Question

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.420 with the floor. If the train is initially moving at a speed of 57.0 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor

Answer 1

Answer:

The distance is  $s= 30.3 \ m$

Explanation:

From the question we are told that

   The  coefficient of static friction is  $\mu_s = 0.42$

    The  initial speed of the train is  $u = 57 \ km /hr = 15.8 \ m/s$

   

For the crate not to slide the friction force must be equal to the force acting on the train i.e

       $-F_f = F$

The negative sign shows that the two forces are acting in opposite direction

=>   $mg * \mu_s = ma$

=>   $-g * \mu_s = a$

=>   $a = -9.8 * 0.420$

=>   $a = -4.116 m/s^2$

From equation of motion

  $v^2 = u^2 + 2as$

Here  v =  0 m/s since it came to a stop

=>   $s= \frac{v^2 - u^2 }{ 2 a}$

 =>   $s= \frac{0 -(15.8)^2 }{ - 2 * 4.116}$

=>     $s= 30.3 \ m$