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.

Answers

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
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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