Question #234656
2. A train of mass 9000 kg is on a horizontal track. Its engine provides a constant driving force
of 4000N. There is a constant air resistance of 400 N.
(a) Find the time taken to reach a speed of 48 m s-1
from rest.
(b) When travelling at 48 m s-1
the train enters a horizontal tunnel 400 m long. In the tunnel
air resistance increases to 1000 N. Find the speed at which the train leaves the tunnel.
1
Expert's answer
2021-09-08T16:14:27-0400

a)a)

m=9000kgm= 9000 kg

Ft=4000N driving force of the trainF_t= 4000N\text{ driving force of the train}

Far=400N air resistance forceF_{ar}=400N\text { air resistance force}

v0=0;v1=48msv_0=0;v_1= 48\frac{m}{s}

F=FtFar=4000400=3600NF = F_t-F_{ar}= 4000-400=3600N

F=maF= ma

a=Fm=36009000=0.4ms2a =\frac{F}{m}=\frac{3600}{9000}=0.4\frac{m}{s^2}

v(t)=v0+atv(t) = v_0+at

v1=v0+at1v_1 = v_0+at_1

t1=v1v0a=4800.4=120st_1 = \frac{v_1-v_0}{a}= \frac{48-0}{0.4}=120s

Answer: 120s\text{Answer: }120s

b)b)

m=9000kgm= 9000 kg

Ft=4000N driving force of the trainF_t= 4000N\text{ driving force of the train}

Far=1000N air resistance forceF_{ar}=1000N\text { air resistance force}

v0=48msv_0=48\frac{m}{s}

s=400ms = 400m

F=FtFar=40001000=3000NF = F_t-F_{ar}= 4000-1000=3000N

F=maF= ma

a=Fm=300090000.33ms2a =\frac{F}{m}=\frac{3000}{9000}\approx0.33\frac{m}{s^2}


s=v12v022as = \frac{v_1^2-v_0^2}{2a}

v1=2sa+v02=24000.33+48250.68msv_1=\sqrt{2sa+v_0^2}=\sqrt{2*400*0.33+48^2}\approx50.68\frac{m}{s}

Answer: 50.68ms\text{Answer: }50.68\frac{m}{s}



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