If a bullet train of mass 1.40x10^8 kg performs 9.25x10^11 Joules of work in moving 4.00x10^3 meters on level ground at constant speed, the coefficient of kinetic friction between the train and rails is _____ ?
I know that the answer is 0.169 but I'm not sure how to get the answer and I've tried several ways.
Can someone please help? I always give five stars!
I know that the answer is 0.169 but I'm not sure how to get the answer and I've tried several ways.
Can someone please help? I always give five stars!
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work = force * distance
9.25 * 10^11 = F * 4*10^3
F = 2.3125 * 10^8 Newtons
so thats the force on the train. since it is at constant velocity, friction is balancing that force, so the frictional force is the same 2.3125 * 10^8 Newtons
Friction force = coefficient of friction * normal force
normal force = weight of train
F = ma
F = 1.40x10^8 * 9.8
F = 1.372 * 10^9 Newtons
back to this
Friction force = coefficient of friction * normal force
2.3125 * 10^8 = coefficient of friction * 1.372 * 10^9
coefficient of friction = 0.169
9.25 * 10^11 = F * 4*10^3
F = 2.3125 * 10^8 Newtons
so thats the force on the train. since it is at constant velocity, friction is balancing that force, so the frictional force is the same 2.3125 * 10^8 Newtons
Friction force = coefficient of friction * normal force
normal force = weight of train
F = ma
F = 1.40x10^8 * 9.8
F = 1.372 * 10^9 Newtons
back to this
Friction force = coefficient of friction * normal force
2.3125 * 10^8 = coefficient of friction * 1.372 * 10^9
coefficient of friction = 0.169