Question

A train with mass 40000 kg moving with a velocity of 60 m\ s comes at rest in 10 sec by applying the brakes. What is the retardation of a train and force on the brake?

Answer 1

Explanation:

mass = 40000kg

initial v. = 60m/s

final v. = 0m/s

time = 10sec

F = ma

a =??

v = u + at ----(1 eq)

0 = 60 + (a)10

-60/10 = a

a = -6 m/s² ------ retardation

F = (40000)(-6)

= -24000N ------- force applied to stop

Answer 2

↬ Given :-

• Mass (m) = 40000 kg

• Initial velocity (u) = 60 m/s

• Time (t) = 10 sec

• Final velocity (v) = 0 m/s

↬To find :-

• Retardation of the train = ?

• Force on the break = ?

↬ Solution :-

$\green{\bigstar}\:\:\underbrace{\bf\pink{{Using\:{1}^{st}\:equation\:of\:motion}}}$

$\orange{\bigstar}\:\:{\underline{\boxed{\bf\red{v=u+at}}}}$

$\rm\longrightarrow\:0=60+a\times\:10$

$\rm\longrightarrow\:0=60+10a$

$\rm\longrightarrow\:-60=10a$

$\rm\longrightarrow\:\cancel\dfrac{-60}{10}=a$

$\rm\longrightarrow\:-6=a$

$\rm\longrightarrow\:a=-6\:m/{s}^{2}$

Hence,the retardation of the train will be -6 m/s².

Now,

we know that,

$\blue{\bigstar}\:\:{\underline{\boxed{\bf\purple{F=ma}}}}$

$\rm\longrightarrow\:f=40000\times(-6)$

$\rm\longrightarrow\:f=-240000\:N$

Hence,the force on the break will be -240000 N.