Question

To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a specified tile surface. A typical athletic shoe has a coefficient of 0.810. In an emergency, what is the minimum time interval in which a person starting from rest can move 3.25 m on a tile surface if she is wearing the athletic shoe?

Answer 1

Answer:

Minimum time interval (t2)=0.90 SECONDS

Explanation:

• coefficient of friction for employees footwear = 0.5

• coefficient of friction for typical athletic shoe = 0.810

• frictional force = coefficient of friction X acceleration due to gravity X mass of body
• Acceleration due to gravity is a constant = 9.81 m/s
• Let frictional force for employee footwear = FF1
• Let frictional force for athletic footwear =FF2

                 FF1 = O.5 X 9.81 X mass of body

                         = 4.905 x mass of body

                  FF2 = 0.810 X 9.81 X mass of body

                          = 7.9461 x mass of body

The body started from rest there by making the initial velocity zero ( u = 0)

From d= ut + 1/2 a x $t^{2}$

•      d = $\frac{1}{2}$ x a x $t^{2}$  .....................................i  

            where d= distance and it is given as 3.25m

•          F =ma  ...................................ii

making acceleration subject of the formula from equation ii

•              a =$\frac{F}{m}$

         Making t subject of formula from equation (i)

• t=$\sqrt{\frac{2d}{(f/m} }$

    where

• $\frac{FF1}{Mass of body}$ = 4.905
• $\frac{FF2}{Mass of body}$ =7.9461

  Let

•            t1 = minimum time taken for frictional force for employee foot wear

•                                 t1 = $\sqrt{\frac{6.5}{4.905} }$ =1.15 seconds

•                                  t2 = $\sqrt{\frac{6.5}{7.9461} }$ = 0.90 seconds

 

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