Forces/Friction?

Question

What is the min. time in which a person starting at rest can move 3 m on a tile surface if he is wearing shoes w/ friction (k) = .5

(3 sig digs) be very detailed

Answer 1

F=Ma, where F is the net force, M is the mass of the person in kg, and a is his acceleration in m/s^2

F in this problem is also the force of friction acting on the person's shoes. The formula for force of friction, F is:

F=uR where u is the coefficient of kinetic friction , and
R is the weight of the person=Mg
Therefore,

Ma=0.5*M*9.8 M cancels out.
a =4.9m/s^2

s=ut+1/2at^2 where s is the distance moved=3m; u the initial velocity of the person=0, because he started from rest; a the acceleration =4.9m/s^2, and t the time in sec.

Substitute known values:

3=0*t+1/2*4.9*t^2
3=2.45t^2
t^2=3/2.45
=1.22
t=1.104 sec.

Answer 2

Normal force times mu is available traction. F = (mass)(g)(mu) is an equation describing this relationship. Term ma from F = ma can be substituted for "F". This gives ma = mg(mu). Mass cancels, giving a = mu(g).

Use equation v = sqrt(2as). Drop in mu(g) in place of "a" and 3 m for "s":

v = sqrt(2mu(g)(3meters))

So square root of two times one half times g times 3 = 5.42 m/s