Question

Two workers are sliding 310 kg crate across the floor. One worker pushes forward on the crate with a force of 380 N while the other pulls in the same direction with a force of 270 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

`0.214`

Step-by-step explanation:

The sketch is in the attachment.

since its speed is constant, the acceleration is zero .

The total force will be:

`F=380N+270N\\F=650N`

mass of the block is:
`m=310kg`

so now we can apply the formula to find the crate's coefficient of kinetic friction.

`F=u_(k) mg\\u_(k) =(F)/(mg) \\u_(k)=(650)/(310*9.81) \\u_(k) =0.214`