Conservation of Energy 3

Question

If a roller coaster begins at rest at the top of a hill, and at the bottom of the hill is moving at 16 m/s, how tall was the hill?

Answer 1

Hello Syd,

Let us assume that we can apply conservation of mechanical energy. (We are assuming that friction and any other nonconservative forces are negligible.) Then

ME_f = ME_i

K_f + U_f = K_i + U_i

In the above equation, K is kinetic energy, U is gravitational potential energy, i stands for "initial" (at the top of the hill), and f stands for "final" (at the bottom of the hill.) Recall that K = (1/2)mv² and U = mgh (near the surface of the earth.) Thus, we may rewrite the above as

(1/2)mv_f² + mgh_f = (1/2)mv_i² + mgh_i

Divide both sides of the equation by m to obtain

(1/2)v_f² + gh_f = (1/2)v_i² + gh_i

From the statement of the problem, we have v_i = 0, and v_f = -16m/s² (if we take down as negative.) Also, for simplicity, we can take the bottom of the hill to be the 0 level for height. That is, h_f = 0. Now the previous equation becomes

(1/2)(-16m/s)² + (9.8m/s²⁾(0) = (1/2)(0)² + (9.8m/s²)h_i

128m²/s² = (9.8m/s²)h_i

which gives

h_i ≅ 13.1 meters.

Hope that help! Let me know if you need any more help.

William