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Komok [63]
3 years ago
5

A5 kg box slides 3 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and t

he box if the box had an initial speed of 3 m / s?
Physics
1 answer:
frutty [35]3 years ago
6 0

Answer:

Coefficient of kinetic friction (Cof. KE) = 0.153

Explanation:

Given:

Mass of box (M) = 5 kg

Distance = 3 m

Initial speed (v) = 3 m/s

Find:

Coefficient of kinetic friction (Cof. KE)

Computation:

v² = u² + 2as

a = v² / 2s

a = 9 / 2(3)

a = 1.5 m/s²

Coefficient of kinetic friction (Cof. KE) = a / g

Coefficient of kinetic friction (Cof. KE) = 1.5 / 9.8

Coefficient of kinetic friction (Cof. KE) = 0.153

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Using the two kinematic equations that can be used for this problem are:
Vf = Vi + at and d=Vit +(1/2)*at^2
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The equations can further be simplified where a is the acceleration, t is the time, Vf is the final velocity which is 70 miles per hour and d is 6 miles

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Substituting equation 1 to equation 2.
6= (1/2)*(70/t)*(t^2)
6= 35t
t= 0.17142 hours or 10.28571 mins or 617.14 sec



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