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a 10kg box putting pushed with a force of 24 Newtons and a fraction of 12N. Determine coefficient of Friction and Determine acceleration of the Box
PLEASE HELP​

Sagot :

LammettHash

Assuming the force is applied horizontally over a flat surface, the net horizontal force is

F (h) = 24 N - 12 N = (10 kg) a

Solve for the acceleration a :

12 N = (10 kg) a

a = 1.2 m/s²

Meanwhile, the net vertical force is zero since the box of only moving horizontally.

F (v) = n - (10 kg) g = 0

where n is the magnitude of the normal force exerted upward by the surface. We see that

n = (10 kg) (9.8 m/s²)

n = 98 N

The friction force has a magnitude that is proportional to the normal force,

12 N = μ (98 N)

where μ is the coefficient of kinetic friction. Solve for μ :

μ = (12 N)/(98 N)

μ ≈ 0.12

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