Two workers are sliding 500 kg crate across the floor. One worker pushes forward on the crate with a force of 440 N while the other pulls in the same direction with a force of 340 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

0.159

Explanation:

Given that

F1 = 440 N

F2 = 340 N

m = 500 kg

acceleration, a = 0 m/s²

Remember that F = m.a

If F - f(f) = 0, then F = f(f)

N = mg

f(f) = μN, substituting for N, we have

f(f) = μmg

The total force, F = F1 + F2

F = 440 + 340

F = 780 N

Recall that we'd already proven that

F = f(f), so F = 780 N = f(f)

And again, f(f) = μmg, if we substitute for all the values, we have

780 = μ * 500 * 9.81

780 = μ * 4905

μ = 780 / 4905

μ = 0.159

Therefore, the coefficient of static friction of the crate on the floor is 0.159

the answer is b ) square root of 5