Asked by Diana princess

A metal block of mass 5kg lies on a rough horizontal platform. If a horizontal force of 8 N applied to the block through its center of mass just slides the block on the platform, then the coefficient of limiting friction between the block and the platform is: A 0.16 B 0.63 C 0.80 D 1.6 E 2.00.
Answered by bobpursley
coefficent= friction/weight= 8/(5*9.8)= ...
Answered by henry2,
M*g = 5*9.8 = 49 N. = Wt. of block = Normal force,

8-u*49 = 5*0.
u = coefficient of friction = 8/49 = 0.16.
Answered by muhammad
answer question which above
Answered by Nneoma iheanyi
The answer is A: 0.16
Answered by sanusi Umar toranka
Yes it is A.0.16
Coefficient of friction =force/mass*acceleration due to gravity. That is 8/5*10 =O.16
Answered by FRIDAY
A METAL BLOCK OF MASS IS 5KG LIES
Answered by Emmanuella
Yes it is 0.16
Answered by Ekeng
Yes it is 0.16
Answered by Jatau
The answer is 0.16
Answered by Momoh abdulqahhar Oluwajuwonlo
Solution
F/mg
8/5*10=0.16
Answered by Emmaunel
Yes it is 0.16
F-uR
8-u*5*10
U-0.16
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