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.

i guess it’s a and b

its a 5.5 n

glad that