Question

Figure shows three blocks that are being pulled along a smooth horizontal surface. The masses of the blocks

Answer 1

Answer:

The pulling force F is 200 N

Explanation:

We note that the tension in a rope between two or three blocks is given by the relation;

Where we have;

m₁ → T₁ → m₂ → T₂ → m₃ → F

$T_1 = \dfrac{m_1}{m_1 + m_2 + m_3} \times F$

$T_2 = \dfrac{m_1 + m_2}{m_1 + m_2 + m_3} \times F$

m₁ = 2 kg

m₂ = 4 kg

m₃ = 4 kg

T₂ = 120 N

Therefore, by substitution, gives;

2 kg → T₁ → 4 kg → 120 N → 4 kg → F

$120 \ N = \dfrac{2 \ kg + 4 \ kg}{2 \ kg+ 4 \ kg+ 4 \ kg} \times F = \dfrac{6 \ kg}{10 \ kg} \times F$

F = 120 N × (10 kg)/(6 kg) = 200 N

The pulling force F = 200 N.