Question

A 500 g block is hanging from a pulley, incline is 20 degree angle. If the coefficient of kinetic friction is 0.24, what is the acceleration of a 250 g block

Answer 1

let the acceleration be "a"

force equation for the hanging block is given as

m₂g - T = m₂ a

T = m₂g - m₂ a eq-1

perpendicular to incline force equation is given as

F_n = m₁ g cos20

frictional force is given as

f = _k F_n

f = (0.24) m₁ g cos20                eq-2          

force equation parallel to incline is given as

T - m₁ g Sin20 - f = m₁ a

using eq-1

m₂g - m₂ a - m₁ g Sin20 - f = m₁ a

using eq-2

m₂g - m₂ a - m₁ g Sin20 - (0.24) m₁ g cos20 = m₁ a

(0.500) (9.8) - (0.500) a - (0.500) (9.8) Sin20 - (0.24) (0.250) (9.8) Cos20 = (0.250) a

a = 3.6 m/s²