Answer:
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.
