Question

Two workers are sliding 450 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 310 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:

Below

Explanation:

First we need to find the normal force acting on the object :

     Fn = mg

     Fn = (450kg)(9.8)

     Fn = 4410 N

Because the crate is sliding at a constant speed, there is no acceleration and therefore there is no applied force/the applied force is balanced with the friction according to Newton's 1st law.

In this case, there is applied force from the workers so the friction would have to equal the applied force in order for it to move at a constant velocity

     Fa = Ff

     450 + 310 = Ff

     750 N = Ff

Now we can plug these values into the formula for the coefficient of friction

     um = Fa / Fn

     um = 750 N / 4410

     um = 0.17

Hope this helps!