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A block of mass 3 kg slides along a horizontal surface while a 20-N force is applied to it at an angle of 25°. If needed, use g = 10 m/s2. For a coefficient of kinetic friction of 0.3 between the block and the surface, the frictional force acting on the block is most nearly

9 N
8.5 N
15 N
6 N
12 N

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  1. friction=mu*mg=.3*3*10 N

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    bobpursley
  2. Is the 25 degree angle of the applied force above or below horizontal? It makes a big difference in the answer.

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  3. If the angle is 25 degree below the horizontal, then the upward component makes the friction lighter (less normal force).

    friction=mu(normal force)
    = .3(mg-20sin25)
    =.3(30-8.45)
    = no answer.

    so assuming the 25 is above the horizontal, friction will be greater..
    friction= = .3(mg+20sin25)
    =.3(38.45)= almost 12 N

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    bobpursley
  4. F(y) = F•sinα =20•sin25º =8.45 N.
    The force may be applied
    \ from top downward
    or / from the bottom upward ,
    then
    F(fr) =μ•N = μ•[mg ± F(y)] =
    = 0.3•{3•10 ± 8.45) =
    = 11.54 N or 6.5 N,

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