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?(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Frictional force

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