What value of F can pull the 5.00 kg box at constant velocity up the incline if uk=0.25. u means coeffiient friction

Question

the angle is 30 degrees and 5.00 kg force is going up

Break up the weight of the box into two components: Perpendicular to the slope, and down the slope. Down the slope the box is mg*SinTheta, and perpendicular to the slope is mg*CosTheta. These two give you all you need here.

Net force up the slope= F - mgSinTheta -friction

where friction = uk*Forceperpendicular.

To pull at constant veloctiy, acceleration is zero, and then Net Force is zero. So..
F - mgSinTheta -friction=0
and you solve for F.