Question

Two workers are sliding 450 kg kg crate across the floor. One worker pushes forward on the crate with a force of 380 NN while the other pulls in the same direction with a force of 230 NN 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

The coefficient of kinetic friction on the floor is 0.138

Step-by-step explanation:

Given;

mass of the crate, m = 450 kg

force applied by the first worker, F₁ = 380 N

force applied by the second worker in the same direction as the first worker, F₁ = 230 N

frictional force opposing the motion of the box = -
`F_k`

Apply Newton's second law of motion;

∑F = ma

`F_1 + F_2 - F_k = ma`

If the crate slides with constant speed, acceleration is zero (0).

`F_1 + F_2 - F_k = ma = 0\\\\F_1 + F_2 - F_k = 0\\\\F_k = F_1 + F_2\\\\\mu _kmg= F_1 + F_2\\\\\mu _k = (F_1 + F_2)/(mg) \\\\\mu _k = (380 + 230)/(450 * 9.8) \\\\\mu _k = 0.138`

Therefore, the coefficient of kinetic friction on the floor is 0.138