Answer:
the safe's coefficient of kinetic friction on the bank floor is [tex]\mathbf{\mu_k =0.2290}[/tex]
Explanation:
GIven that:
Bonnie and Clyde are sliding a 325 kg bank safe across the floor to their getaway car.
So ,let assume they are sliding the bank safe on an horizontal direction
Clyde → Δ(bank safe) → Bonnie
Also; from the above representation; let not forget that the friction force [tex]F_{friction}[/tex] is acting in the opposite direction ←
where;
[tex]F_{friction}[/tex] = [tex]\mu_k mg[/tex]
The safe slides with a constant speed
If Clyde pushes from behind with 377 N of force while Bonnie pulls forward on a rope with 353 N of force.
Thus; since the safe slides with a constant speed if the two conditions are met; then the net force acting on the slide will be equal to zero.
SO;
[tex]F_{net} = F_{Cylde} + F_{Bonnie} - F_{frition}[/tex]
[tex]F_{net} = F_{Cylde} + F_{Bonnie} - \mu_k \ mg[/tex]
Since the net force acting on the slide will be equal to zero.
Then; [tex]F_{net} =0[/tex]
Also; let [tex]F_{Cylde} = F_c[/tex] and [tex]F_{Bonnie} = F_B[/tex]
Then;
[tex]0 = F_c + F_B - \mu_k \ mg[/tex]
[tex]\mu_k \ mg= F_c + F_B[/tex]
[tex]\mu_k = \dfrac{F_c + F_B}{\ mg}[/tex]
where;
[tex]F_c = 377 \ N \\ \\ F_B = 353 \ N \\ \\ mass (m) = 325 \ kg[/tex]
Then;
[tex]\mu_k = \dfrac{377 + 353}{325*9.81}[/tex]
[tex]\mu_k = \dfrac{730}{3188.25}[/tex]
[tex]\mathbf{\mu_k =0.2290}[/tex]
Thus; the safe's coefficient of kinetic friction on the bank floor is [tex]\mathbf{\mu_k =0.2290}[/tex]
