Dr. Joule is pushing a 75 kg refrigerator across the floor at a constant velocity. The coefficient of friction is 0.36 and he moves the refrigerator 3.5 meters. Solve for Force of Friction and pushing Force.

I got force of friction to be 260N and used F_{f}=µ_{k}F_{n} but I can't get the pushing force because it says that there is a constant velocity, which isn't stated, so you can't solve for it nor can you solve for acceleration.

Guest Nov 30, 2018

#1**+1 **

If there is no acceleration, then the forces are balanced....the pushing force is exactly the same in magnitude as the force of the KINETIC friction. 264 N although it is in the OPPOSITE direction as the frictional force.

ElectricPavlov
Nov 30, 2018

#2**+1 **

But if the forces are equal but opposite, doesn't that mean that the object hasn't moved at all?

Guest Nov 30, 2018

#4**+1 **

I think that applies to STATIC friction, but kinetic friction force and force to move object w/out accel are the same as I recall.

ElectricPavlov
Nov 30, 2018

#5**+1 **

I found this with a quick internet search:

Is kinetic friction equal to applied force?

If the externally applied force (F) is equal to the force of kinetic friction, Fk, then the object slides at constant velocity, and the coefficient of friction involved is called the coefficient of kinetic friction, μk.

ElectricPavlov
Nov 30, 2018

#3**+1 **

Dr. Joule is pushing a 75 kg refrigerator across the floor at a constant velocity. The coefficient of friction is 0.36 and he moves the refrigerator 3.5 meters. Solve for Force of Friction \(F_f\) and pushing Force \(F_p\).

**Hello Guest!**

\(F_p=m\times g = 75kg\times 9.81\frac{m}{s^2}=735.75\frac{kgm}{s^2}\times\frac{Ns^2}{kgm}\\ \color{blue }F_p=735.75N\)

\(F_f=F_p\times \mu_k=735.75N\times 0.36\\ \color{blue}F_f=264.87N\)

\(W=F_f\times s=264.87N\times 3.5m \times \frac{J}{Nm}\\ \color{blue}W=927.045J\)

!

asinus
Nov 30, 2018