Question

two workers are sliding 330 kg crate across the floor. one worker pushes forward on the crate with a force of 430 n while the other pulls in the same direction with a force of 210 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

Two workers are sliding 330 kg crate across the floor. The crate's coefficient of kinetic friction on the floor is 0.197.

The proportion of friction force to normal force is known as the coefficient of friction, a dimensionless number.

$F_f = \mu F_n$

$F_f$ = frictional force

$\mu$ = coefficient of kinetic friction

$F_n$ = normal force

$\mu = \frac{F_1 + F_2}{mg}$

m = mass

g = acceleration due to gravity

To calculate the coefficient of friction use above equations :

μ = (430 + 210)/(330 x 9.8)

μ = 640 /3234

μ = 0.197

Two workers are sliding 330 kg crate across the floor. The crate's coefficient of kinetic friction on the floor is 0.197.

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