Question

Two workers are sliding 490 kg crate across the floor. One worker pushes forward on the crate with a force of 410 N while the other pulls in the same direction with a force of 260 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:

μ=0.16

Explanation:

Mass of crate = 490 kg

Pushing force = 410 N

Pulling force = 260 N

Lets coefficient of friction =μ

The normal force,N= mg

N= 490 x 9.81 =4806.9 N

We know that friction force(fr)

fr= μ m g

fr =4806.9 μ

The total horizontal force F

F= 410 + 260

F=670 N

For sliding condition the maximum friction force will be equal to total horizontal force

F=fr

670 = 4806.9 μ

μ=0.16

So the coefficient of friction is 0.16.

Answer 2

Answer:

$\mu = 0.14$

Explanation:

There are two forces acting on the box to slide it over the horizontal surface

1) $F_1 = 410 N$ in forward direction

2) $F_2 = 260 N$ in forward direction

now it is given that box is moving with constant speed so net force on the box must be zero

$F_1 + F_2 - f = 0$

so we will have

$410 + 260 - f = 0$

$f = 670 N$

now by the formula of friction force we know that

$f = \mu mg$

$670 = \mu (490 \times 9.81)$

$\mu = 0.14$