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bchq333
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< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >
A 50,000kg locomotive is traveling at 10m/s on a level track when the engines and brakes both fail. If there is a frictional force of 142N acting to slow it down, how far will it roll before it stops?
a=F/m
t=v/a
d=vt
a=142N/50,000kg
a=2.84*10^-3
t= 10/2.84*10^-3
t=3521 sec
d=10*3521
d=35211 metres
Stopping distance = 35.21 km? This sounds way too long. Can anyone point me in the right direction?
A 50,000kg locomotive is traveling at 10m/s on a level track when the engines and brakes both fail. If there is a frictional force of 142N acting to slow it down, how far will it roll before it stops?
a=F/m
t=v/a
d=vt
a=142N/50,000kg
a=2.84*10^-3
t= 10/2.84*10^-3
t=3521 sec
d=10*3521
d=35211 metres
Stopping distance = 35.21 km? This sounds way too long. Can anyone point me in the right direction?
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