What is the mass of block m2?

Question

A block of mass m1=0.700 kg is connected over a pulley to a mass m2. For an angle of theta=30.0° and a coefficient of kinetic friction of μ= 0.400, an acceleration of magnitude a=0.450 m/s² is observed.

Answer 1

Assuming that m₁ is accelerating up a ramp with acceleration a and block M is accelerating vertically down with acceleration -a, and that the tension in the connecting cable is T,

 

m₁a = T - (m₁g sinθ - μm₁g cosθ) ⇒

 

(i)   m₁a = T - m₁g(sinθ - μcosθ)

 

(ii)  m₂(-a) = T - m₂g

 

Subtract (ii) from (i) to eliminate T.

 

(m₁ + m₂)a = m₂g - m₁g(sinθ - μcosθ) ⇒

 

m₂(a - g) = -m₁g(sinθ - μcosθ) - m₁a

 

m₂ = [-m₁g(sinθ - μcosθ) - m₁a] / (a - g)

 

Just plug in the numbers:

 

m₁ = 0.700 kg

g = 9.8 m/s²

θ = 30°

μ = 0.400

a = 0.450 m/s²

 

You can finish from here.