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

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

Step-by-step 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 = (F_c + F_B)/(\ mg)`

where;

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

Then;

`\mu_k = (377 + 353)/(325*9.81)`

`\mu_k = (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}`