Ask Question
15 April, 14:52

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]

+4
Answers (1)
  1. 15 April, 18:30
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers