Samantha H
New member
What is the magnitude of the force needed to keep a 5.5.-newton rubber block moving across level, dry asphalt in a straight line at a constant speed of 15.6 m/s?
[Include unites and answer to 3 decimal places]
OK....so i used the formula
Ff= Ufn or Force of friction= coefficient of friction x normal force, which i believe is the weight
So Ff= .67(coefficient of rubber on dry asphalt) x 5.5 N
Ff= 3.685
However, it keeps saying that my answer is wrong.....so can someone please explain to me wat I am doing wrong....Please and Thank Yous!!!!
[Include unites and answer to 3 decimal places]
OK....so i used the formula
Ff= Ufn or Force of friction= coefficient of friction x normal force, which i believe is the weight
So Ff= .67(coefficient of rubber on dry asphalt) x 5.5 N
Ff= 3.685
However, it keeps saying that my answer is wrong.....so can someone please explain to me wat I am doing wrong....Please and Thank Yous!!!!