Physics Question: I already have the answer..I just forgot how I got it...

So I'm reviewing my old tests to study for the final ....

A 7.6 kg block on a table (on Earth) is pushed by a force F = 97 N. If the coefficient of kinetic friction between the block and the table is 0.4 the acceleration of the block is ________ m/s^2??

The answer is 4.58 ....but since I did this on a test...I don't have my work. Please help! Thanks!

It depends on the angle of the force being applied. I got an answer closer to 8 or 9, but that was assuming the force is pushing completely horizontally. More information may be required.

(7.6 x 9.8) = weight of 74.48N.
(74.48 x 0.4) = 29.792N. friction.
Net force = (97 - 29.792), = 67.208N.
Acceleration = (f/m), = 67.208/7.6, = 8.84m/sec^2.

Better check your question. Something's wrong.