Question

Three blocks of masses m = 4 kg, m, = 2 kg, m, = 4 kg are connected with ideal strings passing over a smooth,
massless pulley as shown in figure. The acceleration of blocks will be (g = 10 m/s2)
ms
m,
(1) 2 m/s2
(2) 4m/s²
(3) 3 m/s2
(4) 5 m/s2​

Answer 1

Answer :-

$\mathsf{a = 2 m/s^2 }$

Given :-

$m_1 = 4kg \\m_2 = 2 kg \\ m_3= 4kg$

To find :-

The acceleration of blocks.

Solution:-

Let $T_1$ be the tension force between $m_2,m_3$block and $T_2$ be the tension force between $m_2 , m_1$.

• Take $m_1$ as system.

The force acting on it:-

• weight force downward.
• weight force downward. Tension force upward.

→$\mathsf{m_1g - T_1 = m_1a}$

→$\mathsf{ m_1g - m_1a = T_2------1)}$

• For block $m_2$

→ $\mathsf{ T_2 + m_2g - T_1 = m_2a----2) }$

• For block $m_3$

• Tension force upward.
• Tension force upward. weight force downward.

→$\mathsf{ T_1 - m_3g = m_3a}$

→$\mathsf{ T_1 = m_3a + m_3g ----3)}$

Put the value of $T_2 , T_1$ in equation 2.

→$\mathsf{T_2 + m_2g - T_1 = m_2a}$

→$\mathsf{ m_1g -m_1a + m_2g - (m_3a + m_3g) = m_2a }$

→$\mathsf{m_1g - m_1a + m_2g - m_3a - m_3g = m_2a }$

→$\mathsf{m_1g + m_2g -m_3g = m_1a + m_2a + m_3 a}$

→$\mathsf{ g ( m_1 + m_2 - m_3) = a(m_1 + m_2+m_3) }$

→$\mathsf{ a = \left(\dfrac{g(m_1+m_2-m_3)}{(m_1 + m_2 + m_3 )}\right)}$

• Put the value of all masses.

→$\mathsf{a = \left(\dfrac{10(4+2-4)}{(4+2+4)}\right)}$

→$\mathsf{a = \left(\dfrac{10 \times 2}{10}\right)}$

→$\mathsf{a = \dfrac{20}{10}}$

→$\mathsf{a = 2 m/s^2 }$

hence,

The acceleration of the block is 2m/s².

Answer 2

Answer:

for answer view the picture