Answer to Question #5438 in Mechanics | Relativity for Taylor
a 2000 kg truck rolls down a hill which is 10 m high. the kinetic energy at the bottom is found to be 1.5 x 10^5 J. what is the efficiency of the ramp in converting gravitational potential energy into kinetic energy?
Let's make some denotations: M = 2000 kg; Ek = 1.5 x 10^5 J; H = 10m.
Let's find kinetic energy of a truck: Eg = M*g*H = 2000*9.8*10m = 1.96*10^5 J.
The coefficient of efficiency of the ramp is the ratio of potential and kinetic energy: k = Ek/Eg = 1.5*10^5 / 1.96*10^5 = 1.5/1.96 ≈ 0.7653.