A block of mass m1 = 28 kg rests on a wedge of angle θ = 47∘ which is itself attached to a table (the wedge does not move in this problem). An inextensible string is attached to m1, passes over a frictionless pulley at the top of the wedge, and is then attached to another block of mass m2 = 3 kg. The coefficient of kinetic friction between block 1 and the plane is μ = 0.8. The string and wedge are long enough to ensure neither block hits the pulley or the table in this problem, and you may assume that block 1 never reaches the table. take g to be 9.81 m/s2.
The system is released from rest as shown above, at t = 0.
(a) Find the magnitude of the acceleration of block 1 when it is released (in m/s2).
(b) How many cm down the plane will block 1 have traveled when 0.47 s has elapsed?
m₁a=m₁gsinα-T-F(fr)… (1)
0=N-m₁gcosα………….(2)
m₂a=T-m₂g…………….(3)
F(fr)=μN=μm₁gcosα
sum of (1) and (3):
a(m₁+m₂)=m₁gsinα-T-F(fr)+ T-m₂g=
= m₁gsinα - μm₁gcosα - m₂g
a=g(m₁sinα - μm₁cosα - m₂)/(m₁+m₂) =…
s=at₂/2=….