Question

Two blocks of weights 77.0 and 114.0 N are connected by as massless string and slide down an inclined plane making an angle of 45.0 deg. to the horizontal. The cofficient of kinetic friction between the lighter block and the plane is 0.15 and that between the heavier block and the plane is 0.23. Assuming that the lighter block ``leads'', find the magnitude of the heavier block's acceleration.

Answer 1

The blocks move with the same acceleration ‘a’
F(fr)= μN
m₁g=W₁ => m₁=W₁/g= 77/9.8=7.86 kg
m₂g=W₂ => m₂=W₂/g= 114/9.8=11.63 kg

x: m₁a=m₁g•sin α -μ₁N₁-T
y: 0=-m₁g•cos α +N₁
x: m₂a=m₂g•sin α -μ₂N₂+T
y: 0=-m₂g•cos α +N₂
m₁a=m₁g•sin α -μ₁•m₁g•cos α - T
m₂a=m₂g•sin α -μ₂•m₂g•cos α +T
a= g•[sin α(m₁+m₂) - cosα(μ₁•m₁+ μ₂•m₂)]/(m₁+m₂)

Answer 2

how do u find u1?