Block A of mass 5.0 kg and block X are attached to a rope which passes over a pulley, as shown in the figure. An 80-N force P is applied horizontally to block A, keeping it in contact with a rough vertical face. The coefficients of static and kinetic friction between the wall and block are μs = 0.40 and μk = 0.30. The pulley is light and frictionless. The mass of block X is adjusted until block A moves upward with an acceleration of 1.6 m/s2. What is the mass of block X?

According to Newton's second law of motion, the net force (ΣF) on an object is equal to the mass (m) times the acceleration (a). By drawing a free body diagram for each block, we can show all the forces acting on that block.

There are two forces on block X:

Weight force Mg pulling down, where M is the mass of block X
Tension force T pulling up

There are five forces on block A:

Weight force mg pulling down, where m is the mass of block A
Tension force T pulling up
Applied force P pushing left
Normal force N pushing right
Friction force Nμ pushing down, where μ is the coefficient of friction

Block X accelerates downwards. Sum of forces in the -y direction:

∑F = ma

Mg − T = Ma

T = Mg − Ma

Sum of forces on block A in the x direction:

∑F = ma

N − P = 0

N = P

Block A accelerates upwards. Sum of forces in the +y direction:

∑F = ma

T − Nμ − mg = ma

Substitute and solve for mass of block X, M:

Mg − Ma − Pμ − mg = ma

Mg − Ma − Pμ = mg + ma

M (g − a) − Pμ = m (g + a)

M (g − a) = m (g + a) + Pμ

M = [ m (g + a) + Pμ ] / (g − a)

Plug in values. Since block A is moving, use the kinetic friction coefficient.

M = [ 5.0 kg (9.8 m/s² + 1.6 m/s²) + 80 N × 0.30 ] / (9.8 m/s² − 1.6 m/s²)

M = 9.9 kg