The figure below shows a block of mass m resting on a 20¡ã slope. The block has coefficients of friction ¦Ìs = 0.73 and ¦Ìk = 0.53 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.

Answer 1

weight = 2 kg * g accelerating at a down
T = string tension
2 g - T = 2 a
or
T = 2(g-a)

block mass m
slope = 20 deg I think (don't have your font)
mus = .73 and muk = .53

Normal force = m g cos 20
friction force = mu m g cos 20

weight component down slope = m g sin 20

T -m g sin 20 - force of friction down slope = m a

2(g-a) - m g sin 20 - force of friction down slope = m a

max friction down slope without moving up slope = .73 m g cos 20

That is all I can do without knowing what the question is.