Apply the particle under a net force model to the block in the horizontal direction:1) Fx = F cos θ - fk - T = m2ax = m2aApply the particle in equilibrium model to the block in the vertical direction:2) Fy = n + F sin θ - m2g = 0Apply the particle under a net force model to the ball in the vertical direction:3) Fy = T - m1g = m1ay = m1aSolve Equation (2) for n:n = m2g - F sin θSubstitute n into fk = ?kn from the above equation:4) fk = k (m2g - Fsin θ)Substitute Equation (4) and the value of T from Equation (3) into Equation (1):F cos θ - k(m2g - Fsin θ) - m1(a + g) = m2aSolve for a in terms of k, m1, m2, g, and ?:

[F cos θ - k(m2g - Fsin θ) + m1g]/(m1 + m2)

Explanation:

Since F cos θ - k(m2g - Fsin θ) - m1(a + g) = m2a

Expanding the bracket containing a, we have

F cos θ - k(m2g - Fsin θ) - m1a + m1g = m2a

Collecting the terms in a to the right-hand-side of the equation, we have

F cos θ - k(m2g - Fsin θ) + m1g =  m1a + m2a

Factorizing a out, we have

F cos θ - k(m2g - Fsin θ) + m1g = (m1 + m2)a

Dividing both sides by (m1 + m2), we have

[F cos θ - k(m2g - Fsin θ) + m1g]/(m1 + m2) = a

So, a = [F cos θ - k(m2g - Fsin θ) + m1g]/(m1 + m2)

explanation:

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