Question

Two workers are sliding 470 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 330 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

Answer 1

Answer:

0.169

Explanation:

There are three forces acting on the crate along the horizontal direction:

- The pushing force of the first worker, F1 = 450 N forward

- The pushing force of the second worker, F2 = 330 N forward

- The frictional force $F_f$ acting backward

The crate slides with constant speed, so its acceleration is zero: a = 0. This means that we can write Newton's second law as

$\sum F = ma = 0\\F_1 + F_2 - F_f = 0$

The frictional force can be rewritten as

$F_f = \mu mg$

where

$\mu$ is the coefficient of kinetic friction

m = 470 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

Substituting everything into the previous equation, we find:

$F_1 + F_2 - \mu mg = 0\\\mu = \frac{F_1 + F_2}{mg}=\frac{450 N+330 N}{(470 kg)(9.8 m/s^2)}=0.169$