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

Minimum time interval (t2)=0.90 SECONDS

Step-by-step 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 =
  `(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 =
  `(F)/(m)`

Making t subject of formula from equation (i)

• t=
  `\sqrt{(2d)/((f/m) }`

where

• 
  `(FF1)/(Mass of body)` = 4.905
• 
  `(FF2)/(Mass of body)` =7.9461

Let

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

• t1 =
  `\sqrt{(6.5)/(4.905) }` =1.15 seconds

• t2 =
  `\sqrt{(6.5)/(7.9461) }` = 0.90 seconds

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