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 0.870. In an emergency, what is the minimum time interval in which a person starting from rest can move 3.20 m on a tile surface if she is wearing the following footwear? (a) footwear meeting the Postal Service minimum
(b) a typical athletic shoe

(a) 1.14 s

(b) 0.87 s

Explanation:

A person moves by the help of frictional force, as a result of gtound reaction. So, the formula for frictional force is:

F = μR

where,

F = frictional force

μ = coefficient of friction

R = Normal Reaction = Weight of Body = W = mg

Therefore,

F = μmg

but, from Newton's 2nd Law of Motion:

F = ma

Comparing both equations, we get:

μmg = ma

a = μg   equation (1)

Now, to calculate the distance moved by a body, we use 2nd equation of motion:

s = (Vi)(t) + (0.5)at²

using equation (1), we get:

s = (Vi)(t) + (0.5)μgt²

where,

s = distance moved by body

Vi = initial velocity of body

t = time taken to cover the distance

g = acceleration due to gravity

(a)

Vi = 0 m/s

g = 9.8 m/s²

s = 3.2 m

μ = 0.5

t = ?

Therefore,

3.2 m = (0 m/s)(t) + (0.5)(0.5)(9.8 m/s²)t²

t² = 3.2 m/(0.5)(0.5)(9.8 m/s²)

t = √1.30612 s²

t = 1.14 s

(a)

Vi = 0 m/s

g = 9.8 m/s²

s = 3.2 m

μ = 0.5

t = ?

Therefore,

3.2 m = (0 m/s)(t) + (0.5)(0.5)(9.8 m/s²)t²

t² = 3.2 m/(0.5)(0.5)(9.8 m/s²)

t = √1.30612 s²

t = 1.14 s

(a)

Vi = 0 m/s

g = 9.8 m/s²

s = 3.2 m

μ = 0.87

t = ?

Therefore,

3.2 m = (0 m/s)(t) + (0.5)(0.87)(9.8 m/s²)t²

t² = 3.2 m/(0.5)(0.87)(9.8 m/s²)

t = √0.75 s²

t = 0.87 s