Physics: The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.25 with the floor. If the train is initially moving at a speed of 48 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor

The speed of the lander just before it touches the surface is about 4.3 m/sFurther explanationAcceleration is rate of change of velocity.where:a = acceleration ( m/s² )v = final velocity ( m/s )u = initial velocity ( m/s )t = time taken ( s )d = distance ( m )Let us now tackle the problem!Given:position of the lander above the surface = h = 5.5 minitial downward speed of lander = u = 0.8 m/sacceleration due to gravity on the moon = g = 1.6 m/s²Asked:final speed of the lander = v = ?Solution:Conclusion...

The floor of a railroad flatcar is loaded with

Ответ:

stooopid

23.01.2022

Physics

$x_{min}=36.2m$

Explanation:

The forces acting on the crates when the train starts stopping are their weights, the normal force from the train, the static frictional force and the fictional force that is produced by the deceleration of the train. As the gravitational force, this fictional force is equal to the mass of the crates multiplied by the magnitude of the acceleration of the train. So, the equations of motion of the crates will be:

$x: ma_t-f_s=0\implies f_s=ma_t\\\\y: N-mg=0\implies N=mg$

Since the static frictional force is $f_s\leq \mu N$ , we get:

$ma_t\leq \mu mg\\\\\implies a_t\leq \mu g$

So we have a limit to the acceleration of the train. Now, we have to know the distance traveled by the train when it is stopping. Then, we use the kinematic formula:

$x=\frac{v_0^{2}}{2a_t}$