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