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?
1
Expert's answer
2011-12-06T09:31:22-0500
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.
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
Comments
Leave a comment