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3 years ago

A 35 n object is on a 25 degree incline. the force of friction up the incline is 8.0 N.

1 answer:

4 0

The downwards component of weight of the object would be 35sin25 degrees = 14.8

F=14.8 - 8 = 6.8 N                                         m= 35/9.8= 3.57 kg
F=mA
Therefore,
3.57A= 6.8 N
=> A= 6.8/3.57
=> A= 1.902 ms^-2

F(max)= (U)R                       where (U)= coefficient of friction
                                                   and R= Normal reaction force

R= 35cos25
   = 31.72 N
Since, F(max)= 8
          8= (U)* 31.72
      =>(U)= 8/31.72
      =>(U)= 0.25

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lukranit [14]

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U = c*n*R*T

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