A constant force of 85 N accelerates towards a 10 kg box from a speed of 3.0 m/s to a speed of 7.0 m/s as it goes 14 m along a horizontal floor. What is the coefficient of friction between the box and the floor?

Step-by-step explanation:

Let us find the acceleration of box .  

v² = u² + 2as

Putting the values

7² = 3² + 2 a x 14

a = 1.43 m /s²

If coefficient of friction be μ

force of friction = μ mg

= μ x 10 x 9.8

= 98μ

Net force pushing the box

= 85 - 98μ

Applying newton's second law

85 - 98μ = 10 x 1.43

98μ = 85 - 14.3

μ = .72

answer: the anser is 3

step-by-step explanation:

c- 2o is equal to 9

y = 3x+4

step-by-step explanation:

to find the slope

slope = (y2-y1)/ (x2-x1)

    = (-8-4)/ (-4-0)

    = -12/-4

  = 3

using the slope intercept form of a line

y= mx+b

we know the slope is 3

we know the y intercept is 4 ( the point where it crosses the y axis or x=0)

y = 3x+4