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Soloha48 [4]
2 years ago
8

A 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2 , what is t

he coefficient of kinetic friction between the box and the inclined plane?
Physics
1 answer:
koban [17]2 years ago
6 0

Answer:

Explanation:

                            F = ma

mgsinθ - μmgcosθ = ma

      gsinθ - μgcosθ = a

                   μgcosθ = gsinθ - a

                              μ = (gsinθ - a) / gcosθ

                              μ = (9.81sin24 - 0.245) / 9.81cos24

                              μ = 0.4178906...

                              μ = 0.418

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Answer:

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From the question we are told that

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The  frictional torque that must be exerted is mathematically represented as

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substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

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