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4-2.Friction
hard
A block of mass $M$ placed on rough surface of coefficient of friction equal to $3$ . If $F$ is the $(4/5)$ of the minimum force required to just move. Find out the force exerted by ground on the block
A$2.6\ Mg$
B$Mg$
C$4\ Mg$
D$3.4\ Mg$
Solution
The force exerted by ground on the block
=$\sqrt {{N^2} + {f^2}} $
=$\sqrt {{{\left( {Mg} \right)}^2} + {{\left[ {\left( {\frac{4}{5}} \right)\left( {3Mg} \right)} \right]}^2}} $ = $\frac{{13}}{5}Mg$ = $2.6Mg$
=$\sqrt {{N^2} + {f^2}} $
=$\sqrt {{{\left( {Mg} \right)}^2} + {{\left[ {\left( {\frac{4}{5}} \right)\left( {3Mg} \right)} \right]}^2}} $ = $\frac{{13}}{5}Mg$ = $2.6Mg$
Standard 11
Physics
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