Calculating Crate Acceleration: Force, Friction, and Net Force

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To calculate the acceleration of a 210-kg crate pushed with a force of 730N and experiencing a kinetic friction coefficient of 0.19, the force of friction is determined to be 391N. The net force acting on the crate is found by subtracting the frictional force from the applied force, resulting in a net force of 339N. Using Newton's second law, the equation is set up as 339 = 210 * a, leading to an acceleration of approximately 1.6141 m/s². Clarifications were made regarding the correct calculation of friction and the application of Newton's second law. The final acceleration value is confirmed to be accurate based on the corrected formulas.
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A 210-kg crate is pushed horizontally with a force of 730N. If the coefficient of the kinetic friction is 0.19, calculate the acceleration of the crate.

Attempt: FN= 2058

Ff=0.19(391)
Force of friction= 391N

Fnet= Fpush-Ff

ma= Fpush - Ff
2058a = 730 - 391

a=339/2058

can anyone check if my answers right, and if I am wrong explain how to do it?
 
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ok, first, i assume this is just a typo but you wrote Ff=0.19(391), but Ff = .19*2058, which i assume is 391.
now more importantly, you say "ma= Fpush - Ff" which is correct but "2058a = 730 - 391" is not. Remember it's m*a for the left part, while you have FN*a
 


Simple method:

Force by push - force by friction = mass of block *acceleration.

730-.19*(210*9.8) = 210 * a
>338.98= 210 * a
>a= 338.98/210= 1.6141 m/s^2
 

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