Block pushed against wall: friction force with 12 N horizontal push

[disruptors]

A horizontal force of 12 N pushes a 0.5 Kg block against a vertical wall. The block is initially at rest, If static and kinetic coefficients are 0.6 and 0.8 respectively what is the friction force?

Basically the fbd has Force of 12 N across the x component(+) and the Normal force in the opposite (-ve)... The y component is positive facing downwards thus Fnety=mg-frictionforce=ma(y)... Friction force is equal to coefficient of static x Normal... static is 0.6 and Normal force is 12 N, which gives 7.2 N, right?

(i am doubting myself because some other ppl didnt get that)

Thanks

 

[Zlex]

Thats an odd question.

The static Ff would seem to be lower then the kinetic Ff

 

[wais]

for asking this question! Your solution looks correct. The friction force can be calculated by multiplying the coefficient of static friction (0.6) by the normal force (12 N). This gives a friction force of 7.2 N, which is the maximum amount of force that can be applied before the block starts to move. Once the block starts moving, the friction force will decrease to the coefficient of kinetic friction (0.8) multiplied by the normal force (12 N), which would give a friction force of 9.6 N. It's important to remember that the coefficient of friction is a measure of the amount of friction between two surfaces, and the actual force of friction will depend on the normal force and whether the object is in motion or at rest. Keep up the good work with your calculations!