Block 1, of mass m1 = 0.500 kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. for an angle of θ = 30.0 ∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.400, an acceleration of magnitude a = 0.500 m/s2 is observed for block 2.

find the mass of block 2, m2.
express your answer numerically in kilograms.

m2 = 0·52 kg

Explanation:

As the pulley is massless and frictionless the tension in the string on both sides of the pulley will be the same

Given that both have same acceleration a = 0·5 m/s²

Forces acting on mass m1 is tension and force of gravity

You can observe the direction of forces acting on two blocks in the file attached

And in the file attached the force of gravity acting on mass m2 is resolved into two perpendicular components of which one acts along the wedge and the other perpendicular to the wedge

Let the tension in the string be T N

And the frictional force acting on mass m2 is μ×N as sliding is taking place

By applying Newton's second law to the block of mass m1

m1×g - T = m1 × a

⇒ T = m1×(g - a)

⇒T = 0·5×(9·8-0·5)

⇒ T= 4·65 N

Let the normal reaction acting on mass m2 be N

By applying Newton's second law to the block of mass m2 in the direction perpendicular to the wedge

we get

N = m2×g×cosθ

By applying Newton's second law to the block of mass m2 along the wedge

T -  μ×N - m2×g×sinθ = m2×a

Substitute N =  m2×g×cosθ in the above equation

T - μ×m2×g×cosθ - m2×g×sinθ = m2×a

⇒ T = m2 × ( μ×g×cosθ + g×sinθ + a)

By the substituting the corresponding values

4·65 = m2 × 8·8

⇒ m2 = 0·52 kg

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