Answer:
The distance is [tex]s= 30.3 \ m[/tex]
Explanation:
From the question we are told that
The coefficient of static friction is [tex]\mu_s = 0.42[/tex]
The initial speed of the train is [tex]u = 57 \ km /hr = 15.8 \ m/s[/tex]
For the crate not to slide the friction force must be equal to the force acting on the train i.e
[tex]-F_f = F[/tex]
The negative sign shows that the two forces are acting in opposite direction
=> [tex]mg * \mu_s = ma[/tex]
=> [tex]-g * \mu_s = a[/tex]
=> [tex]a = -9.8 * 0.420[/tex]
=> [tex]a = -4.116 m/s^2[/tex]
From equation of motion
[tex]v^2 = u^2 + 2as[/tex]
Here v = 0 m/s since it came to a stop
=> [tex]s= \frac{v^2 - u^2 }{ 2 a}[/tex]
=> [tex]s= \frac{0 -(15.8)^2 }{ - 2 * 4.116}[/tex]
=> [tex]s= 30.3 \ m[/tex]
