Question

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.480 with the floor. If the train is initially moving at a speed of 61.0 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:30.50 m

Explanation:

Given

coefficient of friction between railroad and crates $\mu =0.48$

Train initial velocity (u)$=61 kmph \approx 16.94 m/s$

To get the shortest distance brakes applied should be order of friction force between crates and railroad floor

$a=\mu g=0.48\times 9.8=4.704 m/s^2$

using $v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration or deceleration

s=distance

here v=0 , u=16.94 m/s

$0-(16.94)^2=2\times (-4.704)\times s$

$s=\frac{(16.94)^2}{2\times 4.704}$

s=30.502 m