A 24-N horizontal force is applied to a 40-N block initially at rest on a rough horizontal surface. If the coefficients of friction are μs = 0.5 and μk = 0.4, the magnitude of the frictional force on the block is:
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Answer:
Initially when body was in rest and force of 24N was applied then
Magnitude of static friction will be
F(s)=μN = 0.4×40= 16N
As it's less then 24N so body will start moving
So kinetic friction will come into play
Hence force of friction will be
F(k)= μN = 0.5×40 = 20N
Explanation:
I'm not sure
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0
Answer:
16 N
Explanation:
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