# Answer to Question #5438 in Mechanics | Relativity for Taylor

Question #5438

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?

Expert's answer

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.

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.

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