Compute the resulting acceleration

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To compute the resulting acceleration of a 100 kg trunk being pushed with a force of 500 N at a 30° angle, the normal force is calculated as 1230 N, factoring in the vertical component of the applied force. The kinetic friction force, using a coefficient of 0.3, is determined to be 369 N. The horizontal component of the applied force is 433 N, leading to a net force of 64 N after subtracting friction. Using Newton's second law, the acceleration is calculated as 0.64 m/s², which is noted as incorrect. Further clarification is sought on the application of the formulas to resolve the discrepancy in the acceleration calculation.
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Homework Statement



A 100 kg trunk loaded with old books is to be slid across a floor by a woman who exerts a force of 500 N down and forward at 30° with the horizontal. If μk = 0.3 and μs = 0.4, compute the resulting acceleration.


Homework Equations

and

The Attempt at a Solution



N = mg, of trunk
N = 100kg * 9.8 = 980 N
Push of lady :
N = F sin theta
N = 500N * sin 30 = 250N
Total force being exerted = 980 + 250N = 1230N
friction kinetic force:
f = u*N
f = .3 * 1230N
friction force of moving box = 369N

woman pushes with force:
500N * cos 30 = 433N
So with 433N force of push - 369N friction = 64N of net force on sliding box
Net force = mass * acceleration
64N = 100kg * a
acceleration = 0.64 m/s^2 => INCORRECT => Please help!
 
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we know the formula F=macos theta. and F=mue*mg.from this u can find the force and equal both and find out accelaration
 

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