Answer to Question #234656 in Mechanics | Relativity for Elham

Question

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.

Answer 1

$a)$

$m= 9000 kg$

$F_t= 4000N\text{ driving force of the train}$

$F_{ar}=400N\text { air resistance force}$

$v_0=0;v_1= 48\frac{m}{s}$

$F = F_t-F_{ar}= 4000-400=3600N$

$F= ma$

$a =\frac{F}{m}=\frac{3600}{9000}=0.4\frac{m}{s^2}$

$v(t) = v_0+at$

$v_1 = v_0+at_1$

$t_1 = \frac{v_1-v_0}{a}= \frac{48-0}{0.4}=120s$

$\text{Answer: }120s$

$b)$

$m= 9000 kg$

$F_t= 4000N\text{ driving force of the train}$

$F_{ar}=1000N\text { air resistance force}$

$v_0=48\frac{m}{s}$

$s = 400m$

$F = F_t-F_{ar}= 4000-1000=3000N$

$F= ma$

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

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

$v_1=\sqrt{2sa+v_0^2}=\sqrt{2*400*0.33+48^2}\approx50.68\frac{m}{s}$

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