84. Two blocks, one of mass 5.0 kg and the other of

mass 3.0 kg, are tied together with a massless rope

as in Figure 28. This rope is strung over a massless,

resistance-free pulley. The blocks are released from

rest. Find the following:

a. the tension in the rope

b. the acceleration of the blocks

Hint: You will need to solve two simultaneous equations.

a) 37 N

b) 2.5 m/s²

Explanation:

Let's say m₁ is the 5 kg mass, and m₂ is the 3 kg mass.

Draw a free body diagram for each block. For both, there are two forces: weight down and tension up.

Apply Newton's second law to the 5 kg mass in the y direction (remember it is accelerating downward).

∑F = ma

T - m₁g = m₁ (-a)

m₁g - T = m₁a

Apply Newton's second law to the 3 kg mass in the y direction:

∑F = ma

T - m₂g = m₂a

a) Solve the system of equations. First, solve for a in each equation, then set equal and solve for T.

a = (m₁g - T) / m₁

a = (T - m₂g) / m₂

(m₁g - T) / m₁ = (T - m₂g) / m₂

m₂ (m₁g - T) = m₁ (T - m₂g)

m₁m₂g - Tm₂ = m₁T - m₁m₂g

2m₁m₂g = (m₁ + m₂) T

T = 2m₁m₂g / (m₁ + m₂)

Given m₁ = 5 kg and m₂ = 3 kg:

T = 2 (5 kg) (3 kg) (9.8 m/s²) / (5 kg + 3 kg)

T = 36.75 N

Rounded to two significant figures, the tension is 37 N.

b) Solve the system of equations. First, solve for T in each equation, then set equal and solve for a.

T = m₁g - m₁a

T = m₂g + m₂a

m₁g - m₁a = m₂g + m₂a

m₁g - m₂g = m₁a + m₂a

(m₁ - m₂) g = (m₁ + m₂) a

a = (m₁ - m₂) g / (m₁ + m₂)

Given m₁ = 5 kg and m₂ = 3 kg:

a = (5 kg - 3 kg) (9.8 m/s²) / (5 kg + 3 kg)

a = 2.45 m/s²

Rounded to two significant figures, the magnitude of the acceleration of the blocks is 2.5 m/s².