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

Answer:

The coefficient of kinetic friction on the floor is 0.138

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).

$F_1 + F_2 - F_k = ma = 0\\\\F_1 + F_2 - F_k = 0\\\\F_k = F_1 + F_2\\\\\mu _kmg= F_1 + F_2\\\\\mu _k = \frac{F_1 + F_2}{mg} \\\\\mu _k = \frac{380 + 230}{450 \times 9.8} \\\\\mu _k = 0.138$

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