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 0.825. in an emergency, what is the minimum time interval in which a person starting from rest can move 2.40 m on a tile surface if she is wearing the following footwear?

the minimum time interval is 0.77 seconds

Explanation:

given data

coefficient of static friction = 0.5

coefficient of static friction shoes= 0.825

travel s = 2.40 m

to find out

what is the minimum time interval

solution

we know newton 2nd law

force = mass × acceleration

and

force acting on person due to friction

force = coefficient of static friction shoes × mg

so we can say

coefficient of static friction shoes × mg = ma

so

a = coefficient of static friction shoes × g

and we know g is 9.8 m/s²

so

distance formula by kinematic relation

distance = ut + 0.5 × at²

here put a value and u is zero because initial speed

2.40 = 0 + 0.5  × coefficient of static friction shoes × g× t²

2.40 = 0.5  × 0.825 × 9.8 × t²

t = 0.77 s

the minimum time interval is 0.77 seconds

v=ir = 3x15 =45 v

momentum of soft ball three times less than baskey ball