A box moving on a horizontal surface with an initial velocity of 20 m/s slows to a stop over a time period of 5.0 seconds due solely to the effects of friction. What is the coefficient of kinetic friction between the box and the ground? Answer 0.41

μ = 0.41

Explanation:

The deceleration of the box can be calculated from the equations of motion

u = initial velocity of the box = 20 m/s

v = final velocity of the box = 0 m/s (since it comes to a stop)

t = 5.0 s

a = ?

v = u + at

0 = 20 + 5a

a = - 4 m/s² (minus sign because of deceleration)

The frictional force, Fr, that stops the box is the same force that causes this deceleration

That is,

Fr = ma

But also, Fr = μN = μmg

μmg = ma

μg = a

μ = (a/g)

μ = (4/9.8)

μ = 0.41

Hope this helps!!

object 2 because all the other ones are wrong

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