Acceleration of a box across a rough horizontal plane

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SUMMARY

The discussion focuses on calculating the acceleration of a 50 kg box being pulled across a rough horizontal surface by a force of 150 N at a 30-degree angle. The calculations involve determining the horizontal force (Fx) as 129.9 N and the vertical force (Fy) as 75 N. The normal force (FN) is calculated to be 415.5 N, leading to a kinetic friction force (Fk) of 83.1 N. The net force (Fnet) is found to be 46.8 N, resulting in an acceleration (a) of 0.936 m/s² for the box.

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Confusedbiomedeng
Thread moved from the technical forums, so no Homework Template is shown
Not sure if I'm doing this right would love input

Consider a box m=50kg which is being pulled on a rough, horizontal surface by a person using a rope . Assume that the person is applying a force of T=150N on the block , the rope makes an angle of 30 degrees with the horizontal and the coefficient of kinetic friction between the block and the horizontal surface is 0.2
Determine acceleration of the block if the block remains in full contact with the floor

Fx=FCosθ
Fx=150(Cos30)
Fx=129.9N

Fy=FSinθ
Fy=150(sin30)
Fy=75N

Fk=μFN
FN=mg-fy
FN=50(9.81)-75
FN=415.5

FK=0.2(415.5)
FK=83.1

Fnet=Fx-Fk
Fnet=129.9-83.1
Fnet=46.8=ma=50(a)=46.8
a=0.936m/s
 
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It all looks good to me (except something's missing in the last symbol of the last line :oldsmile: ).
 
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