Question

Block 1, of mass = 0.550 , is connected over an ideal (massless and frictionless) pulley to block 2, of mass , as shown. Assume that the blocks accelerate as shownwith an acceleration of magnitude = 0.250 and that the coefficient of kinetic friction between block 2 and the plane is = 0.250.
Find the mass of block 2, , when = 30.0.

Answer 1

PICTURE:_________________________________________________________________GIVENDATA:Mass of the first block, m₁ = 0.550 kgMass of the second block, m₂ = ?Linear acceleration, a = 0.250 m/s²Angle of the inclined plane, θ = 30⁰Coefficient of friction between block2 and the plane, μ = 0.250Let,Tension in the string = T________________________________________________________________

SOLUTION:For mass m₁:m₁g - T = m₁a.......(1)For mass m₂:T - m₂g sinθ - μ m₂g cosθ = m₂a.......(2)________________________________________________________________

By adding the equations (1) and (2), we getm₁g - m₂ g sinθ - μ m₂ g cosθ = m₁ a + m₂ am₂ ( a + g sinθ + μ g cosθ ) = m₁ ( g - a )m₂[0.250 m/s²+ (9.8 m/s²)(sin30⁰) + (0.250)(9.8m/s²)(cos30⁰)] =(0.550 kg)( 9.8 m/s² - 0.250 m/s²)m₂=(0.550 kg)( 9.8 m/s² - 0.250 m/s²)/ [0.250 m/s²+ (9.8m/s²)(sin30⁰) + (0.250)(9.8 m/s²)(cos30⁰)] = (5.2525 kgm/s²) / [5.15 m/s² +2.12 m/s²] =0.722 kg