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kherson [118]
2 years ago
7

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
Physics
1 answer:
bonufazy [111]2 years ago
7 0

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

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