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 [tex]t^{2}[/tex]
- d = [tex]\frac{1}{2}[/tex] x a x [tex]t^{2}[/tex] .....................................i
where d= distance and it is given as 3.25m
- F =ma ...................................ii
making acceleration subject of the formula from equation ii
- a =[tex]\frac{F}{m}[/tex]
Making t subject of formula from equation (i)
- t=[tex]\sqrt{\frac{2d}{(f/m} }[/tex]
where
- [tex]\frac{FF1}{Mass of body}[/tex] = 4.905
- [tex]\frac{FF2}{Mass of body}[/tex] =7.9461
Let
- t1 = minimum time taken for frictional force for employee foot wear
- t1 = [tex]\sqrt{\frac{6.5}{4.905} }[/tex] =1.15 seconds
- t2 = [tex]\sqrt{\frac{6.5}{7.9461} }[/tex] = 0.90 seconds
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