One More Accelerated Parallel Force

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SUMMARY

The discussion focuses on calculating the parallel force required to accelerate a 5.0 kg object on a horizontal surface at 12 m/s², given coefficients of static friction (μs = 0.15) and kinetic friction (μk = 0.07). The normal force (FN) is determined to be 49 N, leading to a maximum static frictional force of 7.35 N. The total force required combines the frictional force and the force needed for acceleration, resulting in a calculated force of 63.4 N, confirming option D as the correct answer.

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"For a 5.0 kg object on a horizontal surface that has a coefiicient of static friciton where us = 0.15 and a coefficient of kinetic friction where uk = 0.07, what is the parallel force necessary to accelerate the object at 12 m/s^2."

a) 53.6 N
b) 56.6 N
c) 60.0 N
d) 63.4 N
e) 67.4 N

I found that the normal force (FN) is 49 N. The max force is 7.35 N. However, I don't know how I would go about finding a force needed to accelerate an object. ...help...
 
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Which mu should you use if the object is moving? The force in F=ma will be the combination of the frictional force and the force required to accelerate the mass...
 
by using F=ma, 5*12, I get 60. For the uk force, I multiplied 49*.07. then uk force + force = 63.4, answer D
 

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