Question by bassmstr: How far will the locomotive roll before it comes to a stop?
A 4.3×10^4 kg locomotive, with steel wheels, is traveling at 20 m/s on steel rails when its engine and brakes both fail. The coefficient of friction is 2.1×10^−3.
How far will the locomotive roll before it comes to a stop?
Thank you very much
Best answer:
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A 4.3×10^4 kg locomotive, with steel wheels, is traveling at 20 m/s on steel rails when its engine and brakes both fail. The coefficient of friction is 2.1×10^−3.
How far will the locomotive roll before it comes to a stop?
Thank you very much
Best answer:
Answer by Michael Jackson
The horizontal force acting on the locomotive (due to friction) is:
F = m*g*μ
The locomotive's kinetic energy is:
E = (1/2)*m*v^2
The only loss of energy is due to friction, so we set the total kinetic energy equal to the losses due to friction and solve for the distance traveled, x:
(1/2)*m*v^2 = F*x = m*g*μ*x
x = (1/2)*v^2/(g*μ) = (1/2)*(20)^2/(9.8*2.1*10^-3) = 9,718 m
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