph the two bars have the same height and the sum of their height is the initial energy

C) End point highest part of the loop

   starting point, bottom of loop

         Emo = K = ½ m v₀²

from part A of the exercise we saw that it is equal to the elastic energy of the spring

    final point. Highest part of the loop

         Emf = K + U

         Em_f = ½ m + mg (2R)

where R is the radius of the loop

         Em₀ = Em_f

        1/2 m v₀² = 1/2 m v_{f}^2+ mg 2R

        v₀² = v_f^2 + 4gR

In a bar graph there are two bars, one for the  kinetic energy and the other for the gravitational potential energy. The sum of the heights of these bars is the initial energy, so the energy is transformed but not created or destroyed in the process.

the force required to move the cart is divided by three or 33.33 n.