Question

I know that the answer is C, but I do not understand why this is the case.

A 5.00-kg box slides 4.00 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?

A) 0.229 B) 0.267 C) 0.115 D) 0.587 E) 1.13

Answer 1

Initial speed of box = 3.00 m/sec

final speed of box = 0 m/sec

distance traveled by box = 4.00 m

So de-acceleration of box will be, using 3rd kinematic equation

V^2 = U^2 + 2*a*s

a = (V^2 - U^2)/(2*s)

a = (0^2 - 3^2)/(2*4)

a = -1.125 m/sec

Now net force on box will be

Fnet = Fp - Ff = m*a

Fp = applied force = 0 N (Since there is no external force applied)

Ff = uk*N = uk*m*g

-uk*m*g = m*a

uk = -a/g

uk = -(-1.125)/9.81

uk = Coefficient of kinetic friction = 0.115

Correct option is C.

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