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A detailed illustration of a physics experiment. It shows a medium-sized, solid rectangular box at rest on an inclined wooden plane, implying a weight of mass 10kg. The box is shown next to a scale indicating its weight. The surface of the incline plane is visibly rough, illustrating the concept of friction, with indicators to suggest that the coefficients of static and kinetic friction are 0.55 and 0.25 respectively. The angle between the plane and a level surface is variable, with a protractor next to it to emphasize this. No numbers or text are visible in the image.

A rectangular of mass 10kg rest on an inclined plane with a coefficient of static friction of 0.55 and coefficient of kinetic friction 0.25.calculate angle that the box will begin to sliding and it's acceleration

weight normal to plane = m g cos A

so static friction up slope = 0.55 m g cos A
weight component down slope = m g sin A
starts to slide when
0.55 m g cos A = m g sin A
or
tan A = .55
A = 28.8 degrees

net force down slope when sliding = m g sin 28.8 - .25 m g cos 28.8 = m a

so
a = g (sin 28.8 - .25 cos 28.8)

note: who cares what the mass was?

I could not understand the solution

the first person who did this question I did not understand it

I need answer

What a hell? I was looking for solution not explanations

Mk=0.25

Ms=0.55
Mass=10kg
For the mass begin to slide:
Tan -1 0.25=angle
Angle=14.04°

F=M×a
F=Mk×mgCos 14.04°
Ma=Mk×mgCos. 14.04°
Mass ❌ is cancelled
a=0.25×10Cos14.04°
a=2.4m/s^2