Question
A train has a mass of 5.22 multiplied by 106 kg and is moving at 70.0 km/h. The engineer applies the brakes, which results in a net backward force of 1.87 multiplied by 106 N on the train. The brakes are held on for 37.0 s.
(a) What is the new speed of the train?
(b) How far does it travel during this period?
(a) What is the new speed of the train?
(b) How far does it travel during this period?
Answers
bobpursley
A train has a mass of 5.22*106kg? That is a really lightweight train. Think about that. My pickup truck weighs much more than that.
Vf=vi+a t
change vi 70km/h to m/s
a= force/mass= -1.87*106N/(5.22*106)
solve for vf
distance: avg velocity*time= (vi+Vf)*t/2
Vf=vi+a t
change vi 70km/h to m/s
a= force/mass= -1.87*106N/(5.22*106)
solve for vf
distance: avg velocity*time= (vi+Vf)*t/2