Answer : A ) 0.4
B ) Should be greater than 40N .
A) Here in the first question it's given that a block of mass 3kg is pushed with a force of 12N so it just starts to move. This means the maximum value of static friction ( limiting value) is 12N . We know that ,
[tex]\qquad\qquad\sf\longrightarrow\Large{\boxed{\pink{\sf F_{ms}=\mu_s N}}} [/tex]
- Where [tex]\mu_s[/tex] is the coefficient of static friction and N is the normal reaction .
- Since the block is on horizontal , Normal force is equal to the weight of the block that is mg . So ,
[tex]\sf\longrightarrow 12N = \mu_s(mg)\\ [/tex]
[tex]\sf\longrightarrow 12N=\mu_s (3kg)(10m/s^2) \\ [/tex]
[tex]\sf\longrightarrow \mu_s = \dfrac{12N}{30N} \\ [/tex]
[tex]\sf\longrightarrow \underline{\boxed{\bf \mu_{static}= 0.4}} \\ [/tex]
[tex]\rule{200}2[/tex]
B) Now coming to second part of the question which says that a 7kg block is kept on a 3kg block as before . We need to find out the magnitude of force that is required to move the system .
- The total mass of the system = 3kg + 7kg = 10kg .
- Here since it's not mentioned whether the contact surface between the blocks is rough or not ,I am assuming it to be smooth .
- To make the system move the applied force should be greater than the frictional force , from the previous question the coefficient of static friction was 0.4 .
So that ,
[tex]\sf\longrightarrow F_{(applied)}> F_{ms} \\ [/tex]
[tex]\sf\longrightarrow F_{(applied)}> \mu_s mg \\ [/tex]
[tex]\sf\longrightarrow F_{(applied)}> 0.4( 10kg)(10m/s^2) \\ [/tex]
[tex]\sf\longrightarrow \underline{\boxed{\bf F_{(applied)}> 40N }}\\ [/tex]
