A horizontal 20.-newton force is applied to a 5.0-kilogram box to push it across a rough,

horizontal floor at a constant velocity of 3.0 meters per second to the right.

Calculate the coefficient of kinetic friction between the box and the floor. [Show all work, including

the equation and substitution with units]

Question

Ware

Physics

2 February, 10:06

A horizontal 20.-newton force is applied to a 5.0-kilogram box to push it across a rough,

horizontal floor at a constant velocity of 3.0 meters per second to the right.

Calculate the coefficient of kinetic friction between the box and the floor. [Show all work, including

the equation and substitution with units]

+1

Answers (1)

Know the Answer?

Lizbeth Moss · Answer 1

The formula to work out the coefficient of dynamic (kinetic) friction is F=uR where F is the force applied, u is the coefficient of dynamic friction, and R is the reaction force. This is based on the principle the object is moving at a constant velocity which is the case in this question.

Rearranging this we get u=F/R

Substituting in the values we get u=20/50=2/5=0.4

Therefore the coefficient of kinetic friction between the box and the floor is 0.4