Question

Bonnie and Clyde are sliding a 325 kg bank safe across the floor to their getaway car. 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.

Required:
What is the safe's coefficient of kinetic friction on the bank floor?

Answer 1

Answer:

the safe's coefficient of kinetic friction on the bank floor is $\mathbf{\mu_k =0.2290}$

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 $F_{friction}$ is acting in the opposite direction ←

where;

$F_{friction}$  = $\mu_k mg$

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;

$F_{net} = F_{Cylde} + F_{Bonnie} - F_{frition}$

$F_{net} = F_{Cylde} + F_{Bonnie} - \mu_k \ mg$

Since the net force acting on the slide will be equal to zero.

Then; $F_{net} =0$

Also; let $F_{Cylde} = F_c$ and $F_{Bonnie} = F_B$

Then;

$0 = F_c + F_B - \mu_k \ mg$

$\mu_k \ mg= F_c + F_B$

$\mu_k = \dfrac{F_c + F_B}{\ mg}$

where;

$F_c = 377 \ N \\ \\ F_B = 353 \ N \\ \\ mass (m) = 325 \ kg$

Then;

$\mu_k = \dfrac{377 + 353}{325*9.81}$

$\mu_k = \dfrac{730}{3188.25}$

$\mathbf{\mu_k =0.2290}$

Thus; the safe's coefficient of kinetic friction on the bank floor is $\mathbf{\mu_k =0.2290}$